# Re: problems with opening files due to file's path

Discussion in 'Python' started by Alexnb, Jun 10, 2008.

1. ### AlexnbGuest

Gerhard HÃ¤ring wrote:
>
> Alexnb wrote:
>> Okay, so what I want my program to do it open a file, a music file in
>> specific, and for this we will say it is an .mp3. Well, I am using the
>> system() command from the os class. [...]
>>
>> system("\"C:\Documents and Settings\Alex\My Documents\My
>> Music\Rhapsody\Bryanbros\Weezer\(2001)\04 - Island In The Sun.wma\"")
>> [...]

>
> Try os.startfile() instead. It should work better.
>
> -- Gerhard
>
> --
> http://mail.python.org/mailman/listinfo/python-list
>
>

No, it didn't work, but it gave me some interesting feedback when I ran it
in the shell. Heres what it told me:

>>> os.startfile("C:\Documents and Settings\Alex\My Documents\My
>>> Music\Rhapsody\Bryanbros\Jason Mraz\I'm Yours (Single)\01 - I'm
>>> Yours.wma")

Traceback (most recent call last):
File "<pyshell#10>", line 1, in <module>
os.startfile("C:\Documents and Settings\Alex\My Documents\My
Music\Rhapsody\Bryanbros\Jason Mraz\I'm Yours (Single)\01 - I'm Yours.wma")

WindowsError: [Error 2] The system cannot find the file specified:
"C:\\Documents and Settings\\Alex\\My Documents\\My
Music\\Rhapsody\\Bryanbros\\Jason Mraz\\I'm Yours (Single)\x01 - I'm
Yours.wma"

See it made each backslash into two, and the one by the parenthesis and the
0 turned into an x....
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Alexnb, Jun 10, 2008

2. ### Mike DriscollGuest

On Jun 10, 11:45 am, Alexnb <> wrote:
> Gerhard Häring wrote:
>
> > Alexnb wrote:
> >> Okay, so what I want my program to do it open a file, a music file in
> >> specific, and for this we will say it is an .mp3. Well, I am using the
> >> system() command from the os class. [...]

>
> >> system("\"C:\Documents and Settings\Alex\My Documents\My
> >> Music\Rhapsody\Bryanbros\Weezer\(2001)\04 - Island In The Sun.wma\"")
> >> [...]

>
> > Try os.startfile() instead. It should work better.

>
> > -- Gerhard

>
>
> No, it didn't work, but it gave me some interesting feedback when I ran it
> in the shell. Heres what it told me:
>
> >>> os.startfile("C:\Documents and Settings\Alex\My Documents\My
> >>> Music\Rhapsody\Bryanbros\Jason Mraz\I'm Yours (Single)\01 - I'm
> >>> Yours.wma")

>
> Traceback (most recent call last):
>   File "<pyshell#10>", line 1, in <module>
>     os.startfile("C:\Documents and Settings\Alex\My Documents\My
> Music\Rhapsody\Bryanbros\Jason Mraz\I'm Yours (Single)\01 - I'm Yours.wma")
>
> WindowsError: [Error 2] The system cannot find the file specified:
> "C:\\Documents and Settings\\Alex\\My Documents\\My
> Music\\Rhapsody\\Bryanbros\\Jason Mraz\\I'm Yours (Single)\x01 - I'm
> Yours.wma"
>
> See it made each backslash into two, and the one by the parenthesis and the
> 0 turned into an x....
> --
> View this message in context:http://www.nabble.com/problems-with-opening-files-due-to-file's-pat...
> Sent from the Python - python-list mailing list archive at Nabble.com.

Yeah. You need to either double all the backslashes or make it a raw
string by adding an "r" to the beginning, like so:

os.startfile(r'C:\path\to\my\file')

HTH

Mike

Mike Driscoll, Jun 10, 2008

3. ### AlexnbGuest

Hey thanks!, both the raw and the double backslashes worked. You are a
gentleman and a scholar.

Mike Driscoll wrote:
>
> On Jun 10, 11:45 am, Alexnb <> wrote:
>> Gerhard HÃ¤ring wrote:
>>
>> > Alexnb wrote:
>> >> Okay, so what I want my program to do it open a file, a music file in
>> >> specific, and for this we will say it is an .mp3. Well, I am using the
>> >> system() command from the os class. [...]

>>
>> >> system("\"C:\Documents and Settings\Alex\My Documents\My
>> >> Music\Rhapsody\Bryanbros\Weezer\(2001)\04 - Island In The Sun.wma\"")
>> >> [...]

>>
>> > Try os.startfile() instead. It should work better.

