Python: open a file with script?

Question

I have a python script bundled into a application (I'm on a mac) and have the application set to be able to open .zip files. But when I say "open foo.zip with bar.py" how do I access the file that I have passed to it?

Additional info: Using tkinter.

What's a good way to debug this, as there is no terminal to pass info to?

Check the Console application. It will probably have all the informations you're looking for. — zneak
– zneak, Commented Sep 1, 2010 at 2:22
I can't find a python.log or tk.log or anything like that though. Where would those be? — Violet
– Violet, Commented Sep 1, 2010 at 2:38

pyfunc · Accepted Answer · 2010-09-01 02:37:57Z

1

You should be using sys.argv[1]

task = sys.argv[1].decode('utf-8')
if task == u'uppercase':
    pass
elif task == u'openitems':
    item_paths = sys.argv[2:]
    for itempath in item_paths:
        itempath = itempath.decode('utf-8')

answered Sep 1, 2010 at 2:37

pyfunc

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4 Comments

Violet Over a year ago

I'm a little unsure about this. It generates syntax errors on my machine. What is it trying to accomplish?

pyfunc Over a year ago

Idea is to decode the input gathered through «sys.argv» as UTF-8 first. The same we have to do, if we have to send a value to applescript. What will be passed to bar.py is a command, which needs to be decoded. Commands can be subsequently handled as shown in example above.

pyfunc Over a year ago

Yeah .. Thats right , I simply took the code piece that illustrates it, removing the rest.

Violet Over a year ago

sys.argv[1] is filled with -psn_0_****** so that doens't work anyway

Matthew Iselin · Accepted Answer · 2010-09-01 02:29:11Z

0

If I'm not greatly mistaken, it should pass the name of the file as the first argument to the script - sys.argv[1].

answered Sep 1, 2010 at 2:29

Matthew Iselin

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1 Comment

Violet Over a year ago

That's what I would think too... only, regardless of whether I specify a file or not, sys.argv[1] is filled with "-psn_0_#######"

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