Member Avatar for ygarin

Hi everyone!
I wrote a Python 2.5 application, with GUI from Tkinter module. To make it short, the user has to load a file and can start to work on it.
I made it Windows executable with py2exe, and installable with Inno, with a file association setting. So now, if I double-click on a .xxx file, it launches my application.
Now the problem is that I have no clue how to load the double-clicked file into the program. It seems the answer is somewhere in the win32*** modules, but after spending few days looking for it, I didn't find any way to do it.
Any help, lead or solution appreciated!

Thanks,

yan

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  • Latest Post Latest Post by Gribouillis
Member Avatar for Gribouillis

It's only an idea, but don't you think sys.argv contains the path to the double-clicked file ?

Edited by Gribouillis because: n/a
commented: nice indeed +10
Member Avatar for ygarin

Merci Gribouillis!

that was the function I was looking for.
sys.argv returns indeed the path from where the script 's been launched.

Member Avatar for Gribouillis

nice :)

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