if ! output=$(some_command);
then
printf "Error occurred, error output is =\n%s", "$output"
Could you please suggest if this is a good way to do it? I am testing - If the exit status of the command is 1, then only I want to print the contents of the output. If the exit status of the command is 0, then do nothing. Don't print output.
This doesn't allow you to print data before the output of the command, but after:
if some_command | grep .; then
echo "the output was as above"
fi
If you really want to print text before, you can store it and do:
if output=$(some_command | grep .); then
printf "the output is:%s\n" "$output"
fi
Or you can be more explicit and do:
output=$(some_command)
if test -n "$output"; then
printf "the output is:%s\n" "$output"
fi
You're testing whether the command was successful, not whether it returned output.
Assign the variable separately from the if statement.
output=$(some_command)
if [ -n "$output" ]
then
printf "output is =\n%s", "$output"
fi
Do you want to print when there is no output? I read this as if not output print the following
If you want to print when it output is not null the following should work...
More info here: Check if string is neither empty nor space in shell script
temp=$(command)
if [[ -n "${temp// /}" ]];; then
echo "output is $temp"
fi
output="$(some_command)"; if [[ -n $output ]]; then printf "output is =\n%s" "$output"; fi!) is testing the exit status of the command, not the contents of the variable. It is very useful in that aspect, just not as the text of your question describes.