Reading text file with wierd file extension?

Discussion in 'Python' started by Lionel, Feb 2, 2009.

  1. Lionel

    Lionel Guest

    Hi Folks, Python newbie here.

    I'm trying to open (for reading) a text file with the following
    filenaming convension:

    "MyTextFile.slc.rsc"

    My code is as follows:

    Filepath = "C:\\MyTextFile.slc.rsc"
    FileH = open(Filepath)

    The above throws an IOError exception. On a hunch I changed the
    filename (only the filename) and tried again:

    Filepath = "C:\\MyTextFile.txt"
    FileH = open(Filepath)

    The above works well. I am able to open the file and read it's
    contents. I assume to read a file in text file "mode" the parameter is
    scanned for a ".txt" extension, otherwise the Python runtime doesn't
    know what version of "open(...)" to invoke. How do I pass the original
    filename (MyTextFile.slc.rsc) and get Python to open it as a text
    file? Thanks in advance everyone!
    Lionel, Feb 2, 2009
    #1
    1. Advertising

  2. On Feb 2, 12:36 pm, Lionel <> wrote:
    > Hi Folks, Python newbie here.
    >
    > I'm trying to open (for reading) a text file with the following
    > filenaming convension:
    >
    > "MyTextFile.slc.rsc"
    >
    > My code is as follows:
    >
    > Filepath = "C:\\MyTextFile.slc.rsc"
    > FileH = open(Filepath)
    >
    > The above throws an IOError exception. On a hunch I changed the
    > filename (only the filename) and tried again:
    >
    > Filepath = "C:\\MyTextFile.txt"
    > FileH = open(Filepath)
    >
    > The above works well. I am able to open the file and read it's
    > contents. I assume to read a file in text file "mode" the parameter is
    > scanned for a ".txt" extension, otherwise the Python runtime doesn't
    > know what version of "open(...)" to invoke. How do I pass the original
    > filename (MyTextFile.slc.rsc) and get Python to open it as a text
    > file? Thanks in advance everyone!


    The extension shouldn't matter. I tried creating a file with the same
    extension as yours and Python 2.5.2 opened it and read it no problem.
    I tried it in IDLE and with Wing on Windows XP. What are you using?
    What's the complete traceback?

    Mike
    Mike Driscoll, Feb 2, 2009
    #2
    1. Advertising

  3. 2009/2/2 Lionel <>:
    > Hi Folks, Python newbie here.
    >
    > I'm trying to open (for reading) a text file with the following
    > filenaming convension:
    >
    > "MyTextFile.slc.rsc"


    Some kind of a resource fork, perhaps? Where did the file come from?

    Python doesn't do anything magic with filenames, so this must be some
    sort of a fileysytem oddity, I think.

    --
    Cheers,
    Simon B.
    Simon Brunning, Feb 2, 2009
    #3
  4. Lionel

    Lionel Guest

    On Feb 2, 10:41 am, Mike Driscoll <> wrote:
    > On Feb 2, 12:36 pm, Lionel <> wrote:
    >
    >
    >
    >
    >
    > > Hi Folks, Python newbie here.

    >
    > > I'm trying to open (for reading) a text file with the following
    > > filenaming convension:

    >
    > > "MyTextFile.slc.rsc"

    >
    > > My code is as follows:

    >
    > > Filepath = "C:\\MyTextFile.slc.rsc"
    > > FileH = open(Filepath)

    >
    > > The above throws an IOError exception. On a hunch I changed the
    > > filename (only the filename) and tried again:

    >
    > > Filepath = "C:\\MyTextFile.txt"
    > > FileH = open(Filepath)

    >
    > > The above works well. I am able to open the file and read it's
    > > contents. I assume to read a file in text file "mode" the parameter is
    > > scanned for a ".txt" extension, otherwise the Python runtime doesn't
    > > know what version of "open(...)" to invoke. How do I pass the original
    > > filename (MyTextFile.slc.rsc) and get Python to open it as a text
    > > file? Thanks in advance everyone!

    >
    > The extension shouldn't matter. I tried creating a file with the same
    > extension as yours and Python 2.5.2 opened it and read it no problem.
    > I tried it in IDLE and with Wing on Windows XP. What are you using?
    > What's the complete traceback?
    >
    > Mike- Hide quoted text -
    >
    > - Show quoted text -


    Hi Mike,

    maybe it's not a "true" text file? Opening it in Microsoft Notepad
    gives an unformatted view of the file (text with no line wrapping,
    just the end-of-line square box character followed by more text, end-
    of-line character, etc). Wordpad opens it properly i.e. respects the
    end-of-line wrapping. I'm unsure of how these files are being
    generated, I was just given them and told they wanted to be able to
    read them.

    How do I collect the traceback to post it?
    Lionel, Feb 2, 2009
    #4
  5. Lionel

    Lionel Guest

    On Feb 2, 11:20 am, Lionel <> wrote:
    > On Feb 2, 10:41 am, Mike Driscoll <> wrote:
    >
    >
    >
    >
    >
    > > On Feb 2, 12:36 pm, Lionel <> wrote:

    >
    > > > Hi Folks, Python newbie here.

    >
    > > > I'm trying to open (for reading) a text file with the following
    > > > filenaming convension:

    >
    > > > "MyTextFile.slc.rsc"

    >
    > > > My code is as follows:

    >
    > > > Filepath = "C:\\MyTextFile.slc.rsc"
    > > > FileH = open(Filepath)

    >
    > > > The above throws an IOError exception. On a hunch I changed the
    > > > filename (only the filename) and tried again:

    >
    > > > Filepath = "C:\\MyTextFile.txt"
    > > > FileH = open(Filepath)

    >
    > > > The above works well. I am able to open the file and read it's
    > > > contents. I assume to read a file in text file "mode" the parameter is
    > > > scanned for a ".txt" extension, otherwise the Python runtime doesn't
    > > > know what version of "open(...)" to invoke. How do I pass the original
    > > > filename (MyTextFile.slc.rsc) and get Python to open it as a text
    > > > file? Thanks in advance everyone!

    >
    > > The extension shouldn't matter. I tried creating a file with the same
    > > extension as yours and Python 2.5.2 opened it and read it no problem.
    > > I tried it in IDLE and with Wing on Windows XP. What are you using?
    > > What's the complete traceback?