>>
>> > -- Gerhard

>>
>>
>> No, it didn't work, but it gave me some interesting feedback when I ran
>> it
>> in the shell. Heres what it told me:
>>
>> >>> os.startfile("C:\Documents and Settings\Alex\My Documents\My
>> >>> Music\Rhapsody\Bryanbros\Jason Mraz\I'm Yours (Single)\01 - I'm
>> >>> Yours.wma")

>>
>> Traceback (most recent call last):
>> File "<pyshell#10>", line 1, in <module>
>> os.startfile("C:\Documents and Settings\Alex\My Documents\My
>> Music\Rhapsody\Bryanbros\Jason Mraz\I'm Yours (Single)\01 - I'm
>> Yours.wma")
>>
>> WindowsError: [Error 2] The system cannot find the file specified:
>> "C:\\Documents and Settings\\Alex\\My Documents\\My
>> Music\\Rhapsody\\Bryanbros\\Jason Mraz\\I'm Yours (Single)\x01 - I'm
>> Yours.wma"
>>
>> See it made each backslash into two, and the one by the parenthesis and
>> the
>> 0 turned into an x....
>> --
>> View this message in
>> context:http://www.nabble.com/problems-with-opening-files-due-to-file's-pat...
>> Sent from the Python - python-list mailing list archive at Nabble.com.

>
> Yeah. You need to either double all the backslashes or make it a raw
> string by adding an "r" to the beginning, like so:
>
> os.startfile(r'C:\path\to\my\file')
>
> HTH
>
> Mike
> --
> http://mail.python.org/mailman/listinfo/python-list
>
>

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Alexnb, Jun 10, 2008
4. ### AlexnbGuest

Well, now i've hit another problem, this time being that the path will be a
variable, and I can't figure out how to make startfile() make it raw with a
variable, if I put startfile(r variable), it doesn't work and
startfile(rvariable) obviously won't work, do you know how to make that work
or better yet, how to take a regular string that is given and make every
single "\" into a double "\\"?

Mike Driscoll wrote:
>
> On Jun 10, 11:45 am, Alexnb <> wrote:
>> Gerhard HÃ¤ring wrote:
>>
>> > Alexnb wrote:
>> >> Okay, so what I want my program to do it open a file, a music file in
>> >> specific, and for this we will say it is an .mp3. Well, I am using the
>> >> system() command from the os class. [...]

>>
>> >> system("\"C:\Documents and Settings\Alex\My Documents\My
>> >> Music\Rhapsody\Bryanbros\Weezer\(2001)\04 - Island In The Sun.wma\"")
>> >> [...]

>>
>> > Try os.startfile() instead. It should work better.

>>
>> > -- Gerhard

>>
>>
>> No, it didn't work, but it gave me some interesting feedback when I ran
>> it
>> in the shell. Heres what it told me:
>>
>> >>> os.startfile("C:\Documents and Settings\Alex\My Documents\My
>> >>> Music\Rhapsody\Bryanbros\Jason Mraz\I'm Yours (Single)\01 - I'm
>> >>> Yours.wma")

>>
>> Traceback (most recent call last):
>> File "<pyshell#10>", line 1, in <module>
>> os.startfile("C:\Documents and Settings\Alex\My Documents\My
>> Music\Rhapsody\Bryanbros\Jason Mraz\I'm Yours (Single)\01 - I'm
>> Yours.wma")
>>
>> WindowsError: [Error 2] The system cannot find the file specified:
>> "C:\\Documents and Settings\\Alex\\My Documents\\My
>> Music\\Rhapsody\\Bryanbros\\Jason Mraz\\I'm Yours (Single)\x01 - I'm
>> Yours.wma"
>>
>> See it made each backslash into two, and the one by the parenthesis and
>> the
>> 0 turned into an x....
>> --
>> View this message in
>> context:http://www.nabble.com/problems-with-opening-files-due-to-file's-pat...
>> Sent from the Python - python-list mailing list archive at Nabble.com.

>
> Yeah. You need to either double all the backslashes or make it a raw
> string by adding an "r" to the beginning, like so:
>
> os.startfile(r'C:\path\to\my\file')
>
> HTH
>
> Mike
> --
> http://mail.python.org/mailman/listinfo/python-list
>
>

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Alexnb, Jun 10, 2008
5. ### Thomas MortonGuest

maybe try string substitution... not sure if that's really the BEST way to
do it but it should work

startfile(r"%s"%variable)

--------------------------------------------------
From: "Alexnb" <>
Sent: Tuesday, June 10, 2008 7:05 PM
To: <>
Subject: Re: problems with opening files due to file's path

>
> Well, now i've hit another problem, this time being that the path will be
> a
> variable, and I can't figure out how to make startfile() make it raw with
> a
> variable, if I put startfile(r variable), it doesn't work and
> startfile(rvariable) obviously won't work, do you know how to make that
> work
> or better yet, how to take a regular string that is given and make every
> single "\" into a double "\\"?
>
> Mike Driscoll wrote:
>>
>> On Jun 10, 11:45 am, Alexnb <> wrote:
>>> Gerhard HÃ¤ring wrote:
>>>
>>> > Alexnb wrote:
>>> >> Okay, so what I want my program to do it open a file, a music file in
>>> >> specific, and for this we will say it is an .mp3. Well, I am using
>>> >> the
>>> >> system() command from the os class. [...]
>>>
>>> >> system("\"C:\Documents and Settings\Alex\My Documents\My
>>> >> Music\Rhapsody\Bryanbros\Weezer\(2001)\04 - Island In The Sun.wma\"")
>>> >> [...]
>>>
>>> > Try os.startfile() instead. It should work better.
>>>
>>> > -- Gerhard
>>>
>>> > --
>>> >http://mail.python.org/mailman/listinfo/python-list
>>>
>>> No, it didn't work, but it gave me some interesting feedback when I ran
>>> it
>>> in the shell. Heres what it told me:
>>>
>>> >>> os.startfile("C:\Documents and Settings\Alex\My Documents\My
>>> >>> Music\Rhapsody\Bryanbros\Jason Mraz\I'm Yours (Single)\01 - I'm
>>> >>> Yours.wma")
>>>
>>> Traceback (most recent call last):
>>> File "<pyshell#10>", line 1, in <module>
>>> os.startfile("C:\Documents and Settings\Alex\My Documents\My
>>> Music\Rhapsody\Bryanbros\Jason Mraz\I'm Yours (Single)\01 - I'm
>>> Yours.wma")
>>>
>>> WindowsError: [Error 2] The system cannot find the file specified:
>>> "C:\\Documents and Settings\\Alex\\My Documents\\My
>>> Music\\Rhapsody\\Bryanbros\\Jason Mraz\\I'm Yours (Single)\x01 - I'm
>>> Yours.wma"
>>>
>>> See it made each backslash into two, and the one by the parenthesis and
>>> the
>>> 0 turned into an x....
>>> --
>>> View this message in
>>> context:http://www.nabble.com/problems-with-opening-files-due-to-file's-pat...
>>> Sent from the Python - python-list mailing list archive at Nabble.com.

