problems with opening files due to file's path

[Alexnb]

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" <alexnbryan at gmail.com>
> Sent: Tuesday, June 10, 2008 7:05 PM
> To: <python-list at python.org>
> 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 <alexnbr... at gmail.com> 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
>>>>
>>>> > --
>>>> >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%27s-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%27s-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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