I have a Bash command that produces output on STDOUT and can have an exit code of zero or non-zero.
I would like to hide the output if the exit code was non-zero and show the output if the exit code was zero, respectively.
How can this be achieved without running the command twice?
Print the output if it's zero
output="$(command)"
if [[ $? -eq 0 ]] ; then
echo "$output"
fi
IFS.
printf "$output\n" instead.
command, try output=$(command; rv=$?; echo x; exit $rv). Then, echo "${output%x}".The status code is only known when the process terminates. So the only way to do what you are asking, is to run the process having redirected its stdout to a temporary file, and then when it terminates, examine its status code. If it is nonzero, just delete the temporary file, if it is zero, just output the contents of the temporary file to stdout.
Here's a bash script that automates that:
#!/bin/bash
temp_file=$(mktemp)
"$@" > temp_file
if [[ $? == 0 ]]; then
cat "$temp_file"
fi
rm "$temp_file"
If you call this script mute then you can the command foo --bar as follows
mute foo --bar
You can capture an output from a command and check its exit status, without relying on the fragile $?:
if output="$(command)" ; then
echo "$output"
fi
The pattern of decoupling the command and the test is problematic for a few reasons:
if your script is running with set -e, it will exit immediately when the command fails, which may not be what you want. Try it:
#!/bin/bash
set -e
output="$(nonexistent_command)"
echo "Unreachable code"
if [[ $? -eq 0 ]] ; then
echo "$output"
fi
If you insert anything anything between the command and the test, the comparison will be wrong.
That's the gist. See the relevant ShellCheck Wiki for more.
