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

On Jun 10, 1:25 pm, "Thomas Morton" <morton.tho...@xxxxxxxxxxxxxx>
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' %
username)

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
instead.



--------------------------------------------------
From: "Alexnb" <alexnbr...@xxxxxxxxx>
Sent: Tuesday, June 10, 2008 7:05 PM
To: <python-l...@xxxxxxxxxx>
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
.

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