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Hi all,

Is there a way that I can startup my script and pass it a file? For example:

~$ python myscript.py mytext.txt

and then access mytext.txt in myscript.py?

As a long shot, for myscript.py I tried

def __init__(fle):
print fle

expecting the full path to mytext.txt to be printed but that didn't work.

Obviously I've never done this. I hope the above makes sense. any help
will be greatly appreciated.

Thanks,
--
Bradley J. Hintze
Graduate Student
Duke University
School of Medicine
801-712-8799

Bradley Hintze wrote:

> Hi all,
>
> Is there a way that I can startup my script and pass it a file? For
> example:
>
> ~$ python myscript.py mytext.txt
>
> and then access mytext.txt in myscript.py?
>
> As a long shot, for myscript.py I tried
>
> def __init__(fle):
> print fle
>
> expecting the full path to mytext.txt to be printed but that didn't work.
>
> Obviously I've never done this. I hope the above makes sense. any help
> will be greatly appreciated.
>
> Thanks,

You are looking for sys.argv:

$ cat tmp.py
import sys
print sys.argv
$ python tmp.py one two 'many arguments'
['tmp.py', 'one', 'two', 'many arguments']

Around that simple mechanism fancier libraries have been built:

http://docs.python.org/library/argparse.html

Peter