I recently deployed a script
exit_job(){
echo "$1"
exit 0
}
I searched the web and found that the correct exit code. Can someone explain exit 0?
0 is the shell script success code. Thus if you echo something other than this it will be returning error code, and if not handled would break your script.
Related
I have a bash script that run some tool that on some use cases returns an error code of 137.
I use this script as part of my CI pipeline and and as far as i am concerned this is suceessful state.
My question is how can i catch this exit code and return an exit code of 0 instead(like trap command, only the I do not think it support custom exit codes, only predefined one).
Thanks in advance,
Alon
After you run the command/tool in your script:
command
You can see what the exit code and do some logic based on that return:
command
return=$?
if [ $return -eq 137 ]; then
exit 0
else # #chepner's very good suggestion
exit $return
fi
That will end your bash script with an exit code of 0 if the command you ran (your tool) returns an exit code of 137.
I have read about unix exit commands but please can someone tell me how they work exactly.
I mean what is their purpose and how can they be used.
Also i see people talking about success = 0 or something and i dont have a clue what they mean by this.
the "exit" command exits the shell script
echo "A"
exit 1
echo "B"
In the above example 'echo "B"' is not executed because of the exit statement.
It's like a return statement in normal progemming languages. The expression after the exit is the return value. Convention is that 0 means "Success" other values means an error.
So if the above script is called q.sh, than this script can be called from an other script:
sh ./q.sh
echo $?
The code "$?" means "exit" value of the last shell script.
Above script prints "1"
I have a bash script I want to self-destruct on execution. So far it works great but I'd like some final check that if no errors have occurred (any output to stderr), then go ahead and self destruct. Otherwise, I'd like to leave the script in tact. I have the code for everything except the error check. Not sure if I can just output err to a file and check if file is empty. I'm sure it's a simple solution.
How could I do this?
Thanks for any help.
You can try this out. $? contains the return code for the process last executed by command. Moreover standard nix derivatives demarcate 0 as (no error) and 1 - 255 as some kind of errors that happened. Note that this will report errors that do not necessarily have any stderr output.
command
if [ "$?" -ne 0 ]; then
echo "command failed";
# your termination logic here
exit 1;
fi
Assuming that your script returns the value 0 on success, a value from 1 to 255 if an error occur you can use the following command
if /path/to/myscript; then
echo success
else
echo failed
fi
you can also use the following (shorter) command
[[ /path/to/myscript ]] && echo success || echo failed
The exit command when executed from a file being sourced doesn't terminate the execution of the program where it was being sourced, how to do this? Consider this files for a clearer explanation:
a.fish:
source b.fish
echo "This should never run!"
b.fish:
echo "Failing now"
exit 1
This will result in this (undesired output):
Failing now
This should never run!
And the exit status is 0! Is there a solution for B to terminate execution of A as if exit was written in A itself?
It's perfectly working the way you want with bash and zsh. Still, I found a solution for fish:
source b.fish; or exit 1
This will exit a.fish if b.fish exited with exit 1, and will continue otherwise.
I'm start to playing around with ShellScript Unix, so maybe it's a silly question. Apologies if that's the case.
I was trying to handle the exit codes to properly address adverse situations in my code, and for this, I created a code snippet to understand the unix exit behavior. Here it is:
#!/usr/bin/bash
RES=1
if [ $RES -eq 0 ]
then
echo "Finishing with success!"
exit 0
else
echo "Finishing with error!"
exit 1
fi
I understood that, once the code is called (and terminated) I'd go back to bash prompt. However, it seems the exit instruction is also leaving bash. Is it normal? Maybe it's something related to my development environment?
Here are the messages...
bash-3.00$ . errtest.sh
Finishing with error!
$ echo $?
1
$ bash
bash-3.00$ which bash
/usr/bin/bash
For reference, I've added the return and the bash location. Hope it helps.
Thanks in advance!
This is because you're sourcing the script in your current environment (by using the . command). Try executing the script with either:
bash ./errtest.sh
or by giving the necessary permissions to the script file and executing it like this:
chmod u+x ./errtest.sh
./errtest.sh