How do you stop a function in bash?
The exit code of the function (within the function) is set by using return . So when in a function return 0 is run, the function execution terminates, giving an exit code of 0.
How do I quit shell?
To end a shell script and set its exit status, use the exit command. Give exit the exit status that your script should have. If it has no explicit status, it will exit with the status of the last command run.
How to return an exit code in Bash?
I’d like to return an exit code from a BASH script that is called within another script, but could also be called directly. It roughly looks like this: #!/bin/bash dq2-get $1 if [ $? -ne 0 ]; then echo “ERROR: …” # EXIT HERE fi # extract, do some stuff # Now in the line EXIT HERE the script should exit and return exit code 1.
How can I exit a script without exiting the invoking shell?
The return bash builtin will exit the sourced script without stopping the calling (parent/sourcing) script. Causes a function to stop executing and return the value specified by n to its caller. If n is omitted, the return status is that of the last command executed in the function body. …
Is the true command never executed in Bash?
The true command is never executed but is used to build a compound command that is not exited by the -e shell flag. That sets the exit code without exiting the shell (nor a sourced script). For the more complex question of exiting (with an specific exit code) either if executed or sourced:
Is there any way to exit a script and then stay in the terminal?
When I use exit command in a shell script, the script will terminate the terminal (the prompt). Is there any way to terminate a script and then staying in the terminal? My script run.sh is expected to execute by directly being sourced, or sourced from another script. EDIT: To be more specific, there are two scripts run2.sh as