problems with opening files due to file's path

Thomas Morton morton.thomas at googlemail.com
Tue Jun 10 20:25:57 CEST 2008


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 




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