Bash check if command returns output, and print it-CodePudding

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.

CodePudding user response：

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

CodePudding user response：

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

CodePudding user response：

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