Call multiple python scripts sequentially from a bash script using arguments

Question

I would like to make a bash script that runs some number of python scripts in a row based on arguments given in the terminal (the order in which they run doesn't matter).

Something like:

kwargs = kwarg1 kwarg2 kwarg3

if kwarg1: python script1.py
if kwarg2: python name2000.py
if kwarg3: python program30.py

But with the option to take in any number of keywords as long as they matched up to a script.

How would this be done?

Answer 1

There's a lot you can do to add to this, but here's the simplest of what you're asking for:

#!/bin/bash
for script in $*; do
    python $script.py
done

Put that in a new file, e.g. "run_python.bash", and make it executable (chmod a+w run_python.bash).

Then you run it by: run_python.bash one two three, and it will execute one.py, then two.py, then three.py .

If you want to use the if structure in the for loop (i.e. if your keywords don't match the script name), the simple version starts like this:

#!/bin/bash
for script in $*; do
    if [ $script == 'kwarg1' ]; then
        python script1.py
    fi
    if [ $script == 'kwarg2' ]; then
        python name2000.py
    fi
    if [ $script == 'kwarg3' ]; then
        python program30.py
    fi
done

This will execute whatever you tell it to, based on the order of the keywords you put on the command line.

p.s. This has nothing to do with python. The script works the same, whether you're invoking python, or ls, or anything else that you can run in bash.

Answer 2

you need shell script.

script.sh:

#!/bin/bash

if [ "$1" == "kwarg1" ]
then
    python script1.py
elif [ "$1" == "kwarg2" ]
then
    python name2000.py
fi

chmod +x script.sh

./script.sh kwarg1