Issue with my code

Discussion in 'Python' started by maiden129, Feb 5, 2013.

  1. maiden129

    maiden129 Guest

    Hi,

    I'm trying to create this program that counts the occurrences of each digit in a string which the user have to enter.

    Here is my code:

    s=input("Enter a string, eg(4856w23874): ")
    s=list(s)

    checkS=['0','1','2','3','4','5','6','7','8','9']

    for i in s:
    if i in checkS:
    t=s.count(i)
    if t>1:
    for k in range(1,t):
    s=s.remove(i)
    print(i, "occurs", t,"times.")

    elif t==1:
    print(i,"occurs 1 time.")
    else: pass

    but it keeps showing this error:

    t=s.count(i)
    AttributeError: 'NoneType' object has no attribute 'count'

    I wanted to show like this:

    Example:

    Enter a string: 3233456

    3 occurs 3
    2 occurs 1
    4 occurs 1
    5 occurs 1
    6 occurs 1
     
    maiden129, Feb 5, 2013
    #1
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  2. maiden129

    maiden129 Guest

    Also I’m using Python 3.2.3.

    On Tuesday, February 5, 2013 1:38:55 PM UTC-5, maiden129 wrote:
    > Hi,
    >
    >
    >
    > I'm trying to create this program that counts the occurrences of each digit in a string which the user have to enter.
    >
    >
    >
    > Here is my code:
    >
    >
    >
    > s=input("Enter a string, eg(4856w23874): ")
    >
    > s=list(s)
    >
    >
    >
    > checkS=['0','1','2','3','4','5','6','7','8','9']
    >
    >
    >
    > for i in s:
    >
    > if i in checkS:
    >
    > t=s.count(i)
    >
    > if t>1:
    >
    > for k in range(1,t):
    >
    > s=s.remove(i)
    >
    > print(i, "occurs", t,"times.")
    >
    >
    >
    > elif t==1:
    >
    > print(i,"occurs 1 time.")
    >
    > else: pass
    >
    >
    >
    > but it keeps showing this error:
    >
    >
    >
    > t=s.count(i)
    >
    > AttributeError: 'NoneType' object has no attribute 'count'
    >
    >
    >
    > I wanted to show like this:
    >
    >
    >
    > Example:
    >
    >
    >
    > Enter a string: 3233456
    >
    >
    >
    > 3 occurs 3
    >
    > 2 occurs 1
    >
    > 4 occurs 1
    >
    > 5 occurs 1
    >
    > 6 occurs 1
     
    maiden129, Feb 5, 2013
    #2
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  3. maiden129

    marduk Guest

    On Tue, Feb 5, 2013, at 01:38 PM, maiden129 wrote:
    > Hi,
    >
    > I'm trying to create this program that counts the occurrences of each
    > digit in a string which the user have to enter.
    >
    > Here is my code:
    >
    > s=input("Enter a string, eg(4856w23874): ")
    > s=list(s)
    >
    > checkS=['0','1','2','3','4','5','6','7','8','9']
    >
    > for i in s:
    > if i in checkS:
    > t=s.count(i)
    > if t>1:
    > for k in range(1,t):
    > s=s.remove(i)
    > print(i, "occurs", t,"times.")
    >
    > elif t==1:
    > print(i,"occurs 1 time.")
    > else: pass
    >
    > but it keeps showing this error:
    >
    > t=s.count(i)
    > AttributeError: 'NoneType' object has no attribute 'count'


    s=s.remove(i) does not return a new list but modifies the list in
    place.

    So you probably just want

    >>> s.remove(i)


    Also, there are various inefficiencies in your code, but that is the
    main issue with the AttributeError.
     
    marduk, Feb 5, 2013
    #3
  4. maiden129

    MRAB Guest

    On 2013-02-05 18:38, maiden129 wrote:
    > Hi,
    >
    > I'm trying to create this program that counts the occurrences of each digit in a string which the user have to enter.
    >
    > Here is my code:
    >
    > s=input("Enter a string, eg(4856w23874): ")
    > s=list(s)
    >
    > checkS=['0','1','2','3','4','5','6','7','8','9']
    >
    > for i in s:
    > if i in checkS:
    > t=s.count(i)
    > if t>1:
    > for k in range(1,t):
    > s=s.remove(i)


    The 'remove' method changes the list itself and then returns None, so
    after executing this line the first time, s will be None.