    >
    > > Mike- Hide quoted text -

    >
    > > - Show quoted text -

    >
    > Hi Mike,
    >
    > maybe it's not a "true" text file? Opening it in Microsoft Notepad
    > gives an unformatted view of the file (text with no line wrapping,
    > just the end-of-line square box character followed by more text, end-
    > of-line character, etc). Wordpad opens it properly i.e. respects the
    > end-of-line wrapping. I'm unsure of how these files are being
    > generated, I was just given them and told they wanted to be able to
    > read them.
    >
    > How do I collect the traceback to post it?- Hide quoted text -
    >
    > - Show quoted text -


    Sorry, I forgot to mention I'm using Python 2.5 and the Python IDLE.
    Lionel, Feb 2, 2009
    #5
  6. > Hi Mike,
    >
    > maybe it's not a "true" text file? Opening it in Microsoft Notepad
    > gives an unformatted view of the file (text with no line wrapping,
    > just the end-of-line square box character followed by more text, end-
    > of-line character, etc). Wordpad opens it properly i.e. respects the
    > end-of-line wrapping. I'm unsure of how these files are being
    > generated, I was just given them and told they wanted to be able to
    > read them.



    Sounds like a unix-line-ended textfile. Python has no problems reading
    these, and frankly I'm a bit wondering why you have that IOError -but
    then, on windows *anything* is possible.

    You can try to open the file using the "rb" flag, which opens it in binary.

    However, Python doesn't care about any extensions, that must be
    something else you did different.

    > How do I collect the traceback to post it?


    Copy and paste from the console or wherever you run the script?

    Diez
    Diez B. Roggisch, Feb 2, 2009
    #6
  7. On Feb 2, 1:20 pm, Lionel <> wrote:
    > On Feb 2, 10:41 am, Mike Driscoll <> wrote:
    >
    >
    >
    > > On Feb 2, 12:36 pm, Lionel <> wrote:

    >
    > > > Hi Folks, Python newbie here.

    >
    > > > I'm trying to open (for reading) a text file with the following
    > > > filenaming convension:

    >
    > > > "MyTextFile.slc.rsc"

    >
    > > > My code is as follows:

    >
    > > > Filepath = "C:\\MyTextFile.slc.rsc"
    > > > FileH = open(Filepath)

    >
    > > > The above throws an IOError exception. On a hunch I changed the
    > > > filename (only the filename) and tried again:

    >
    > > > Filepath = "C:\\MyTextFile.txt"
    > > > FileH = open(Filepath)

    >
    > > > The above works well. I am able to open the file and read it's
    > > > contents. I assume to read a file in text file "mode" the parameter is
    > > > scanned for a ".txt" extension, otherwise the Python runtime doesn't
    > > > know what version of "open(...)" to invoke. How do I pass the original
    > > > filename (MyTextFile.slc.rsc) and get Python to open it as a text
    > > > file? Thanks in advance everyone!

    >
    > > The extension shouldn't matter. I tried creating a file with the same
    > > extension as yours and Python 2.5.2 opened it and read it no problem.
    > > I tried it in IDLE and with Wing on Windows XP. What are you using?
    > > What's the complete traceback?

    >
    > > Mike- Hide quoted text -

    >
    > > - Show quoted text -

    >
    > Hi Mike,
    >
    > maybe it's not a "true" text file? Opening it in Microsoft Notepad
    > gives an unformatted view of the file (text with no line wrapping,
    > just the end-of-line square box character followed by more text, end-
    > of-line character, etc). Wordpad opens it properly i.e. respects the
    > end-of-line wrapping. I'm unsure of how these files are being
    > generated, I was just given them and told they wanted to be able to
    > read them.
    >
    > How do I collect the traceback to post it?


    The traceback should look something like this fake one:

    Traceback (most recent call last):
    File "<pyshell#3>", line 1, in <module>
    raise IOError
    IOError

    Just copy and paste it in your next message. The other guys are
    probably right in that it is a line ending issue, but as they and I
    have said, Python shouldn't care (and doesn't on my machine).

    Mike
    Mike Driscoll, Feb 2, 2009
    #7
  8. Lionel

    Lionel Guest

    On Feb 2, 12:10 pm, Mike Driscoll <> wrote:
    > On Feb 2, 1:20 pm, Lionel <> wrote:
    >
    >
    >
    >
    >
    > > On Feb 2, 10:41 am, Mike Driscoll <> wrote:

    >
    > > > On Feb 2, 12:36 pm, Lionel <> wrote:

    >
    > > > > Hi Folks, Python newbie here.

    >
    > > > > I'm trying to open (for reading) a text file with the following
    > > > > filenaming convension:

    >
    > > > > "MyTextFile.slc.rsc"

    >
    > > > > My code is as follows:

    >
    > > > > Filepath = "C:\\MyTextFile.slc.rsc"
    > > > > FileH = open(Filepath)

    >
    > > > > The above throws an IOError exception. On a hunch I changed the
    > > > > filename (only the filename) and tried again:

    >
    > > > > Filepath = "C:\\MyTextFile.txt"
    > > > > FileH = open(Filepath)

    >
    > > > > The above works well. I am able to open the file and read it's
    > > > > contents. I assume to read a file in text file "mode" the parameter is
    > > > > scanned for a ".txt" extension, otherwise the Python runtime doesn't
    > > > > know what version of "open(...)" to invoke. How do I pass the original
    > > > > filename (MyTextFile.slc.rsc) and get Python to open it as a text
    > > > > file? Thanks in advance everyone!

    >
    > > > The extension shouldn't matter. I tried creating a file with the same
    > > > extension as yours and Python 2.5.2 opened it and read it no problem.
    > > > I tried it in IDLE and with Wing on Windows XP. What are you using?
    > > > What's the complete traceback?

    >
    > > > Mike- Hide quoted text -

    >
    > > > - Show quoted text -

    >
    > > Hi Mike,

    >
    > > maybe it's not a "true" text file? Opening it in Microsoft Notepad
    > > gives an unformatted view of the file (text with no line wrapping,
    > > just the end-of-line square box character followed by more text, end-
    > > of-line character, etc). Wordpad opens it properly i.e. respects the
    > > end-of-line wrapping. I'm unsure of how these files are being
    > > generated, I was just given them and told they wanted to be able to
    > > read them.

    >
    > > How do I collect the traceback to post it?

