Passing arguments to python interpreter from bash script

Question

Sorry this is a very newbie question, but I just can't seem to get it to work.

in my bash script, I have

python=/path/to/python
script=$1
exec $python $script "$@"

How would I pass an argument, say -O to the python interpreter? I have tried:

exec $python -O $script "$@"

And have tried changing python variable to "/path/to/python -O", as well as passing -O to the script, but every time i do any of these three, I get import errors for modules that succeed when I remove the -O.

So my question is how to tell the python interpreter to run with -O argument from a bash script?

Thanks.

While your script does not appear to solve any useful problem (and lacks a shift) the obvious and correct solution is what you have already tried, i.e. python -O scriptname.py "$@". Find out why those modules don't work with -O instead. Posting an error message here, or perhaps better yet posting a new question - and perhaps deleting this one - seems like the road forward. — tripleee
– tripleee, Commented Sep 1, 2013 at 21:24

konsolebox · Accepted Answer · 2013-09-01 21:21:38Z

1

You should shift your positional parameters to the left by 1 to exclude your script which is in the first arguments from being included to the arguments for python.

#!/bin/sh
python=/path/to/python
script=$1; shift
exec "$python" -O "$script" "$@"

Then run the script as bash script.sh your_python_script arg1 arg2 ... or sh script.sh your_python_script arg1 arg2 ....

answered Sep 1, 2013 at 21:21

konsolebox

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