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Re: [Question #185801]: Start a file on Windows

 

Question #185801 on Sikuli changed:
https://answers.launchpad.net/sikuli/+question/185801

RaiMan posted a new comment:
Thanks for the feedback.

To be complete with the start command (since it has some more
interesting options):

It seems, that you have to use the /D parameter to position the working
directory and only the plain filename:

This works for me:

cmd = r'start /D "C:\Documents and Settings\myuser\Mes documents" test3.pdf'
os.popen(cmd)

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