>>
>> Yeah. You need to either double all the backslashes or make it a raw
>> string by adding an "r" to the beginning, like so:
>>
>> os.startfile(r'C:\path\to\my\file')
>>
>> HTH
>>
>> Mike
>> --
>> http://mail.python.org/mailman/listinfo/python-list
>>
>>

>
> --
> View this message in context:
> http://www.nabble.com/problems-with-opening-files-due-to-file's-path-tp17759531p17761338.html
> Sent from the Python - python-list mailing list archive at Nabble.com.
>
> --
> http://mail.python.org/mailman/listinfo/python-list

Thomas Morton, Jun 10, 2008
6. ### AlexnbGuest

No this time it perhaps gave me the worst of all heres what I entered, and
the output

>>> startfile(r"%s"%full) ***full is the path***

startfile(r"%s"%full)

WindowsError: [Error 2] The system cannot find the file specified:
'"C:\\Documents and Settings\\Alex\\My Documents\\My
Music\\Rhapsody\\Bryanbros\\Jason Mraz\\I\'m Yours (Single)\x01 - I\'m
Yours.wma"'

Thomas Morton wrote:
>
> maybe try string substitution... not sure if that's really the BEST way to
> do it but it should work
>
> startfile(r"%s"%variable)
>
> --------------------------------------------------
> From: "Alexnb" <>
> Sent: Tuesday, June 10, 2008 7:05 PM
> To: <>
> Subject: Re: problems with opening files due to file's path
>
>>
>> Well, now i've hit another problem, this time being that the path will be
>> a
>> variable, and I can't figure out how to make startfile() make it raw with
>> a
>> variable, if I put startfile(r variable), it doesn't work and
>> startfile(rvariable) obviously won't work, do you know how to make that
>> work
>> or better yet, how to take a regular string that is given and make every
>> single "\" into a double "\\"?
>>
>> Mike Driscoll wrote:
>>>
>>> On Jun 10, 11:45 am, Alexnb <> wrote:
>>>> Gerhard HÃ¤ring wrote:
>>>>
>>>> > Alexnb wrote:
>>>> >> Okay, so what I want my program to do it open a file, a music file
>>>> in
>>>> >> specific, and for this we will say it is an .mp3. Well, I am using
>>>> >> the
>>>> >> system() command from the os class. [...]
>>>>
>>>> >> system("\"C:\Documents and Settings\Alex\My Documents\My
>>>> >> Music\Rhapsody\Bryanbros\Weezer\(2001)\04 - Island In The
>>>> Sun.wma\"")
>>>> >> [...]
>>>>
>>>> > Try os.startfile() instead. It should work better.
>>>>
>>>> > -- Gerhard
>>>>
>>>> > --
>>>> >http://mail.python.org/mailman/listinfo/python-list
>>>>
>>>> No, it didn't work, but it gave me some interesting feedback when I ran
>>>> it
>>>> in the shell. Heres what it told me:
>>>>
>>>> >>> os.startfile("C:\Documents and Settings\Alex\My Documents\My
>>>> >>> Music\Rhapsody\Bryanbros\Jason Mraz\I'm Yours (Single)\01 - I'm
>>>> >>> Yours.wma")
>>>>
>>>> Traceback (most recent call last):
>>>> File "<pyshell#10>", line 1, in <module>
>>>> os.startfile("C:\Documents and Settings\Alex\My Documents\My
>>>> Music\Rhapsody\Bryanbros\Jason Mraz\I'm Yours (Single)\01 - I'm
>>>> Yours.wma")
>>>>
>>>> WindowsError: [Error 2] The system cannot find the file specified:
>>>> "C:\\Documents and Settings\\Alex\\My Documents\\My
>>>> Music\\Rhapsody\\Bryanbros\\Jason Mraz\\I'm Yours (Single)\x01 - I'm
>>>> Yours.wma"
>>>>
>>>> See it made each backslash into two, and the one by the parenthesis and
>>>> the
>>>> 0 turned into an x....
>>>> --
>>>> View this message in
>>>> context:http://www.nabble.com/problems-with-opening-files-due-to-file's-pat...
>>>> Sent from the Python - python-list mailing list archive at Nabble.com.
>>>
>>> Yeah. You need to either double all the backslashes or make it a raw
>>> string by adding an "r" to the beginning, like so:
>>>
>>> os.startfile(r'C:\path\to\my\file')
>>>
>>> HTH
>>>
>>> Mike
>>> --
>>> http://mail.python.org/mailman/listinfo/python-list
>>>
>>>