    > print(i, "occurs", t,"times.")
    >
    > elif t==1:
    > print(i,"occurs 1 time.")
    > else: pass
    >
    > but it keeps showing this error:
    >
    > t=s.count(i)
    > AttributeError: 'NoneType' object has no attribute 'count'
    >
    > I wanted to show like this:
    >
    > Example:
    >
    > Enter a string: 3233456
    >
    > 3 occurs 3
    > 2 occurs 1
    > 4 occurs 1
    > 5 occurs 1
    > 6 occurs 1
    >

    You shouldn't add or remove items from a collection, such as a list,
    over which you're iterating. Is it even necessary in this case? No.

    Have a look at the Counter class in the collections module. That'll let
    you eliminate most of your code! :)
     
    MRAB, Feb 5, 2013
    #4
  5. maiden129

    maiden129 Guest

    On Tuesday, February 5, 2013 1:56:55 PM UTC-5, marduk wrote:
    > On Tue, Feb 5, 2013, at 01:38 PM, maiden129 wrote:
    >
    > > Hi,

    >
    > >

    >
    > > I'm trying to create this program that counts the occurrences of each

    >
    > > digit in a string which the user have to enter.

    >
    > >

    >
    > > Here is my code:

    >
    > >

    >
    > > s=input("Enter a string, eg(4856w23874): ")

    >
    > > s=list(s)

    >
    > >

    >
    > > checkS=['0','1','2','3','4','5','6','7','8','9']

    >
    > >

    >
    > > for i in s:

    >
    > > if i in checkS:

    >
    > > t=s.count(i)

    >
    > > if t>1:

    >
    > > for k in range(1,t):

    >
    > > s=s.remove(i)

    >
    > > print(i, "occurs", t,"times.")

    >
    > >

    >
    > > elif t==1:

    >
    > > print(i,"occurs 1 time.")

    >
    > > else: pass

    >
    > >

    >
    > > but it keeps showing this error:

    >
    > >

    >
    > > t=s.count(i)

    >
    > > AttributeError: 'NoneType' object has no attribute 'count'

    >
    >
    >
    > s=s.remove(i) does not return a new list but modifies the list in
    >
    > place.
    >
    >
    >
    > So you probably just want
    >
    >
    >
    > >>> s.remove(i)

    >
    >
    >
    > Also, there are various inefficiencies in your code, but that is the
    >
    > main issue with the AttributeError.


    when I removed "s.remove(i), it starts to repeat the number of occurrences too

    many times like this:

    2 occurs 3 times.
    2 occurs 3 times.
    3 occurs 3 times.
    3 occurs 3 times.
    2 occurs 3 times.
    2 occurs 3 times.
    5 occurs 1 time.
    3 occurs 3 times.
    3 occurs 3 times.
    4 occurs 1 time.
    3 occurs 3 times.
    3 occurs 3 times.
    1 occurs 1 time.
    2 occurs 3 times.
    2 occurs 3 times.

    How can I stop this?
     
    maiden129, Feb 5, 2013
    #5
  6. maiden129

    maiden129 Guest

    On Tuesday, February 5, 2013 1:56:55 PM UTC-5, marduk wrote:
    > On Tue, Feb 5, 2013, at 01:38 PM, maiden129 wrote:
    >
    > > Hi,

    >
    > >

    >
    > > I'm trying to create this program that counts the occurrences of each

    >
    > > digit in a string which the user have to enter.

    >
    > >

    >
    > > Here is my code:

    >
    > >

    >
    > > s=input("Enter a string, eg(4856w23874): ")

    >
    > > s=list(s)

    >
    > >

    >
    > > checkS=['0','1','2','3','4','5','6','7','8','9']

    >
    > >

    >
    > > for i in s:

    >
    > > if i in checkS:

    >
    > > t=s.count(i)

    >
    > > if t>1:

    >
    > > for k in range(1,t):

    >
    > > s=s.remove(i)

    >
    > > print(i, "occurs", t,"times.")

    >
    > >

    >
    > > elif t==1:

    >
    > > print(i,"occurs 1 time.")

