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

On Jun 10, 11:45 am, Alexnb <alexnbr...@xxxxxxxxx> 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....
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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
.

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