>>
>> --
>> View this message in context:
>> http://www.nabble.com/problems-with-opening-files-due-to-file's-path-tp17759531p17761338.html
>> Sent from the Python - python-list mailing list archive at Nabble.com.
>>
>> --
>> http://mail.python.org/mailman/listinfo/python-list

>
> --
> http://mail.python.org/mailman/listinfo/python-list
>

--
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Alexnb, Jun 10, 2008
7. ### Mike DriscollGuest

On Jun 10, 1:25 pm, "Thomas Morton" <>
wrote:
> maybe try string substitution... not sure if that's really the BEST way to
> do it but it should work
>
> startfile(r"%s"%variable)

I concur. That should work. A slightly more in depth example (assuming
Windows):

os.startfile(r'C:\Documents and Settings\%s\Desktop\myApp.exe' %

or

os.startfile(r'C:\Program Files\%s' % myApp)

Hopefully this is what you are talking about. If you were referring to
passing in arguments, than you'll want to use the subprocess module

>
> --------------------------------------------------
> From: "Alexnb" <>
> Sent: Tuesday, June 10, 2008 7:05 PM
> To: <>
> Subject: Re: problems with opening files due to file's path
>
>
>
> > Well, now i've hit another problem, this time being that the path will be
> > a
> > variable, and I can't figure out how to make startfile() make it raw with
> > a
> > variable, if I put startfile(r variable), it doesn't work and
> > startfile(rvariable) obviously won't work, do you know how to make that
> > work
> > or better yet, how to take a regular string that is given and make every
> > single "\" into a double "\\"?

>

<snip>

Mike

Mike Driscoll, Jun 10, 2008
8. ### AlexnbGuest

That would work, but not for what I want. See the file could be anywhere on
the user's system and so the entire path will be unique, and that didn't
work with a unique path. What is the subprocess module you are talking

Mike Driscoll wrote:
>
> On Jun 10, 1:25Â pm, "Thomas Morton" <>
> wrote:
>> maybe try string substitution... not sure if that's really the BEST way
>> to
>> do it but it should work
>>
>> startfile(r"%s"%variable)

>
>
> I concur. That should work. A slightly more in depth example (assuming
> Windows):
>
> os.startfile(r'C:\Documents and Settings\%s\Desktop\myApp.exe' %
>
> or
>
> os.startfile(r'C:\Program Files\%s' % myApp)
>
> Hopefully this is what you are talking about. If you were referring to
> passing in arguments, than you'll want to use the subprocess module
>
>
>>
>> --------------------------------------------------
>> From: "Alexnb" <>
>> Sent: Tuesday, June 10, 2008 7:05 PM
>> To: <>
>> Subject: Re: problems with opening files due to file's path
>>
>>
>>
>> > Well, now i've hit another problem, this time being that the path will

>> be
>> > a
>> > variable, and I can't figure out how to make startfile() make it raw

>> with
>> > a
>> > variable, if I put startfile(r variable), it doesn't work and
>> > startfile(rvariable) obviously won't work, do you know how to make that
>> > work
>> > or better yet, how to take a regular string that is given and make

>> every
>> > single "\" into a double "\\"?

>>

>
> <snip>
>
> Mike
> --
> http://mail.python.org/mailman/listinfo/python-list
>
>

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Alexnb, Jun 10, 2008
9. ### Thomas MortonGuest

Just tested locally in IDLE (I know I know!) and it works for me like this:

>>> test = os.path.join(os.getcwd(),"NEWS.txt")
>>> test

'D:\\Python25\\NEWS.txt'
>>> os.startfile(r"%s"%test)
>>>

And the file opens...

Does the file definitely exist?

Tom
--------------------------------------------------
From: "Alexnb" <>
Sent: Tuesday, June 10, 2008 7:37 PM
To: <>
Subject: Re: problems with opening files due to file's path

>
> No this time it perhaps gave me the worst of all heres what I entered, and
> the output
>
>>>> startfile(r"%s"%full) ***full is the path***

