[bash] How to check the exit status using an if statement

I was wondering what would be the best way to check the exit status in an if statement in order to echo a specific output.

I'm thinking of it being

if [ $? -eq 1 ]
then
   echo "blah blah blah"
fi

The issue I am also having is that the exit statement is before the if statement simply because it has to have that exit code. Also, I know I'm doing something wrong since the exit would obviously exit the program.

Every command that runs has an exit status.

That check is looking at the exit status of the command that finished most recently before that line runs.

If you want your script to exit when that test returns true (the previous command failed) then you put exit 1 (or whatever) inside that if block after the echo.

That being said if you are running the command and wanting to test its output using the following is often more straight-forward.

if some_command; then
    echo command returned true
else
    echo command returned some error
fi

Or to turn that around use ! for negation

if ! some_command; then
    echo command returned some error
else
    echo command returned true
fi

Note though that neither of those cares what the error code is. If you know you only care about a specific error code then you need to check $? manually.

For the record, if the script is run with set -e (or #!/bin/bash -e) and you therefore cannot check $? directly (since the script would terminate on any return code other than zero), but want to handle a specific code, @gboffis comment is great:

/some/command || error_code=$?
if [ "${error_code}" -eq 2 ]; then
   ...

Note that exit codes != 0 are used to report error. So, it's better to do:

retVal=$?
if [ $retVal -ne 0 ]; then
    echo "Error"
fi
exit $retVal

instead of

# will fail for error codes > 1
retVal=$?
if [ $retVal -eq 1 ]; then
    echo "Error"
fi
exit $retVal

If you are writing a function, which is always preferred, you should propagate the error like this:

function() {
    if some_command; then
        echo worked
    else
        return $?
fi
}

This will propagate the error to the caller, so that he can do things like function && next as expected.

You could try to find out if $? is empty by using the said script below:

# Code that may fail here
if [ -z "$?" ]
then
      echo "Code if not failed"
else
      echo "Code if failed"
fi

Just to add to the helpful and detailed answer:

If you have to check the exit code explicitly, it is better to use the arithmetic operator, (( ... )), this way:

run_some_command
(($? != 0)) && { printf '%s\n' "Command exited with non-zero"; exit 1; }

Or, use a case statement:

run_some_command; ec=$?  # grab the exit code into a variable so that it can
                         # be reused later, without the fear of being overwritten
case $ec in
    0) ;;
    1) printf '%s\n' "Command exited with non-zero"; exit 1;;
    *) do_something_else;;
esac

Related answer about error handling in Bash:

• Raise error in a Bash script

Alternative to explicit if statement

Minimally:

test $? -eq 0 || echo "something bad happened"

Complete:

EXITCODE=$?
test $EXITCODE -eq 0 && echo "something good happened" || echo "something bad happened"; 
exit $EXITCODE

Using zsh you can simply use:

if [[ $(false)? -eq 1 ]]; then echo "yes" ;fi

When using bash & set -e is on you can use:

false || exit_code=$?
if [[ ${exit_code} -ne 0 ]]; then echo ${exit_code}; fi

$? is a parameter like any other. You can save its value to use before ultimately calling exit.

exit_status=$?
if [ $exit_status -eq 1 ]; then
    echo "blah blah blah"
fi
exit $exit_status