How to pass path names to Python script by “dropping” files/folders over script icon

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粉色の甜心
粉色の甜心 2021-02-15 23:40

I am working in Mac OS X and have been writing simple file/folder copy scripts in Python. Is there a way to drag and drop a folder on top of a Python script icon and pass the fi

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  • 2021-02-16 00:17

    Just use the "Build Applet" utility:

    /Developer/Applications/Utilities/MacPython\ 2.5/Build\ Applet.app
    

    and the dropped file paths will be available thru sys.argv.

    Note that you may have to use Python2.5 (or a patched version) -- See this note: https://bitbucket.org/ronaldoussoren/py2app/issue/16/argv-emulation-code-needs-rewrite

    Quick example -- edit this file and put it on your desktop:

    #!/usr/bin/python2.5
    import sys
    print sys.argv
    

    Control-click on it, and select open with "Build Applet (2.5.4)"

    App icon will appear on desktop.

    Open Console Utility & clear display.

    Drop some files onto App icon -- you'll see the print in the console window.

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  • 2021-02-16 00:19

    What you might really like is Mac OS Services. They are Automator workflows which can nicely integrate into the operating system in a context-specifc manner: e.g. you can make your script appear in Finder's context menu when you select a folder.

    You can make a service from python script in following way:

    1. Open Automator.app and create a new Service;
    2. On top of workflow block you select what kind of input your application expects (folders is your choice if I understand correctly);
    3. Drag "Run Shell Script" block from left pane into workflow;
    4. Now you can either use default bash shell (/bin/bash) to call your script:

      /full/path/to/your/python /full/path/to/your/script.py $@
      

    5. Or use /usr/bin/python (default python) and paste your code directly into text block;

    6. Don't forget to set Pass input: as arguments in top right corner of the block.

    It's a little bit tricky to debug such workflows (as you won't see stdout & stderr). Possible workaround for debugging is to setup custom excepthook and output all exceptions into some plain text file:

    import sys, traceback
    
    def excepthook(type, exc, tb):
        with open("error.log", "a") as f:
            traceback.print_exc(file=f)
    
    sys.excepthook = excepthook
    
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