    >
    > > else: pass

    >
    > >

    >
    > > but it keeps showing this error:

    >
    > >

    >
    > > t=s.count(i)

    >
    > > AttributeError: 'NoneType' object has no attribute 'count'

    >
    >
    >
    > s=s.remove(i) does not return a new list but modifies the list in
    >
    > place.
    >
    >
    >
    > So you probably just want
    >
    >
    >
    > >>> s.remove(i)

    >
    >
    >
    > Also, there are various inefficiencies in your code, but that is the
    >
    > main issue with the AttributeError.


    when I removed "s.remove(i), it starts to repeat the number of occurrences too

    many times like this:

    2 occurs 3 times.
    2 occurs 3 times.
    3 occurs 3 times.
    3 occurs 3 times.
    2 occurs 3 times.
    2 occurs 3 times.
    5 occurs 1 time.
    3 occurs 3 times.
    3 occurs 3 times.
    4 occurs 1 time.
    3 occurs 3 times.
    3 occurs 3 times.
    1 occurs 1 time.
    2 occurs 3 times.
    2 occurs 3 times.

    How can I stop this?
     
    maiden129, Feb 5, 2013
    #6
  7. maiden129

    Dave Angel Guest

    On 02/05/2013 02:20 PM, maiden129 wrote:
    > On Tuesday, February 5, 2013 1:56:55 PM UTC-5, marduk wrote:
    >> On Tue, Feb 5, 2013, at 01:38 PM, maiden129 wrote:
    >>

    <Snipping double-spaced googlegroups trash>
    >
    > when I removed "s.remove(i), it starts to repeat the number of occurrences too
    >
    > many times like this:
    >
    > 2 occurs 3 times.
    > 2 occurs 3 times.
    > 3 occurs 3 times.
    > 3 occurs 3 times.
    > 2 occurs 3 times.
    > 2 occurs 3 times.
    > 5 occurs 1 time.
    > 3 occurs 3 times.
    > 3 occurs 3 times.
    > 4 occurs 1 time.
    > 3 occurs 3 times.
    > 3 occurs 3 times.
    > 1 occurs 1 time.
    > 2 occurs 3 times.
    > 2 occurs 3 times.
    >
    > How can I stop this?
    >


    As MRAB pointed out, don't delete items from a list you're iterating
    over. It can make the iterator go nuts. He suggests the collections
    module.

    But if you want to do it by hand, one approach is to reverse the two
    loops. Iterate over the characters in CheckS list, examining the entire
    s list for each one and figuring out how many times the character occurs.

    Another approach is to build a dict, or a defaultdict, to keep counts
    for each of the characters in CheckS.

    --
    DaveA
     
    Dave Angel, Feb 5, 2013
    #7
  8. maiden129

    maiden129 Guest

    How to reverse the two loops?

    On Tuesday, February 5, 2013 2:43:47 PM UTC-5, Dave Angel wrote:
    > On 02/05/2013 02:20 PM, maiden129 wrote:
    >
    > > On Tuesday, February 5, 2013 1:56:55 PM UTC-5, marduk wrote:

    >
    > >> On Tue, Feb 5, 2013, at 01:38 PM, maiden129 wrote:

    >
    > >>

    >
    > <Snipping double-spaced googlegroups trash>
    >
    > >

    >
    > > when I removed "s.remove(i), it starts to repeat the number of occurrences too

    >
    > >

    >
    > > many times like this:

    >
    > >

    >
    > > 2 occurs 3 times.

    >
    > > 2 occurs 3 times.

    >
    > > 3 occurs 3 times.

    >
    > > 3 occurs 3 times.

    >
    > > 2 occurs 3 times.

    >
    > > 2 occurs 3 times.

    >
    > > 5 occurs 1 time.

    >
    > > 3 occurs 3 times.

    >
    > > 3 occurs 3 times.

    >
    > > 4 occurs 1 time.

    >
    > > 3 occurs 3 times.

    >
    > > 3 occurs 3 times.

    >
    > > 1 occurs 1 time.

    >
    > > 2 occurs 3 times.

    >
    > > 2 occurs 3 times.

    >
    > >

    >
    > > How can I stop this?