>
> startfile(r"%s"%full)
>
> WindowsError: [Error 2] The system cannot find the file specified:
> '"C:\\Documents and Settings\\Alex\\My Documents\\My
> Music\\Rhapsody\\Bryanbros\\Jason Mraz\\I\'m Yours (Single)\x01 - I\'m
> Yours.wma"'
>
>
> Thomas Morton wrote:
>>
>> maybe try string substitution... not sure if that's really the BEST way
>> to
>> do it but it should work
>>
>> startfile(r"%s"%variable)
>>
>> --------------------------------------------------
>> From: "Alexnb" <>
>> Sent: Tuesday, June 10, 2008 7:05 PM
>> To: <>
>> Subject: Re: problems with opening files due to file's path
>>
>>>
>>> Well, now i've hit another problem, this time being that the path will
>>> be
>>> a
>>> variable, and I can't figure out how to make startfile() make it raw
>>> with
>>> a
>>> variable, if I put startfile(r variable), it doesn't work and
>>> startfile(rvariable) obviously won't work, do you know how to make that
>>> work
>>> or better yet, how to take a regular string that is given and make every
>>> single "\" into a double "\\"?
>>>
>>> Mike Driscoll wrote:
>>>>
>>>> On Jun 10, 11:45 am, Alexnb <> wrote:
>>>>> Gerhard HÃ¤ring wrote:
>>>>>
>>>>> > Alexnb wrote:
>>>>> >> Okay, so what I want my program to do it open a file, a music file
>>>>> in
>>>>> >> specific, and for this we will say it is an .mp3. Well, I am using
>>>>> >> the
>>>>> >> system() command from the os class. [...]
>>>>>
>>>>> >> system("\"C:\Documents and Settings\Alex\My Documents\My
>>>>> >> Music\Rhapsody\Bryanbros\Weezer\(2001)\04 - Island In The
>>>>> Sun.wma\"")
>>>>> >> [...]
>>>>>
>>>>> > Try os.startfile() instead. It should work better.
>>>>>
>>>>> > -- Gerhard
>>>>>
>>>>> > --
>>>>> >http://mail.python.org/mailman/listinfo/python-list
>>>>>
>>>>> No, it didn't work, but it gave me some interesting feedback when I
>>>>> ran
>>>>> it
>>>>> in the shell. Heres what it told me:
>>>>>
>>>>> >>> os.startfile("C:\Documents and Settings\Alex\My Documents\My
>>>>> >>> Music\Rhapsody\Bryanbros\Jason Mraz\I'm Yours (Single)\01 - I'm
>>>>> >>> Yours.wma")
>>>>>
>>>>> Traceback (most recent call last):
>>>>> File "<pyshell#10>", line 1, in <module>
>>>>> os.startfile("C:\Documents and Settings\Alex\My Documents\My
>>>>> Music\Rhapsody\Bryanbros\Jason Mraz\I'm Yours (Single)\01 - I'm
>>>>> Yours.wma")
>>>>>
>>>>> WindowsError: [Error 2] The system cannot find the file specified:
>>>>> "C:\\Documents and Settings\\Alex\\My Documents\\My
>>>>> Music\\Rhapsody\\Bryanbros\\Jason Mraz\\I'm Yours (Single)\x01 - I'm
>>>>> Yours.wma"
>>>>>
>>>>> See it made each backslash into two, and the one by the parenthesis
>>>>> and
>>>>> the
>>>>> 0 turned into an x....
>>>>> --
>>>>> View this message in
>>>>> context:http://www.nabble.com/problems-with-opening-files-due-to-file's-pat...
>>>>> Sent from the Python - python-list mailing list archive at Nabble.com.
>>>>
>>>> Yeah. You need to either double all the backslashes or make it a raw
>>>> string by adding an "r" to the beginning, like so:
>>>>
>>>> os.startfile(r'C:\path\to\my\file')
>>>>
>>>> HTH
>>>>
>>>> Mike
>>>> --
>>>> http://mail.python.org/mailman/listinfo/python-list
>>>>
>>>>
>>>
>>> --
>>> View this message in context:
>>> http://www.nabble.com/problems-with-opening-files-due-to-file's-path-tp17759531p17761338.html
>>> Sent from the Python - python-list mailing list archive at Nabble.com.
>>>
>>> --
>>> http://mail.python.org/mailman/listinfo/python-list

>>
>> --
>> http://mail.python.org/mailman/listinfo/python-list
>>

>
> --
> View this message in context:
> http://www.nabble.com/problems-with-opening-files-due-to-file's-path-tp17759531p17761946.html
> Sent from the Python - python-list mailing list archive at Nabble.com.
>
> --
> http://mail.python.org/mailman/listinfo/python-list

Thomas Morton, Jun 10, 2008
10. ### Carsten HaeseGuest

Alexnb wrote:
> No this time it perhaps gave me the worst of all heres what I entered, and
> the output
>
>>>> startfile(r"%s"%full) ***full is the path***

>
> startfile(r"%s"%full)
>
> WindowsError: [Error 2] The system cannot find the file specified:
> '"C:\\Documents and Settings\\Alex\\My Documents\\My
> Music\\Rhapsody\\Bryanbros\\Jason Mraz\\I\'m Yours (Single)\x01 - I\'m
> Yours.wma"'

Contrary to what other posters have asserted, doing the above can't make
a difference. Putting 'r' in front of a string literal tells Python not
to give backslashes in the string literal any special treatment. Since
there are no backslashes in "%s", the 'r' does nothing.

Therefore, (r"%s"%full) is the same as ("%s"%full), which is the same as
(full), assuming that full is the name of a string.

The real answer lies in fixing the code where you're assigning the
pathname to 'full', which you haven't posted. Please post the code where
you're assigning the pathname, or better yet, post the complete code
you're running.