    >
    > The traceback should look something like this fake one:
    >
    > Traceback (most recent call last):
    >   File "<pyshell#3>", line 1, in <module>
    >     raise IOError
    > IOError
    >
    > Just copy and paste it in your next message. The other guys are
    > probably right in that it is a line ending issue, but as they and I
    > have said, Python shouldn't care (and doesn't on my machine).
    >
    > Mike- Hide quoted text -
    >
    > - Show quoted text -


    Okay, I think I see what's going on. My class takes a single parameter
    when it is instantiated...the file path of the data file the user
    wants to open. This is of the form "someFile.slc". In the same
    directory of "someFile.slc" is a resource file that (like a file
    header) contains a host of parameters associated with the data file.
    The resource file is of the form "someFile.slc.rsc". So, when the user
    creates an instance of my class which, it invokes the __init__ method
    where I add the ".rsc" extension to the original filename/path
    parameter that was passed to the class "constructor". For example:

    Note: try-catch blocks ommitted.

    class MyUtilityClass:
    def __init__(self, DataFilepath):
    Resourcepath = DataFilepath + ".rsc"
    DataFileH = open(DataFilepath)
    ResourceFileH = open(Resourcepath)


    Invoking this from the Python shell explicitly is no problem i.e.
    "TestH = open("C:\\TestResourceFile.slc.rsc") works. BUT...something
    is lost when I append the ".rsc" extension to the DataFilePath
    parameter as above. When the __init__ method is invoked, Python will
    open the data file but generates the exception with the "open
    (Resourcepath)" instruction. I think it is somehow related to the
    backslashes but I'm not entirely sure of this. Any ideas?

    Thanks for the help folks.
    Lionel, Feb 2, 2009
    #8
  9. Lionel schrieb:
    > On Feb 2, 12:10 pm, Mike Driscoll <> wrote:
    >> On Feb 2, 1:20 pm, Lionel <> wrote:
    >>
    >>
    >>
    >>
    >>
    >>> On Feb 2, 10:41 am, Mike Driscoll <> wrote:
    >>>> On Feb 2, 12:36 pm, Lionel <> wrote:
    >>>>> Hi Folks, Python newbie here.
    >>>>> I'm trying to open (for reading) a text file with the following
    >>>>> filenaming convension:
    >>>>> "MyTextFile.slc.rsc"
    >>>>> My code is as follows:
    >>>>> Filepath = "C:\\MyTextFile.slc.rsc"
    >>>>> FileH = open(Filepath)
    >>>>> The above throws an IOError exception. On a hunch I changed the
    >>>>> filename (only the filename) and tried again:
    >>>>> Filepath = "C:\\MyTextFile.txt"
    >>>>> FileH = open(Filepath)
    >>>>> The above works well. I am able to open the file and read it's
    >>>>> contents. I assume to read a file in text file "mode" the parameter is
    >>>>> scanned for a ".txt" extension, otherwise the Python runtime doesn't
    >>>>> know what version of "open(...)" to invoke. How do I pass the original
    >>>>> filename (MyTextFile.slc.rsc) and get Python to open it as a text
    >>>>> file? Thanks in advance everyone!
    >>>> The extension shouldn't matter. I tried creating a file with the same
    >>>> extension as yours and Python 2.5.2 opened it and read it no problem.
    >>>> I tried it in IDLE and with Wing on Windows XP. What are you using?
    >>>> What's the complete traceback?
    >>>> Mike- Hide quoted text -
    >>>> - Show quoted text -
    >>> Hi Mike,
    >>> maybe it's not a "true" text file? Opening it in Microsoft Notepad
    >>> gives an unformatted view of the file (text with no line wrapping,
    >>> just the end-of-line square box character followed by more text, end-
    >>> of-line character, etc). Wordpad opens it properly i.e. respects the
    >>> end-of-line wrapping. I'm unsure of how these files are being
    >>> generated, I was just given them and told they wanted to be able to
    >>> read them.
    >>> How do I collect the traceback to post it?

    >> The traceback should look something like this fake one:
    >>
    >> Traceback (most recent call last):
    >> File "<pyshell#3>", line 1, in <module>
    >> raise IOError
    >> IOError
    >>
    >> Just copy and paste it in your next message. The other guys are
    >> probably right in that it is a line ending issue, but as they and I
    >> have said, Python shouldn't care (and doesn't on my machine).
    >>
    >> Mike- Hide quoted text -
    >>
    >> - Show quoted text -

    >
    > Okay, I think I see what's going on. My class takes a single parameter
    > when it is instantiated...the file path of the data file the user
    > wants to open. This is of the form "someFile.slc". In the same
    > directory of "someFile.slc" is a resource file that (like a file
    > header) contains a host of parameters associated with the data file.
    > The resource file is of the form "someFile.slc.rsc". So, when the user
    > creates an instance of my class which, it invokes the __init__ method
    > where I add the ".rsc" extension to the original filename/path
    > parameter that was passed to the class "constructor". For example:
    >
    > Note: try-catch blocks ommitted.
    >
    > class MyUtilityClass:
    > def __init__(self, DataFilepath):
    > Resourcepath = DataFilepath + ".rsc"
    > DataFileH = open(DataFilepath)
    > ResourceFileH = open(Resourcepath)
    >
    >
    > Invoking this from the Python shell explicitly is no problem i.e.
    > "TestH = open("C:\\TestResourceFile.slc.rsc") works. BUT...something
    > is lost when I append the ".rsc" extension to the DataFilePath
    > parameter as above. When the __init__ method is invoked, Python will
    > open the data file but generates the exception with the "open
    > (Resourcepath)" instruction. I think it is somehow related to the
    > backslashes but I'm not entirely sure of this. Any ideas?


    This is written very slowly, so you can read it better:

    Please post the traceback.

    Diez
    Diez B. Roggisch, Feb 2, 2009
    #9
  10. Lionel

    Lionel Guest

    On Feb 2, 1:07 pm, "Diez B. Roggisch" <> wrote:

    This is written very slowly, so you can read it better:

    Please post without sarcasm.


