How to check the exit status using an 'if' statement
BashShellIf StatementError HandlingExit CodeBash Problem Overview
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.
Bash Solutions
Solution 1 - Bash
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 are wanting to test its output, using the following is often more straightforward.
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 $?
manually.
Solution 2 - Bash
Note that exit codes != 0 are used to report errors. So, it's better to do:
retVal=$?
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
Solution 3 - Bash
An alternative to an 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
Solution 4 - Bash
$?
is a parameter like any other. You can save its value to use before ultimately calling exit
.
exit_status=$?
if [ $exit_status -eq 1 ]; then
echo "blah blah blah"
fi
exit $exit_status
Solution 5 - Bash
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 [ "${error_code}" -eq 2 ]; then ...
Solution 6 - Bash
Just to add to the helpful and detailed answer:
If you have to check the exit code explicitly, it is better to use the arithmetic operator, (( ... ))
, this way:
run_some_command
(($? != 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:
Solution 7 - Bash
If you are writing a function – which is always preferred – you can propagate the error like this:
function()
{
if <command>; then
echo worked
else
return
fi
}
Now, the caller can do things like function && next
as expected! This is useful if you have a lot of things to do in the if
block, etc. (otherwise there are one-liners for this). It can easily be tested using the false
command.
Solution 8 - Bash
Using Z shell (zsh
) you can simply use:
if [[ $(false)? -eq 1 ]]; then echo "yes" ;fi
When using Bash and set -e
is on, you can use:
false || exit_code=$?
if [[ ${exit_code} -ne 0 ]]; then echo ${exit_code}; fi