    >
    > >

    >
    >
    >
    > As MRAB pointed out, don't delete items from a list you're iterating
    >
    > over. It can make the iterator go nuts. He suggests the collections
    >
    > module.
    >
    >
    >
    > But if you want to do it by hand, one approach is to reverse the two
    >
    > loops. Iterate over the characters in CheckS list, examining the entire
    >
    > s list for each one and figuring out how many times the character occurs.
    >
    >
    >
    > Another approach is to build a dict, or a defaultdict, to keep counts
    >
    > for each of the characters in CheckS.
    >
    >
    >
    > --
    >
    > DaveA
     
    maiden129, Feb 5, 2013
    #8
  9. maiden129

    maiden129 Guest

    How to reverse the two loops?

    On Tuesday, February 5, 2013 2:43:47 PM UTC-5, Dave Angel wrote:
    > On 02/05/2013 02:20 PM, maiden129 wrote:
    >
    > > On Tuesday, February 5, 2013 1:56:55 PM UTC-5, marduk wrote:

    >
    > >> On Tue, Feb 5, 2013, at 01:38 PM, maiden129 wrote:

    >
    > >>

    >
    > <Snipping double-spaced googlegroups trash>
    >
    > >

    >
    > > when I removed "s.remove(i), it starts to repeat the number of occurrences too

    >
    > >

    >
    > > many times like this:

    >
    > >

    >
    > > 2 occurs 3 times.

    >
    > > 2 occurs 3 times.

    >
    > > 3 occurs 3 times.

    >
    > > 3 occurs 3 times.

    >
    > > 2 occurs 3 times.

    >
    > > 2 occurs 3 times.

    >
    > > 5 occurs 1 time.

    >
    > > 3 occurs 3 times.

    >
    > > 3 occurs 3 times.

    >
    > > 4 occurs 1 time.

    >
    > > 3 occurs 3 times.

    >
    > > 3 occurs 3 times.

    >
    > > 1 occurs 1 time.

    >
    > > 2 occurs 3 times.

    >
    > > 2 occurs 3 times.

    >
    > >

    >
    > > How can I stop this?

    >
    > >

    >
    >
    >
    > As MRAB pointed out, don't delete items from a list you're iterating
    >
    > over. It can make the iterator go nuts. He suggests the collections
    >
    > module.
    >
    >
    >
    > But if you want to do it by hand, one approach is to reverse the two
    >
    > loops. Iterate over the characters in CheckS list, examining the entire
    >
    > s list for each one and figuring out how many times the character occurs.
    >
    >
    >
    > Another approach is to build a dict, or a defaultdict, to keep counts
    >
    > for each of the characters in CheckS.
    >
    >
    >
    > --
    >
    > DaveA
     
    maiden129, Feb 5, 2013
    #9
  10. maiden129

    darnold Guest

    On Feb 5, 2:19 pm, maiden129 <> wrote:
    > How to reverse the two loops?
    >


    s=input("Enter a string, eg(4856w23874): ")

    checkS=['0','1','2','3','4','5','6','7','8','9']

    for digit in checkS:
    t = s.count(digit)
    if t == 0:
    pass
    elif t == 1:
    print(digit,"occurs 1 time.")
    else:
    print(digit, "occurs", t,"times.")


    >>>

    Enter a string, eg(4856w23874): 23493049weee2039412367
    0 occurs 2 times.
    1 occurs 1 time.
    2 occurs 3 times.
    3 occurs 4 times.
    4 occurs 3 times.
    6 occurs 1 time.
    7 occurs 1 time.
    9 occurs 3 times.
    >>>


    HTH,
    Don
     
    darnold, Feb 5, 2013
    #10
  11. maiden129

    marduk Guest

    On Tue, Feb 5, 2013, at 04:37 PM, darnold wrote:
    > On Feb 5, 2:19 pm, maiden129 <> wrote:
    > > How to reverse the two loops?
    > >

    >
    > s=input("Enter a string, eg(4856w23874): ")
    >
    > checkS=['0','1','2','3','4','5','6','7','8','9']
    >
    > for digit in checkS:
    > t = s.count(digit)
    > if t == 0:
    > pass
    > elif t == 1:
    > print(digit,"occurs 1 time.")
    > else:
    > print(digit, "occurs", t,"times.")
    >
    >
    > >>>

    > Enter a string, eg(4856w23874): 23493049weee2039412367
    > 0 occurs 2 times.
    > 1 occurs 1 time.
    > 2 occurs 3 times.
    > 3 occurs 4 times.
    > 4 occurs 3 times.
    > 6 occurs 1 time.
    > 7 occurs 1 time.
    > 9 occurs 3 times.
    > >>>


    Although that implementation also scans the string 10 times (s.count()),
    which may not be as efficient (although it is happening in C, so perhaps
    not).