    This is the output from my Python shell:

    >>> DatafilePath = "C:\\C8Example1.slc"
    >>> ResourcefilePath = DatafilePath + ".rsc"
    >>> DatafileFH = open(DatafilePath)
    >>> ResourceFh = open(ResourcefilePath)
    >>> DatafilePath

    'C:\\C8Example1.slc'
    >>> ResourcefilePath

    'C:\\C8Example1.slc.rsc'

    It seems to run without trouble. However, here is the offending code
    in my class (followed by console output):

    class C8DataType:

    def __init__(self, DataFilepath):

    try:
    DataFH = open(DataFilepath, "rb")

    except IOError, message:
    # Error opening file.
    print(message)
    return None

    ResourceFilepath = DataFilepath + ".src"

    print(DataFilepath)
    print(ResourceFilepath)

    # Try to open resource file, catch exception:
    try:
    ResourceFH = open(ResourceFilepath)

    except IOError, message:
    # Error opening file.
    print(message)
    print("Error opening " + ResourceFilepath)
    DataFH.close()
    return None

    Console output when invoking as "someObject = C8DataType("C:\
    \C8Example1.slc")" :

    C:\C8Example1.slc
    C:\C8Example1.slc.src
    [Errno 2] No such file or directory: 'C:\\C8Example1.slc.src'
    Error opening C:\C8Example1.slc.src


    Thank you


    > Lionel schrieb:
    >
    >
    >
    >
    >
    > > On Feb 2, 12:10 pm, Mike Driscoll <> wrote:
    > >> On Feb 2, 1:20 pm, Lionel <> wrote:

    >
    > >>> On Feb 2, 10:41 am, Mike Driscoll <> wrote:
    > >>>> On Feb 2, 12:36 pm, Lionel <> wrote:
    > >>>>> Hi Folks, Python newbie here.
    > >>>>> I'm trying to open (for reading) a text file with the following
    > >>>>> filenaming convension:
    > >>>>> "MyTextFile.slc.rsc"
    > >>>>> My code is as follows:
    > >>>>> Filepath = "C:\\MyTextFile.slc.rsc"
    > >>>>> FileH = open(Filepath)
    > >>>>> The above throws an IOError exception. On a hunch I changed the
    > >>>>> filename (only the filename) and tried again:
    > >>>>> Filepath = "C:\\MyTextFile.txt"
    > >>>>> FileH = open(Filepath)
    > >>>>> The above works well. I am able to open the file and read it's
    > >>>>> contents. I assume to read a file in text file "mode" the parameter is
    > >>>>> scanned for a ".txt" extension, otherwise the Python runtime doesn't
    > >>>>> know what version of "open(...)" to invoke. How do I pass the original
    > >>>>> filename (MyTextFile.slc.rsc) and get Python to open it as a text
    > >>>>> file? Thanks in advance everyone!
    > >>>> The extension shouldn't matter. I tried creating a file with the same
    > >>>> extension as yours and Python 2.5.2 opened it and read it no problem..
    > >>>> I tried it in IDLE and with Wing on Windows XP. What are you using?
    > >>>> What's the complete traceback?
    > >>>> Mike- Hide quoted text -
    > >>>> - Show quoted text -
    > >>> Hi Mike,
    > >>> maybe it's not a "true" text file? Opening it in Microsoft Notepad
    > >>> gives an unformatted view of the file (text with no line wrapping,
    > >>> just the end-of-line square box character followed by more text, end-
    > >>> of-line character, etc). Wordpad opens it properly i.e. respects the
    > >>> end-of-line wrapping. I'm unsure of how these files are being
    > >>> generated, I was just given them and told they wanted to be able to
    > >>> read them.
    > >>> How do I collect the traceback to post it?
    > >> The traceback should look something like this fake one:

    >
    > >> Traceback (most recent call last):
    > >>   File "<pyshell#3>", line 1, in <module>
    > >>     raise IOError
    > >> IOError

    >
    > >> Just copy and paste it in your next message. The other guys are
    > >> probably right in that it is a line ending issue, but as they and I
    > >> have said, Python shouldn't care (and doesn't on my machine).

    >
    > >> Mike- Hide quoted text -

    >
    > >> - Show quoted text -

    >
    > > Okay, I think I see what's going on. My class takes a single parameter
    > > when it is instantiated...the file path of the data file the user
    > > wants to open. This is of the form "someFile.slc". In the same
    > > directory of "someFile.slc" is a resource file that (like a file
    > > header) contains a host of parameters associated with the data file.
    > > The resource file is of the form "someFile.slc.rsc". So, when the user
    > > creates an instance of my class which, it invokes the __init__ method
    > > where I add the ".rsc" extension to the original filename/path
    > > parameter that was passed to the class "constructor". For example:

    >
    > > Note: try-catch blocks ommitted.

    >
    > > class MyUtilityClass:
    > >     def __init__(self, DataFilepath):
    > >         Resourcepath  = DataFilepath + ".rsc"
    > >         DataFileH     = open(DataFilepath)
    > >         ResourceFileH = open(Resourcepath)

    >
    > > Invoking this from the Python shell explicitly is no problem i.e.
    > > "TestH = open("C:\\TestResourceFile.slc.rsc") works. BUT...something
    > > is lost when I append the ".rsc" extension to the DataFilePath
    > > parameter as above. When the __init__ method is invoked, Python will
    > > open the data file but generates the exception with the "open
    > > (Resourcepath)" instruction. I think it is somehow related to the
    > > backslashes but I'm not entirely sure of this. Any ideas?

    >
    > This is written very slowly, so you can read it better:
    >
    > Please post the traceback.
    >
    > Diez- Hide quoted text -
    >
    > - Show quoted text -
    Lionel, Feb 2, 2009
    #10
  11. Lionel

    John Machin Guest

    On Feb 3, 8:07 am, "Diez B. Roggisch" <> wrote:

    >
    > This is written very slowly, so you can read it better:
    >
    > Please post the traceback.


    *AND* please post the text of the IOError message

    *AND* please do yourself a favour and move your files out of the root
    directory into a directory with a meaningful name
    John Machin, Feb 2, 2009
    #11
  12. On Feb 2, 3:43 pm, Lionel <> wrote:
    > On Feb 2, 1:07 pm, "Diez B. Roggisch" <> wrote:
    >
    > This is written very slowly, so you can read it better:
    >
    > Please post without sarcasm.
    >
    > This is the output from my Python shell:
    >
    > >>> DatafilePath = "C:\\C8Example1.slc"
    > >>> ResourcefilePath = DatafilePath + ".rsc"
    > >>> DatafileFH = open(DatafilePath)
    > >>> ResourceFh = open(ResourcefilePath)
    > >>> DatafilePath

