bash 脚本键盘中断移交给 Python 脚本

Question

I have a bash script that starts my python script. Point of this is, that I hand over a lot of (sometimes changing) arguments to the python script. So I found it useful to start my python script with a bash script where I "save" my argument list.

#!/bin/bash
cd $(dirname $0)
python3 script.py [arg0] [arg1]

In my Python script I have the KeyboardInterrupt-Exception implemented which would save some open files and then exit the python script. Now my question: When I run the shell script I have to at least press 3-times Strg+C, get some python errors and it will stop.

Am i guessing right, that my Strg+C is not recognized by python, but by the shell script instead? Is there a way to hand over the Keyboard Interrupt from the shell script to the python script running in it?

btw: the python script is running an infinite loop (if this is important)

Python script looks like this for the exception. As pointed out, it runs an infinite loop.

    while True:
        try:
            #doing stuff here

        except KeyboardInterrupt:
            for file in files:
                file.close()
            break

Answer 1

try "trap" command to assign custom action on signals. For Ctrl-C you need to ignore SIGINT, also you can ignore SIGTERM:

function ignore_ctrlc() {
        echo "ignored"
}

trap ignore_ctrlc SIGINT SIGTERM
declare -i counter=0
while true; do
  echo "sleep for 1 sec"
  sleep 1
  counter+=1
  [[ $counter -gt 10 ]] && exit 0
done

In the script above, Ctrl-C will interrupt executing command (most probability it will be sleep 1), but will not interrupt the script itself. In your case, your python script should ignore Ctrl-C, so I think it will be ignored at all.

Answer 2

If that's your entire script, why not tail-call python ? Bash will get completely out of the way.

#!/bin/bash
cd $(dirname $0)
exec python3 script.py [arg0] [arg1]