How start next bash file only if previous was success?

Question

In bash file myBig.sh I call another two bash files like this:

/bin/bash myBuildShortServer.sh
/bin/bash myStart.sh

It's work... but has one problem. I want to start myStart.sh only after success execute myBuildShortServer.sh.

Content of myBuildShortServer.sh is this:

mvn install -Pruntime -DskipTests=true -f pom-server.xml

The bash file myBuildShortServer.sh just start maven tasks (goals). And when success finish tasks print in console smt like this:

[INFO] ------------------------------------------------------------------------
[INFO] BUILD SUCCESS
[INFO] ------------------------------------------------------------------------
[INFO] Total time:  30.254 s
[INFO] Finished at: 2022-06-06T12:05:43+03:00
[INFO] ------------------------------------------------------------------

So I need to start myStart.sh ONLY after myBuildShortServer.sh success finish.

Is it possible?

Answer 1

If the first script is only this line you can try something like:

 myBuildShortServer.sh && myStart.sh

Of course you should make the scripts executable:

chmod +x <name of the scripts>

Answer 2

Add to myBuildShortServer.sh a code to set the error code of the script with the bash exit command, to propagate the error code returned by mvn, like this:

mvn install ...
rc=$?
if [ $rc -ne 0 ] ; then
  echo Could not perform mvn install, exit code [$rc]; exit $rc
fi
exit 0

You could then test the error code of the first script and avoid executing the following script if it's non-zero.