Bash print stderr only, not stdout [duplicate]

Answer 1

Just send the stdout to null:

cmd > /dev/null

This retains stderr, but suppresses stdout.

Answer 2

If there are no errors, it prints a "No errors" message to stdout. If there are errors, it prints a detailed message to stderr. I want only this one.

This would print the error message if one is sent or No error. if empty.

ERROR=$(exec php -l "foo/bar.php" 2>&1 >/dev/null)
[[ -z $ERROR ]] && ERROR='No error.'
echo "$ERROR"

Answer 3

You can use:

php -l "foo/bar.php" 2>&1 > /dev/null

i.e. redirect stderr to stdout and stdout to /dev/null in order to get only stderr on your terminal.