    >
    > 'C:\\C8Example1.slc'>>> ResourcefilePath
    >
    > 'C:\\C8Example1.slc.rsc'
    >
    > It seems to run without trouble. However, here is the offending code
    > in my class (followed by console output):
    >
    > class C8DataType:
    >
    >     def __init__(self, DataFilepath):
    >
    >         try:
    >             DataFH = open(DataFilepath, "rb")
    >
    >         except IOError, message:
    >             # Error opening file.
    >             print(message)
    >             return None
    >
    >         ResourceFilepath = DataFilepath + ".src"
    >
    >         print(DataFilepath)
    >         print(ResourceFilepath)
    >
    >         # Try to open resource file, catch exception:
    >         try:
    >             ResourceFH = open(ResourceFilepath)
    >
    >         except IOError, message:
    >             # Error opening file.
    >             print(message)
    >             print("Error opening " + ResourceFilepath)
    >             DataFH.close()
    >             return None
    >
    > Console output when invoking as "someObject = C8DataType("C:\
    > \C8Example1.slc")" :
    >
    > C:\C8Example1.slc
    > C:\C8Example1.slc.src
    > [Errno 2] No such file or directory: 'C:\\C8Example1.slc.src'
    > Error opening C:\C8Example1.slc.src
    >
    > Thank you
    >
    > > Lionel schrieb:

    >
    > > > On Feb 2, 12:10 pm, Mike Driscoll <> wrote:
    > > >> On Feb 2, 1:20 pm, Lionel <> wrote:

    >
    > > >>> On Feb 2, 10:41 am, Mike Driscoll <> wrote:
    > > >>>> On Feb 2, 12:36 pm, Lionel <> wrote:
    > > >>>>> Hi Folks, Python newbie here.
    > > >>>>> I'm trying to open (for reading) a text file with the following
    > > >>>>> filenaming convension:
    > > >>>>> "MyTextFile.slc.rsc"
    > > >>>>> My code is as follows:
    > > >>>>> Filepath = "C:\\MyTextFile.slc.rsc"
    > > >>>>> FileH = open(Filepath)
    > > >>>>> The above throws an IOError exception. On a hunch I changed the
    > > >>>>> filename (only the filename) and tried again:
    > > >>>>> Filepath = "C:\\MyTextFile.txt"
    > > >>>>> FileH = open(Filepath)
    > > >>>>> The above works well. I am able to open the file and read it's
    > > >>>>> contents. I assume to read a file in text file "mode" the parameter is
    > > >>>>> scanned for a ".txt" extension, otherwise the Python runtime doesn't
    > > >>>>> know what version of "open(...)" to invoke. How do I pass the original
    > > >>>>> filename (MyTextFile.slc.rsc) and get Python to open it as a text
    > > >>>>> file? Thanks in advance everyone!
    > > >>>> The extension shouldn't matter. I tried creating a file with the same
    > > >>>> extension as yours and Python 2.5.2 opened it and read it no problem.
    > > >>>> I tried it in IDLE and with Wing on Windows XP. What are you using?
    > > >>>> What's the complete traceback?
    > > >>>> Mike- Hide quoted text -
    > > >>>> - Show quoted text -
    > > >>> Hi Mike,
    > > >>> maybe it's not a "true" text file? Opening it in Microsoft Notepad
    > > >>> gives an unformatted view of the file (text with no line wrapping,
    > > >>> just the end-of-line square box character followed by more text, end-
    > > >>> of-line character, etc). Wordpad opens it properly i.e. respects the
    > > >>> end-of-line wrapping. I'm unsure of how these files are being
    > > >>> generated, I was just given them and told they wanted to be able to
    > > >>> read them.
    > > >>> How do I collect the traceback to post it?
    > > >> The traceback should look something like this fake one:

    >
    > > >> Traceback (most recent call last):
    > > >>   File "<pyshell#3>", line 1, in <module>
    > > >>     raise IOError
    > > >> IOError

    >
    > > >> Just copy and paste it in your next message. The other guys are
    > > >> probably right in that it is a line ending issue, but as they and I
    > > >> have said, Python shouldn't care (and doesn't on my machine).

    >
    > > >> Mike- Hide quoted text -

    >
    > > >> - Show quoted text -

    >
    > > > Okay, I think I see what's going on. My class takes a single parameter
    > > > when it is instantiated...the file path of the data file the user
    > > > wants to open. This is of the form "someFile.slc". In the same
    > > > directory of "someFile.slc" is a resource file that (like a file
    > > > header) contains a host of parameters associated with the data file.
    > > > The resource file is of the form "someFile.slc.rsc". So, when the user
    > > > creates an instance of my class which, it invokes the __init__ method
    > > > where I add the ".rsc" extension to the original filename/path
    > > > parameter that was passed to the class "constructor". For example:

    >
    > > > Note: try-catch blocks ommitted.

    >
    > > > class MyUtilityClass:
    > > >     def __init__(self, DataFilepath):
    > > >         Resourcepath  = DataFilepath + ".rsc"
    > > >         DataFileH     = open(DataFilepath)
    > > >         ResourceFileH = open(Resourcepath)

    >
    > > > Invoking this from the Python shell explicitly is no problem i.e.
    > > > "TestH = open("C:\\TestResourceFile.slc.rsc") works. BUT...something
    > > > is lost when I append the ".rsc" extension to the DataFilePath
    > > > parameter as above. When the __init__ method is invoked, Python will
    > > > open the data file but generates the exception with the "open
    > > > (Resourcepath)" instruction. I think it is somehow related to the
    > > > backslashes but I'm not entirely sure of this. Any ideas?

    >
    > > This is written very slowly, so you can read it better:

    >
    > > Please post the traceback.

    >
    > > Diez- Hide quoted text -

    >
    > > - Show quoted text -


    Well, if I understand your code correctly, you pass in a filename that
    exists, then you append an extension to it and expect that to change
    the file's name. This doesn't work, as you only changed the string,
    not the filename in the file system itself. You can probably use
    os.rename() to do it. Then it should be able to open it.

    Mike
    Mike Driscoll, Feb 2, 2009
    #12
  13. Lionel

    John Machin Guest

    On Feb 3, 8:43 am, Lionel <> wrote:

    >         ResourceFilepath = DataFilepath + ".src"


    Don't you mean ".rsc"?
    John Machin, Feb 2, 2009
    #13
  14. Lionel

    Lionel Guest

    On Feb 2, 2:01 pm, Mike Driscoll <> wrote:
    > On Feb 2, 3:43 pm, Lionel <> wrote:
    >
    >
    >
    >
    >
    > > On Feb 2, 1:07 pm, "Diez B. Roggisch" <> wrote:

    >
    > > This is written very slowly, so you can read it better:

    >
    > > Please post without sarcasm.

