how to handle an application when executed by opening its 'file type'

hello,

I have an application that reads a file, then does some math stuff. I used py2exe to create and executable and also inno setup to create the installation program. so it all works quite well, except how do I handle the case when I double click on my file (windows knows that this file extension is for my application) and windows launches my application?. how can I know what file the user double clicked so that I can process it? is the file name passed to my application somehow?

thanks.
Jeff

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sys.argv, that is

Karsten

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On Thu, Aug 31, 2006 at 10:18:54PM -0700, Robin Dunn wrote:

Yep. If you've got the association setup right then it will put the
filename on the command line. You can get at it with sys.path.

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Karsten Hilbert wrote:

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On Thu, Aug 31, 2006 at 10:18:54PM -0700, Robin Dunn wrote:

Yep. If you've got the association setup right then it will put the filename on the command line. You can get at it with sys.path.

sys.argv, that is

Oops, yes of course. Must have been a little too late last night when I wrote that.

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Robin Dunn
Software Craftsman
http://wxPython.org Java give you jitters? Relax with wxPython!

it works well! thanks for the help!

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Robin Dunn robin@alldunn.com wrote:

Karsten Hilbert wrote:

On Thu, Aug 31, 2006 at 10:18:54PM -0700, Robin Dunn wrote:

Yep. If you’ve got the association setup right then it will put the
filename on the command line. You can get at it with sys.path.

sys.argv, that is

Oops, yes of course. Must have been a little too late last night when I
wrote that.


Robin Dunn
Software Craftsman
http://wxPython.org Java give you jitters? Relax with wxPython!


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