Popping up the "Open With" dialog window

Discussion in 'Python' started by Micah, Sep 14, 2004.

  1. Micah

    Micah Guest

    I know it's possible to use os.startfile() to open/execute a file based on
    its extension, but when a file is not recognized, I get a WindowsError with
    Errno 1155.

    Is there any way (using wxPython) to popup the "Open With" dialog that
    appears when trying to open an unrecognized file in Windows Explorer?

    Thanks,
    Micah

    --
    =================================
    Micah Z. Wedemeyer
    Research Scientist I, ELSYS
    Georgia Tech Research Institute
    Atlanta, GA 30332
    678.428.1283
    =================================
    Micah, Sep 14, 2004
    #1
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  2. Micah

    Cliff Wells Guest

    On Tue, 2004-09-14 at 10:41 -0400, Micah wrote:
    > I know it's possible to use os.startfile() to open/execute a file based on
    > its extension, but when a file is not recognized, I get a WindowsError with
    > Errno 1155.
    >
    > Is there any way (using wxPython) to popup the "Open With" dialog that
    > appears when trying to open an unrecognized file in Windows Explorer?


    No. You'll probably have to look to the native win32all package for
    that.

    Regards,
    Cliff

    --
    Cliff Wells <>
    Cliff Wells, Sep 15, 2004
    #2
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