    A better solution involves only scanning the string once.
     
    marduk, Feb 5, 2013
    #11
  12. On Tue, 5 Feb 2013 10:38:55 -0800 (PST), maiden129
    <> declaimed the following in
    gmane.comp.python.general:

    > Hi,
    >
    > I'm trying to create this program that counts the occurrences of each digit in a string which the user have to enter.
    >
    > Here is my code:
    >
    > s=input("Enter a string, eg(4856w23874): ")
    > s=list(s)
    >
    > checkS=['0','1','2','3','4','5','6','7','8','9']
    >
    > for i in s:
    > if i in checkS:
    > t=s.count(i)


    Let's see... for each character in the input, see if that character
    is in the check list, and if it is, then reverse and count the
    occurences in the input string. (Oh, that's after you converted the
    input to a list of single characters)


    > if t>1:
    > for k in range(1,t):
    > s=s.remove(i)
    > print(i, "occurs", t,"times.")



    Oh Oh.... Besides the fact that .remove() does its work in-place
    (and doesn't return a value -- hence the "s=" is binding None to s), you
    are trying to modify the string that you are looping over... And that is
    a no-no -- it will cause you to skip over items.

    -=-=-=-=-=-
    >>> s = list("123456789")
    >>> for c in s:

    .... if c == "3":
    .... s.remove(c)
    .... print "c: %s\tREMOVED" % (c,)
    .... else:
    .... print "c: %s\ts: %s" % (c, s)
    ....
    c: 1 s: ['1', '2', '3', '4', '5', '6', '7', '8', '9']
    c: 2 s: ['1', '2', '3', '4', '5', '6', '7', '8', '9']
    c: 3 REMOVED
    c: 5 s: ['1', '2', '4', '5', '6', '7', '8', '9']
    c: 6 s: ['1', '2', '4', '5', '6', '7', '8', '9']
    c: 7 s: ['1', '2', '4', '5', '6', '7', '8', '9']
    c: 8 s: ['1', '2', '4', '5', '6', '7', '8', '9']
    c: 9 s: ['1', '2', '4', '5', '6', '7', '8', '9']
    >>>

    -=-=-=-=-=-

    Note how there is no line for c=4. When you remove the "3", all the
    rest shift left to fill in -- but the next round of the loop is going to
    "increment" to the character after the position that "3" was at... That
    is now the "5".

    Rather than working from the input list, why not turn it around...

    -=-=-=-=-=-
    >>> inpt = "4856w2304874"
    >>> digits = "0123456789"
    >>> for d in digits:

    .... o = inpt.count(d)
    .... if o:
    .... print "%s occurs %s times" % (d, o)
    ....
    0 occurs 1 times
    2 occurs 1 times
    3 occurs 1 times
    4 occurs 3 times
    5 occurs 1 times
    6 occurs 1 times
    7 occurs 1 times
    8 occurs 2 times
    >>>

    -=-=-=-=-=-
    --
    Wulfraed Dennis Lee Bieber AF6VN
    HTTP://wlfraed.home.netcom.com/
     
    Dennis Lee Bieber, Feb 5, 2013
    #12
  13. maiden129

    darnold Guest

    On Feb 5, 4:05 pm, marduk <> wrote:
    >
    > Although that implementation also scans the string 10 times (s.count()),
    > which may not be as efficient (although it is happening in C, so perhaps
    > not).
    >
    > A better solution involves only scanning the string once.


    agreed. i was specifically showing how to reverse the loop.
    using the much-better-suited Counter class:


    from collections import Counter

    s=input("Enter a string, eg(4856w23874): ")

    checkS=['0','1','2','3','4','5','6','7','8','9']

    cnt = Counter()

    for char in s:
    cnt[char] += 1

    for char, tally in sorted(cnt.items()):
    if char in checkS and tally > 0:
    if tally == 1:
    print(char,"occurs 1 time.")
    else:
    print(char, "occurs", tally,"times.")