--
Carsten Haese
http://informixdb.sourceforge.net

Carsten Haese, Jun 10, 2008
11. ### Mike DriscollGuest

On Jun 10, 2:09 pm, Carsten Haese <> wrote:
> Alexnb wrote:
> > No this time it perhaps gave me the worst of all heres what I entered, and
> > the output

>
> >>>> startfile(r"%s"%full)    ***full is the path***

>
> > startfile(r"%s"%full)

>
> > WindowsError: [Error 2] The system cannot find the file specified:
> > '"C:\\Documents and Settings\\Alex\\My Documents\\My
> > Music\\Rhapsody\\Bryanbros\\Jason Mraz\\I\'m Yours (Single)\x01 - I\'m
> > Yours.wma"'

>
> Contrary to what other posters have asserted, doing the above can't make
> a difference. Putting 'r' in front of a string literal tells Python not
> to give backslashes in the string literal any special treatment. Since
> there are no backslashes in "%s", the 'r' does nothing.

I assumed the OP was trying to do the string substitution within a
path. If the OP is instead doing as you think, then you are quite
correct.

>
> Therefore, (r"%s"%full) is the same as ("%s"%full), which is the same as
> (full), assuming that full is the name of a string.
>
> The real answer lies in fixing the code where you're assigning the
> pathname to 'full', which you haven't posted. Please post the code where
> you're assigning the pathname, or better yet, post the complete code
> you're running.
>
> --
> Carsten Haesehttp://informixdb.sourceforge.net

Sometimes I get too eager to help and don't do enough mental

Mike

Mike Driscoll, Jun 10, 2008
12. ### Mike DriscollGuest

On Jun 10, 1:57 pm, Alexnb <> wrote:
> That would work, but not for what I want. See the file could be anywhere on
> the user's system and so the entire path will be unique, and that didn't
> work with a unique path. What is the subprocess module you are talking
>

<snip>

As Carsten pointed out, we don't really know what you're doing. Or at
least, I don't. Why do you even want to do string substitution? How
does the user navigate to the files that the user wants to open? Are
you using a GUI or a command line interface?

you this:

http://docs.python.org/lib/module-subprocess.html

Mike

Mike Driscoll, Jun 10, 2008
13. ### AlexnbGuest

I am using GUI, Tkinter to be exact. But regardless of how the path gets
there, it needs to opened correctly. The problem I am running into is that
the program receives a path of a file, either .wma or .mp3 and is supposed
to open it. I run into problems when there is either a ")" or a number next
to the backslash "\" in the file path. I am looking for a way to make it
work with a variable, I can make it work when I physically type it in, but
not with a variable that holds the path.

Mike Driscoll wrote:
>
> On Jun 10, 1:57Â pm, Alexnb <> wrote:
>> That would work, but not for what I want. See the file could be anywhere
>> on
>> the user's system and so the entire path will be unique, and that didn't
>> work with a unique path. What is the subprocess module you are talking
>>

>
>
> <snip>
>
> As Carsten pointed out, we don't really know what you're doing. Or at
> least, I don't. Why do you even want to do string substitution? How
> does the user navigate to the files that the user wants to open? Are
> you using a GUI or a command line interface?
>
> you this:
>
> http://docs.python.org/lib/module-subprocess.html
>
> Mike
> --
> http://mail.python.org/mailman/listinfo/python-list
>
>

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Alexnb, Jun 11, 2008
14. ### AlexnbGuest

Okay, I don't understand how it is too vague, but here:

>>> path = "C:\Documents and Settings\Alex\My Documents\My
>>> Music\Rhapsody\Bryanbros\Jason Mraz\I'm Yours (Single)\01 - I'm
>>> Yours.wma"
>>> os.startfile(path)

Traceback (most recent call last):
File "<pyshell#39>", line 1, in <module>
os.startfile(path)
WindowsError: [Error 2] The system cannot find the file specified:
"C:\\Documents and Settings\\Alex\\My Documents\\My
Music\\Rhapsody\\Bryanbros\\Jason Mraz\\I'm Yours (Single)\x01 - I'm
Yours.wma"

Here's another way:

>>> os.startfile(r"%s"%path)

Traceback (most recent call last):
File "<pyshell#40>", line 1, in <module>
os.startfile(r"%s"%path)
WindowsError: [Error 2] The system cannot find the file specified:
"C:\\Documents and Settings\\Alex\\My Documents\\My
Music\\Rhapsody\\Bryanbros\\Jason Mraz\\I'm Yours (Single)\x01 - I'm
Yours.wma"

Same output, however if I personally input it like so:

>>> os.startfile("C:\\Documents and Settings\\Alex\\My Documents\\My
>>> Music\\Rhapsody\\Bryanbros\\Jason Mraz\\I'm Yours (Single)\\01 - I'm
>>> Yours.wma")

It works out fine because I can make each backslash doubles so it doesn't
mess stuff up. So if I could take the path varible and make ever "\" into a
"\\" then it would also work.

Did I clarify?

Grant Edwards wrote:
>
> On 2008-06-11, Alexnb <> wrote:
>
>> I am using GUI, Tkinter to be exact. But regardless of how the
>> path gets there, it needs to opened correctly. The problem I
>> am running into is that the program receives a path of a file,
>> either .wma or .mp3 and is supposed to open it. I run into
>> problems when there is either a ")" or a number next to the
>> backslash "\" in the file path. I am looking for a way to make
>> it work with a variable, I can make it work when I physically
>> type it in, but not with a variable that holds the path.