    >
    > > This is the output from my Python shell:

    >
    > > >>> DatafilePath = "C:\\C8Example1.slc"
    > > >>> ResourcefilePath = DatafilePath + ".rsc"
    > > >>> DatafileFH = open(DatafilePath)
    > > >>> ResourceFh = open(ResourcefilePath)
    > > >>> DatafilePath

    >
    > > 'C:\\C8Example1.slc'>>> ResourcefilePath

    >
    > > 'C:\\C8Example1.slc.rsc'

    >
    > > It seems to run without trouble. However, here is the offending code
    > > in my class (followed by console output):

    >
    > > class C8DataType:

    >
    > >     def __init__(self, DataFilepath):

    >
    > >         try:
    > >             DataFH = open(DataFilepath, "rb")

    >
    > >         except IOError, message:
    > >             # Error opening file.
    > >             print(message)
    > >             return None

    >
    > >         ResourceFilepath = DataFilepath + ".src"

    >
    > >         print(DataFilepath)
    > >         print(ResourceFilepath)

    >
    > >         # Try to open resource file, catch exception:
    > >         try:
    > >             ResourceFH = open(ResourceFilepath)

    >
    > >         except IOError, message:
    > >             # Error opening file.
    > >             print(message)
    > >             print("Error opening " + ResourceFilepath)
    > >             DataFH.close()
    > >             return None

    >
    > > Console output when invoking as "someObject = C8DataType("C:\
    > > \C8Example1.slc")" :

    >
    > > C:\C8Example1.slc
    > > C:\C8Example1.slc.src
    > > [Errno 2] No such file or directory: 'C:\\C8Example1.slc.src'
    > > Error opening C:\C8Example1.slc.src

    >
    > > Thank you

    >
    > > > Lionel schrieb:

    >
    > > > > On Feb 2, 12:10 pm, Mike Driscoll <> wrote:
    > > > >> On Feb 2, 1:20 pm, Lionel <> wrote:

    >
    > > > >>> On Feb 2, 10:41 am, Mike Driscoll <> wrote:
    > > > >>>> On Feb 2, 12:36 pm, Lionel <> wrote:
    > > > >>>>> Hi Folks, Python newbie here.
    > > > >>>>> I'm trying to open (for reading) a text file with the following
    > > > >>>>> filenaming convension:
    > > > >>>>> "MyTextFile.slc.rsc"
    > > > >>>>> My code is as follows:
    > > > >>>>> Filepath = "C:\\MyTextFile.slc.rsc"
    > > > >>>>> FileH = open(Filepath)
    > > > >>>>> The above throws an IOError exception. On a hunch I changed the
    > > > >>>>> filename (only the filename) and tried again:
    > > > >>>>> Filepath = "C:\\MyTextFile.txt"
    > > > >>>>> FileH = open(Filepath)
    > > > >>>>> The above works well. I am able to open the file and read it's
    > > > >>>>> contents. I assume to read a file in text file "mode" the parameter is
    > > > >>>>> scanned for a ".txt" extension, otherwise the Python runtime doesn't
    > > > >>>>> know what version of "open(...)" to invoke. How do I pass the original
    > > > >>>>> filename (MyTextFile.slc.rsc) and get Python to open it as a text
    > > > >>>>> file? Thanks in advance everyone!
    > > > >>>> The extension shouldn't matter. I tried creating a file with the same
    > > > >>>> extension as yours and Python 2.5.2 opened it and read it no problem.
    > > > >>>> I tried it in IDLE and with Wing on Windows XP. What are you using?
    > > > >>>> What's the complete traceback?
    > > > >>>> Mike- Hide quoted text -
    > > > >>>> - Show quoted text -
    > > > >>> Hi Mike,
    > > > >>> maybe it's not a "true" text file? Opening it in Microsoft Notepad
    > > > >>> gives an unformatted view of the file (text with no line wrapping,
    > > > >>> just the end-of-line square box character followed by more text, end-
    > > > >>> of-line character, etc). Wordpad opens it properly i.e. respects the
    > > > >>> end-of-line wrapping. I'm unsure of how these files are being
    > > > >>> generated, I was just given them and told they wanted to be able to
    > > > >>> read them.
    > > > >>> How do I collect the traceback to post it?
    > > > >> The traceback should look something like this fake one:

    >
    > > > >> Traceback (most recent call last):
    > > > >>   File "<pyshell#3>", line 1, in <module>
    > > > >>     raise IOError
    > > > >> IOError

    >
    > > > >> Just copy and paste it in your next message. The other guys are
    > > > >> probably right in that it is a line ending issue, but as they and I
    > > > >> have said, Python shouldn't care (and doesn't on my machine).

    >
    > > > >> Mike- Hide quoted text -

    >
    > > > >> - Show quoted text -

    >
    > > > > Okay, I think I see what's going on. My class takes a single parameter
    > > > > when it is instantiated...the file path of the data file the user
    > > > > wants to open. This is of the form "someFile.slc". In the same
    > > > > directory of "someFile.slc" is a resource file that (like a file
    > > > > header) contains a host of parameters associated with the data file..
    > > > > The resource file is of the form "someFile.slc.rsc". So, when the user
    > > > > creates an instance of my class which, it invokes the __init__ method
    > > > > where I add the ".rsc" extension to the original filename/path
    > > > > parameter that was passed to the class "constructor". For example:

    >
    > > > > Note: try-catch blocks ommitted.

    >
    > > > > class MyUtilityClass:
    > > > >     def __init__(self, DataFilepath):
    > > > >         Resourcepath  = DataFilepath + ".rsc"
    > > > >         DataFileH     = open(DataFilepath)
    > > > >         ResourceFileH = open(Resourcepath)

    >
    > > > > Invoking this from the Python shell explicitly is no problem i.e.
    > > > > "TestH = open("C:\\TestResourceFile.slc.rsc") works. BUT...something
    > > > > is lost when I append the ".rsc" extension to the DataFilePath
    > > > > parameter as above. When the __init__ method is invoked, Python will
    > > > > open the data file but generates the exception with the "open
    > > > > (Resourcepath)" instruction. I think it is somehow related to the
    > > > > backslashes but I'm not entirely sure of this. Any ideas?