    >>>

    Enter a string, eg(4856w23874): 192398209asdfbc12903348955
    0 occurs 2 times.
    1 occurs 2 times.
    2 occurs 3 times.
    3 occurs 3 times.
    4 occurs 1 time.
    5 occurs 2 times.
    8 occurs 2 times.
    9 occurs 5 times.
    >>>


    HTH,
    Don
     
    darnold, Feb 5, 2013
    #13
  14. maiden129

    Terry Reedy Guest

    On 2/5/2013 1:38 PM, maiden129 wrote:
    > Hi,
    >
    > I'm trying to create this program that counts the occurrences
    > of each digit in a string which the user have to enter.
    >
    > Here is my code:
    >
    > s=input("Enter a string, eg(4856w23874): ")
    > s=list(s)


    Unnecessary conversion.

    > checkS=['0','1','2','3','4','5','6','7','8','9']


    checks = '0123456789' is much easier to type.
    >
    > for i in s:
    > if i in checkS:


    Loop through checks, not s

    > t=s.count(i)
    > if t>1:
    > for k in range(1,t):
    > s=s.remove(i)
    > print(i, "occurs", t,"times.")
    >
    > elif t==1:
    > print(i,"occurs 1 time.")
    > else: pass


    Replace everything with

    s = input("Enter a string of digits: ")
    for d in '0123456789':
    c = s.count(d)
    if c:
    print("{} occurs {} time{}".format(d, c, '' if c == 1 else 's'))

    Enter a string of digits: 12233344499
    1 occurs 1 time
    2 occurs 2 times
    3 occurs 3 times
    4 occurs 3 times
    9 occurs 2 times

    --
    Terry Jan Reedy
     
    Terry Reedy, Feb 6, 2013
    #14
  15. maiden129

    rusi Guest

    On Feb 5, 11:38 pm, maiden129 <> wrote:
    > Hi,
    >
    > I'm trying to create this program that counts the occurrences of each digit in a string which the user have to enter.
    >
    > Here is my code:
    >
    > s=input("Enter a string, eg(4856w23874): ")
    > s=list(s)
    >
    > checkS=['0','1','2','3','4','5','6','7','8','9']
    >
    > for i in s:
    >     if i in checkS:
    >         t=s.count(i)
    >         if t>1:
    >             for k in range(1,t):
    >                 s=s.remove(i)
    >                 print(i, "occurs", t,"times.")
    >
    >         elif t==1:
    >             print(i,"occurs 1 time.")
    >     else: pass
    >
    > but it keeps showing this error:
    >
    >  t=s.count(i)
    > AttributeError: 'NoneType' object has no attribute 'count'
    >
    > I wanted to show like this:
    >
    > Example:
    >
    > Enter a string: 3233456
    >
    > 3 occurs 3
    > 2 occurs 1
    > 4 occurs 1
    > 5 occurs 1
    > 6 occurs 1


    Pythons 2.7 and later have dictionary comprehensions. So you can do
    this:


    >>> {item: s.count(item) for item in set(s)}

    {'a': 1, 'b': 1, '1': 2, '3': 1, '2': 2, '4': 1}

    Which gives counts for all letters. To filter out the digit-counts
    only:

    >>> digits="0123456789"
    >>> {item: s.count(item) for item in set(s) if item in dig}

    {'1': 2, '3': 1, '2': 2, '4': 1}

    You can then print out the values in d in any which way you want.
    [Starting with printing is usually a bad idea]
     
    rusi, Feb 6, 2013
    #15
  16. maiden129

    rusi Guest

    > Pythons 2.7 and later have dictionary comprehensions. So you can do
    > this:
    >
    > >>> {item: s.count(item) for item in set(s)}

    >
    > {'a': 1, 'b': 1, '1': 2, '3': 1, '2': 2, '4': 1}
    >
    > Which gives counts for all letters. To filter out the digit-counts
    > only:
    >
    > >>> digits="0123456789"
    > >>> {item: s.count(item) for item in set(s) if item in dig}

    >
    > {'1': 2, '3': 1, '2': 2, '4': 1}
    >
    > You can then print out the values in d in any which way you want.
    > [Starting with printing is usually a bad idea]


    Sorry cut-paste slip-up.
    1. use dig or digits (or none, just inline the "0123456789")
    2. I assumed
    >>> s = "12ab3412"
     
    rusi, Feb 6, 2013
    #16
    1. Advertising

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