Setting pipefail for a single piped command

I need to execute a number of piped shell commands from a non-BASH script (namely PHP script) like these:

command1 | command2 | command3 

so that, if command1 fails with a non-zero exit code, each other command fails too. So far, what I've come up with is:

set -o pipefail && command1 | command2 | command3 

But even though it runs fine from the terminal, it produces this if executed from the script:

sh: 1: set: Illegal option -o pipefail

3 Answers

From the Bash command line you would need to invoke a subshell to avoid pipefail being set afterwards:

$ (set -o pipefail && command1 | command2 | command3) 

This would limit the effect of the pipefail option to the subshell create by the parentheses (...).

A real example:

$ (set -o pipefail && false | true) && echo pipefail inactive || echo pipefail active pipefail active 

If you use an explicit shell call with the -c option you do not need a subshell, either with bash or with an sh alias to bash:

$ bash -c "set -o pipefail && false | true" && echo pipefail inactive || echo pipefail active pipefail active $ sh -c "set -o pipefail && false | true" && echo pipefail inactive || echo pipefail active pipefail active 

Since your sh does not accept the pipefail option, I would have to assume that it is either some older or modified version of bash - or that it is actually some other shell completely.

Not sure why it wasn't mentioned above, but it is possible to explicitly unset pipefail, using set +o pipefail.

set -o pipefail command1 | command2 | command3 set +o pipefail 

If you are executing a fragment, and unsure about the pipefail already set, you can use this with the subshell as previously suggested:

# explicitly execute with pipefail set (set -o pipefail ; command1 | command2 | command3 ) # explicitly execute with pipefail unset (set +o pipefail ; command1 | command2 | command3 ) 

You can use $(command1) combined with $? :

a=$(echo "a") [ $? -eq 0 ] && b=$(echo $a"b") [ $? -eq 0 ] && c=$(echo $b"c") echo $c 

Prints "abc"

a=$(ls unexitingDir) [ $? -eq 0 ] && b=$(echo $a"b") [ $? -eq 0 ] && c=$(echo $b"c") echo $c 

Prints "".

"[ $? -eq 0 ] &&" means that the following command is executed only if the previous one succed.

That dosen't answer your question but it's a solution to your problem.