I have 2 python scripts a.py and b.py and I want to write a bash script that will load a.py and not run b.py until a.py is done doing it's thing. simplistically
#!/usr/bin/env bash
python a.py
python b.py
but this is naive, a check to see if a.py is done... how do I do that?
This by default will already run one after the other.
To check that python a.py completed successfully as a required condition for running python b.py, you can do:
#!/usr/bin/env bash
python a.py && python b.py
Conversely, attempt to run python a.py, and ONLY run 'python b.py' if python a.py did not terminate successfully:
#!/usr/bin/env bash
python a.py || python b.py
To run them at the same time as background processes:
#!/usr/bin/env bash
python a.py &
python b.py &
(Responding to comment) - You can chain this for several commands in a row, for example:
python a.py && python b.py && python c.py && python d.py
prompt_err() {
echo -e "\E[31m[ERROR]\E[m"
}
prompt_ok() {
echo -e "\E[32m[OK]\E[m"
}
status() {
if [ $1 -eq 0 ]; then
prompt_ok
else
prompt_err
exit -1
fi
}
a.py
status
b.py
You can use the check code above.
If 'a.py' is done only then it will process 'b.py', otherwise it will exit with an 'Error'.