    >
    > > > This is written very slowly, so you can read it better:

    >
    > > > Please post the traceback.

    >
    > > > Diez- Hide quoted text -

    >
    > > > - Show quoted text -

    >
    > Well, if I understand your code correctly, you pass in a filename that
    > exists, then you append an extension to it and expect that to change
    > the file's name. This doesn't work, as you only changed the string,
    > not the filename in the file system itself. You can probably use
    > os.rename() to do it. Then it should be able to open it.
    >
    > Mike- Hide quoted text -
    >
    > - Show quoted text -


    Hello Mike,

    I'm sorry, I'm not making myself clear. I have two files in my root
    directory: "C8Example1.slc" and "C8Example1.slc.rsc". (They are just
    in the root directory for testing for the time being). The ".slc" file
    contains the data and the ".slc.rsc" file contains the info necessary
    to extract it (datatype, width, height, etc). I know a priori that the
    header file will always be in the same directory and have the same
    name as the data file with the exception that it will have ".rsc"
    appended to its name. Therefore, when the user passes the data file
    name and path as a string to the class, I check the parameter by
    seeing if I can read to that filepath. If so, I simply add ".rsc" to
    the end of the parameter string and try to read to that new filepath/
    name in order to acquire the header information. This works when
    typing it directly into the shell script (see above), but not when
    coded in my class.

    The IOError is printed above, but being a newbie I'm unsure of how to
    see the traceback. I've only been using these tools for a couple of
    days.
    Lionel, Feb 2, 2009
    #14
  15. Lionel

    Lionel Guest

    On Feb 2, 2:07 pm, John Machin <> wrote:
    > On Feb 3, 8:43 am, Lionel <> wrote:
    >
    > >         ResourceFilepath = DataFilepath + ".src"

    >
    > Don't you mean ".rsc"?


    Good Grief!!! That's It!! I've been staring at it all day and I didn't
    see it.

    I'm sorry I've wasted everyone's time. This is bloody embarassing.
    Amateur night.

    Thank you for the help everyone.

    -L
    Lionel, Feb 2, 2009
    #15
  16. Lionel

    MRAB Guest

    Lionel wrote:
    > On Feb 2, 1:07 pm, "Diez B. Roggisch" <> wrote:
    >
    > This is written very slowly, so you can read it better:
    >
    > Please post without sarcasm.
    >
    >
    > This is the output from my Python shell:
    >
    >>>> DatafilePath = "C:\\C8Example1.slc"
    >>>> ResourcefilePath = DatafilePath + ".rsc"
    >>>> DatafileFH = open(DatafilePath)
    >>>> ResourceFh = open(ResourcefilePath)
    >>>> DatafilePath

    > 'C:\\C8Example1.slc'
    >>>> ResourcefilePath

    > 'C:\\C8Example1.slc.rsc'
    >

    Here the extension is '.rsc' ...

    > It seems to run without trouble. However, here is the offending code
    > in my class (followed by console output):
    >
    > class C8DataType:
    >
    > def __init__(self, DataFilepath):
    >
    > try:
    > DataFH = open(DataFilepath, "rb")
    >
    > except IOError, message:
    > # Error opening file.
    > print(message)
    > return None
    >
    > ResourceFilepath = DataFilepath + ".src"
    >

    .... but here the extension is '.src'.

    Is that the problem.

    > print(DataFilepath)
    > print(ResourceFilepath)
    >
    > # Try to open resource file, catch exception:
    > try:
    > ResourceFH = open(ResourceFilepath)
    >
    > except IOError, message:
    > # Error opening file.
    > print(message)
    > print("Error opening " + ResourceFilepath)
    > DataFH.close()
    > return None
    >
    > Console output when invoking as "someObject = C8DataType("C:\
    > \C8Example1.slc")" :
    >
    > C:\C8Example1.slc
    > C:\C8Example1.slc.src
    > [Errno 2] No such file or directory: 'C:\\C8Example1.slc.src'
    > Error opening C:\C8Example1.slc.src
    >

    [snip]
    It can't find "C:\C8Example1.slc.src", but I think you intended
    "C:\C8Example1.slc.rsc".
    MRAB, Feb 2, 2009
    #16
  17. Lionel

    Denis Kasak Guest

    On Mon, Feb 2, 2009 at 10:43 PM, Lionel <> wrote:

    <snip>

    > >>> ResourcefilePath

    > 'C:\\C8Example1.slc.rsc'


    <snip>

    > C:\C8Example1.slc.src


    The extension you used in the interactive shell differs from the one
    you used in the class code (i.e. "rsc" vs "src").

    --
    Denis Kasak
    Denis Kasak, Feb 3, 2009
    #17
  18. Lionel

    Lionel Guest

    On Feb 2, 4:50 pm, Denis Kasak <> wrote:
    > On Mon, Feb 2, 2009 at 10:43 PM, Lionel <> wrote:
    >
    > <snip>
    >
    > > >>> ResourcefilePath

    > > 'C:\\C8Example1.slc.rsc'

    >
    > <snip>
    >
    > > C:\C8Example1.slc.src

    >
    > The extension you used in the interactive shell differs from the one
    > you used in the class code (i.e. "rsc" vs "src").
    >
    > --
    > Denis Kasak


    Yes, I see the problem now. Thank you everyone!

    -L
    Lionel, Feb 3, 2009
    #18
  19. Lionel

    Rhodri James Guest

    [Quoting restored for reduced

    On Mon, 02 Feb 2009 22:33:50 -0000, Lionel <> wrote:

    > On Feb 2, 2:01 pm, Mike Driscoll <> wrote:
    >> On Feb 2, 3:43 pm, Lionel <> wrote:
    >> > On Feb 2, 1:07 pm, "Diez B. Roggisch" <> wrote:

    >>
    >> >> This is written very slowly, so you can read it better:

    >>
    >> > Please post without sarcasm.


    On Usenet? You'll be wanting single unequivocal answers next!

    Seriously though, you had been asked several times for the traceback,
    so that we could stop guessing and tell you for sure what was going
    on, and you hadn't provided it. Diez's mild sarcasm was not uncalled-
    for. The fact that you didn't have a traceback partially excuses you,
    but it would have helped if you'd said so.

