bash: using function's exit code in conditional expressions

Question

I have next code:

...
docker_test () {

        local masterip=$1
        nc -zv $masterip 2377
}
...
        until $(docker_test $masterip); do
           echo "First Master not initialized yet..."
           sleep 30
        done
...

The question is - is it correct approach in bash to use loops or if/else in a such way?

Another example could be:

if $(docker_test $masterip); then
    echo "Passed"
fi

What is the issue here, if your intention is run nc -zv.. until it is successful (returning a code 0) this looks fine — Inian
– Inian, Commented Mar 23, 2017 at 12:43
@Inian Thanks. Yes, it works. But I have doubts in exactly until $(funcname); then construction. Will be until [ "$(docker_test $masterip)" == "0" ] more correct way here? Or it can be safly simplified to the same code, as mine is? — setevoy
– setevoy, Commented Mar 23, 2017 at 12:46
@setevoy Do you want the standard output or the exit code of the function? If the latter, just drop the command substitution. until docker_test "$master_ip"; do. Otherwise, until [ "$(docker_test "$master_ip")" = 0 ] (=, not ==) is correct. — chepner
– chepner, Commented Mar 23, 2017 at 13:07

odradek · Accepted Answer · 2017-03-23 13:49:10Z

1

i wouldn't risk the chance of the script hanging indefinitely if there's a problem with master. i would do something like:

#!/bin/bash -eu

readonly TRIES = 5  # attempts before failing
readonly SLEEP = 30  # sleep in seconds

for try in $(seq 1 $TRIES); do
    if docker_test "$master_ip"; then
        echo "Passed"
        break
    else
        if [ $try -eq $TRIES ]; then
            echo "Failed"
            exit 1  # or whatever you need
        fi
        echo "Retrying. Try $try of $TRIES"
        sleep $SLEEP
    fi
done

answered Mar 23, 2017 at 13:49

odradek

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