>
> You're going to have to show us code and example input and
> output. Your description of the problem is far too vague for
>
> --
> Grant Edwards grante Yow! With YOU, I can
> be
> at MYSELF... We don't NEED
> visi.com Dan Rather...
> --
> http://mail.python.org/mailman/listinfo/python-list
>
>

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Alexnb, Jun 11, 2008
15. ### Carsten HaeseGuest

Alexnb wrote:
> Okay, I don't understand how it is too vague, but here:
>
> [snip a bunch of irrelevant examples...]
>
> Did I clarify?

No. Earlier you wrote:

>> On 2008-06-11, Alexnb <> wrote:
>>> I am using GUI, Tkinter to be exact. But regardless of how the
>>> path gets there, it needs to opened correctly.

This implies that the file doesn't get opened correctly if the file name
is entered/chosen in the GUI. Yet, the examples you posted don't contain
any GUI code whatsoever. They merely demonstrate that you don't have a
firm grasp on how backslashes in string literals are treated.

So, this begs the question, do you actually have any GUI code that is
failing, or are you just worried, given the problems you had with string
literals, that the GUI code you have yet to write will fail in the same way?

If this is the case, you should just write the GUI code and try it. It
might just work. Backslashes entered into a GUI text box are not treated
the same as backslashes in a Python string literal.

If, on the other hand, you do have some GUI code for getting the file
name from the user, and that code is failing, then please, show us THAT
CODE and show us how it's failing.

--
Carsten Haese
http://informixdb.sourceforge.net

Carsten Haese, Jun 11, 2008
16. ### AlexnbGuest

I don't think you understand it doesn't matter how the variable gets there,
the same code is run regardless, I have no problem with the GUI, but you
asked, and so I told you. the code os.startfile(.... is run if there is a
GUI or it is a console app.

Carsten Haese-2 wrote:
>
> Alexnb wrote:
>> Okay, I don't understand how it is too vague, but here:
>>
> > [snip a bunch of irrelevant examples...]
>>
>> Did I clarify?

>
> No. Earlier you wrote:
>
>>> On 2008-06-11, Alexnb <> wrote:
>>>> I am using GUI, Tkinter to be exact. But regardless of how the
>>>> path gets there, it needs to opened correctly.

>
> This implies that the file doesn't get opened correctly if the file name
> is entered/chosen in the GUI. Yet, the examples you posted don't contain
> any GUI code whatsoever. They merely demonstrate that you don't have a
> firm grasp on how backslashes in string literals are treated.
>
> So, this begs the question, do you actually have any GUI code that is
> failing, or are you just worried, given the problems you had with string
> literals, that the GUI code you have yet to write will fail in the same
> way?
>
> If this is the case, you should just write the GUI code and try it. It
> might just work. Backslashes entered into a GUI text box are not treated
> the same as backslashes in a Python string literal.
>
> If, on the other hand, you do have some GUI code for getting the file
> name from the user, and that code is failing, then please, show us THAT
> CODE and show us how it's failing.
>
> --
> Carsten Haese
> http://informixdb.sourceforge.net
> --
> http://mail.python.org/mailman/listinfo/python-list
>
>

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Alexnb, Jun 11, 2008
17. ### Carsten HaeseGuest

Alexnb wrote:
> I don't think you understand it doesn't matter how the variable gets there

But it *does* matter. Compare this:

py> filename = "C:\Somewhere\01 - Some Song.mp3"
py> print filename
C:\Somewhere - Some Song.mp3

To this:

py> filename = raw_input("Enter the filename: ")
Enter the filename: C:\Somewhere\01 - Some Song.mp3
py> print filename
C:\Somewhere\01 - Some Song.mp3

Note that the "\01" in the first case seems to have disappeared, whereas
in the second case it's preserved.

description of the ACTUAL problem.

--
Carsten Haese
http://informixdb.sourceforge.net

Carsten Haese, Jun 11, 2008
18. ### LieGuest

On Jun 11, 10:07 am, Alexnb <> wrote:
> I don't think you understand it doesn't matter how the variable gets there,
> the same code is run regardless, I have no problem with the GUI, but you
> asked, and so I told you. the code os.startfile(.... is run if there is a
> GUI or it is a console app.

(snip)

I think I know why you get confused by this, clearly, you have no idea
of what an escape character and escape sequence is.

Python (and C/C++ and some other languages) treats \ (the backspace
character) inside a string specially, it makes the character after the
backspace lose their special meaning OR get a special meaning, you
might probably be used to seeing something like this: 'First line
\nSecond Line', which when printed, would give:

>>> print 'First Line\nSecond Line'

First Line
Second Line

The second behavior of the escape sequence is to make special
character (generally the backspace itself), lose their special
meaning:
>>> print 'path\\file.txt'

path\file.txt

In some cases, you might sometimes want a path like this: 'path
\nick.txt'
if you do this:
>>> print 'path\nick.txt'

path
ick.txt

because the \n is considered as a newline.