    >>
    >> > This is the output from my Python shell:

    >>
    >> > >>> DatafilePath = "C:\\C8Example1.slc"
    >> > >>> ResourcefilePath = DatafilePath + ".rsc"
    >> > >>> DatafileFH = open(DatafilePath)
    >> > >>> ResourceFh = open(ResourcefilePath)
    >> > >>> DatafilePath

    >>
    >> > 'C:\\C8Example1.slc'>>> ResourcefilePath

    >>
    >> > 'C:\\C8Example1.slc.rsc'

    >>
    >> > It seems to run without trouble. However, here is the offending code
    >> > in my class (followed by console output):

    >>
    >> > class C8DataType:

    >>
    >> >     def __init__(self, DataFilepath):

    >>
    >> >         try:
    >> >             DataFH = open(DataFilepath, "rb")

    >>
    >> >         except IOError, message:
    >> >             # Error opening file.
    >> >             print(message)
    >> >             return None


    You're catching the IOError, presumably so that you can fail
    gracefully elsewhere. This may not be a particularly good
    idea, and in any case it stops the exception reaching the
    console where it would cause the traceback to be displayed.
    More on this later.

    >> >         ResourceFilepath = DataFilepath + ".src"


    As other people have pointed out, you've got a typo here.

    >> >         print(DataFilepath)
    >> >         print(ResourceFilepath)

    >>
    >> >         # Try to open resource file, catch exception:
    >> >         try:
    >> >             ResourceFH = open(ResourceFilepath)

    >>
    >> >         except IOError, message:
    >> >             # Error opening file.
    >> >             print(message)
    >> >             print("Error opening " + ResourceFilepath)
    >> >             DataFH.close()
    >> >             return None


    [Huge amounts of text trimmed]

    Fair enough, you're catching the IOError so that you can
    ensure that DataFH is closed. Unfortunately this concealed
    the traceback information, which would have made it more
    obvious to you what people were talking about. Given that
    this has rather stuffed your C8DataType instance, you
    might want to think about re-raising the exception after
    you've closed DataFH and letting the outer layers deal
    with it in a more appropriate fashion.

    Incidentally, this code isn't going to do anything useful
    for you anyway even after you've fixed the typo. DataFH
    and ResourceFH are both local variables to __init__ and
    will be tossed away when it finishes executing. If you
    want to use them later, make them self.data_fh and
    self.resource_fh respectively.

    (PEP 8 recommends that you use lower_case_with_underscores
    for variable or attribute names, and leave MixedCase for
    class names.)

    --
    Rhodri James *-* Wildebeeste Herder to the Masses
    Rhodri James, Feb 3, 2009
    #19
  20. Lionel

    Lionel Guest

    On Feb 2, 5:40 pm, "Rhodri James" <> wrote:
    > [Quoting restored for reduced
    >
    > On Mon, 02 Feb 2009 22:33:50 -0000, Lionel <> wrote:
    > > On Feb 2, 2:01 pm, Mike Driscoll <> wrote:
    > >> On Feb 2, 3:43 pm, Lionel <> wrote:
    > >> > On Feb 2, 1:07 pm, "Diez B. Roggisch" <> wrote:

    >
    > >> >> This is written very slowly, so you can read it better:

    >
    > >> > Please post without sarcasm.

    >
    > On Usenet?  You'll be wanting single unequivocal answers next!
    >
    > Seriously though, you had been asked several times for the traceback,
    > so that we could stop guessing and tell you for sure what was going
    > on, and you hadn't provided it.  Diez's mild sarcasm was not uncalled-
    > for.  The fact that you didn't have a traceback partially excuses you,
    > but it would have helped if you'd said so.
    >
    >
    >
    >
    >
    >
    >
    > >> > This is the output from my Python shell:

    >
    > >> > >>> DatafilePath = "C:\\C8Example1.slc"
    > >> > >>> ResourcefilePath = DatafilePath + ".rsc"
    > >> > >>> DatafileFH = open(DatafilePath)
    > >> > >>> ResourceFh = open(ResourcefilePath)
    > >> > >>> DatafilePath

    >
    > >> > 'C:\\C8Example1.slc'>>> ResourcefilePath

    >
    > >> > 'C:\\C8Example1.slc.rsc'

    >
    > >> > It seems to run without trouble. However, here is the offending code
    > >> > in my class (followed by console output):

    >
    > >> > class C8DataType:

    >
    > >> >     def __init__(self, DataFilepath):

    >
    > >> >         try:
    > >> >             DataFH = open(DataFilepath, "rb")

    >
    > >> >         except IOError, message:
    > >> >             # Error opening file.
    > >> >             print(message)
    > >> >             return None

    >
    > You're catching the IOError, presumably so that you can fail
    > gracefully elsewhere.  This may not be a particularly good
    > idea, and in any case it stops the exception reaching the
    > console where it would cause the traceback to be displayed.
    > More on this later.
    >
    > >> >         ResourceFilepath = DataFilepath + ".src"

    >
    > As other people have pointed out, you've got a typo here.
    >
    > >> >         print(DataFilepath)
    > >> >         print(ResourceFilepath)

    >
    > >> >         # Try to open resource file, catch exception:
    > >> >         try:
    > >> >             ResourceFH = open(ResourceFilepath)

    >
    > >> >         except IOError, message:
    > >> >             # Error opening file.
    > >> >             print(message)
    > >> >             print("Error opening " + ResourceFilepath)
    > >> >             DataFH.close()
    > >> >             return None

    >
    > [Huge amounts of text trimmed]
    >
    > Fair enough, you're catching the IOError so that you can
    > ensure that DataFH is closed.  Unfortunately this concealed
    > the traceback information, which would have made it more
    > obvious to you what people were talking about.  Given that
    > this has rather stuffed your C8DataType instance, you
    > might want to think about re-raising the exception after
    > you've closed DataFH and letting the outer layers deal
    > with it in a more appropriate fashion.
    >
    > Incidentally, this code isn't going to do anything useful
    > for you anyway even after you've fixed the typo.  DataFH
    > and ResourceFH are both local variables to __init__ and
    > will be tossed away when it finishes executing.  If you
    > want to use them later, make them self.data_fh and
    > self.resource_fh respectively.
    >
    > (PEP 8 recommends that you use lower_case_with_underscores
    > for variable or attribute names, and leave MixedCase for
    > class names.)
    >
    > --
    > Rhodri James *-* Wildebeeste Herder to the Masses- Hide quoted text -
    >
    > - Show quoted text -


    Very good comments. I'll be implementing some of your suggestions, to
    be sure. Thanks Rhodri.
    Lionel, Feb 3, 2009
    #20
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