>>> print 'path\\nick.txt'

path\nick.txt

or
>>> print r'path\nick.txt'

path\nick.txt

the r'' string is raw string, most of the magics of a regular string
'' is lost for an r'' string. It allows you to avoid the need to
escape the special characters. Raw string is usually used for re
(regular expressions) and paths in Windows both of which uses the
backslash character regularly.

you first case of:
system("\"C:\Documents and Settings\Alex\My Documents\My Music\Rhapsody
\Bryanbros\Weezer\(2001)\04 - Island In The Sun.wma\"")

is interpreted by python as this:
"C:\Documents and Settings\Alex\My Documents\My Music\Rhapsody
\Bryanbros\Weezer\(2001) - Island In The Sun.wma"

Notice that the '\0' is substituted into '', because \0 is the escape
sequence for null character.
(Personally I think python should raise errors if any escape character
that isn't followed by a valid escape sequence is found (such as \D,
\A, \M, \M, \R, \B, \W, \(), but perhaps they're trying not to be too
mean for newbies.)

that should be correctly written like this:
system(r'"C:\Documents and Settings\Alex\My Documents\My Music\Rhapsody
\Bryanbros\Weezer\(2001)\04 - Island In The Sun.wma"')
or:
system('"C:\\Documents and Settings\\Alex\\My Documents\\My Music\
\Rhapsody\\Bryanbros\\Weezer\\(2001)\\04 - Island In The Sun.wma"')

Now, to the next question:
How if I want the path to come from a variable:
path = "C:\Documents and Settings\Alex\My Documents\My Music\Rhapsody
\Bryanbros\Jason Mraz\I'm Yours (Single)\01 - I'm Yours.wma"
os.startfile(path)

I can see in a glance why it doesn't work, the string is escaped by
python, it should be written like this.
path = r"C:\Documents and Settings\Alex\My Documents\My Music\Rhapsody
\Bryanbros\Jason Mraz\I'm Yours (Single)\01 - I'm Yours.wma"
os.startfile(path)

OR:
path = "C:\\Documents and Settings\\Alex\\My Documents\\My Music\
\Rhapsody\\Bryanbros\\Jason Mraz\\I'm Yours (Single)\\01 - I'm
Yours.wma"
os.startfile(path)

In most GUI toolkits (including Tkinter) and raw_input() function,
when you input a string (using the textbox, a.k.a Entry widget) it
would automatically be escaped for you, so when you input 'path\path
\file.txt', the GUI toolkit would convert it into 'path\\path\
\file.txt'.

Lie, Jun 11, 2008
19. ### Carsten HaeseGuest

Lie wrote:
> In most GUI toolkits (including Tkinter) and raw_input() function,
> when you input a string (using the textbox, a.k.a Entry widget) it
> would automatically be escaped for you, so when you input 'path\path
> \file.txt', the GUI toolkit would convert it into 'path\\path\
> \file.txt'.

That's incorrect. If you enter text into a text box or in raw_input(),
*no* conversion of backslashes is happening. A backslash entered in
raw_input is just a backslash. A backslash entered in a textbox is just
a backslash. A backslash read from a file is just a backslash.

A "conversion" happens when you print the repr() of a string that was
obtained from raw_input or from a text box, because repr() tries to show
the string literal that would result in the contents, and in a string
literal, a backslash is not (always) a backslash, so repr() escapes the
backslashes:

py> text = raw_input("Enter some text: ")
Enter some text: This is a backslash: \
py> print text
This is a backslash: \
py> print repr(text)
'This is a backslash: \\'

As you can see, I entered a single backslash, and the string ends up
containing a single backslash. Only when I ask Python for the repr() of
the string does the backslash get doubled up.

--
Carsten Haese
http://informixdb.sourceforge.net

Carsten Haese, Jun 11, 2008
20. ### LieGuest

On Jun 11, 9:14 pm, Carsten Haese <> wrote:
> Lie wrote:
> > In most GUI toolkits (including Tkinter) and raw_input() function,
> > when you input a string (using the textbox, a.k.a Entry widget) it
> > would automatically be escaped for you, so when you input 'path\path
> > \file.txt', the GUI toolkit would convert it into 'path\\path\
> > \file.txt'.

>
> That's incorrect. If you enter text into a text box or in raw_input(),
> *no* conversion of backslashes is happening. A backslash entered in
> raw_input is just a backslash. A backslash entered in a textbox is just
> a backslash. A backslash read from a file is just a backslash.

I know, but I thought it'd be easier for him to understand it like
that and discover the real 'how-it-works' later. My guilt is that I
forget to put a "this is not how it actually works behind the scene,
but you can think of it like this at least for now since this model is

> A "conversion" happens when you print the repr() of a string that was
> obtained from raw_input or from a text box, because repr() tries to show
> the string literal that would result in the contents, and in a string
> literal, a backslash is not (always) a backslash, so repr() escapes the
> backslashes:
>
> py> text = raw_input("Enter some text: ")
> Enter some text: This is a backslash: \
> py> print text
> This is a backslash: \
> py> print repr(text)
> 'This is a backslash: \\'
>
> As you can see, I entered a single backslash, and the string ends up
> containing a single backslash. Only when I ask Python for the repr() of
> the string does the backslash get doubled up.
>
> --
> Carsten Haesehttp://informixdb.sourceforge.net

Lie, Jun 11, 2008