enhance JobLauncherTestUtils.launchStep(...) to be able to test a local slave step from a partition step [BATCH-2225]

[spring-projects-issues]

Michael Pralow opened BATCH-2225 and commented

With version 2.2.6 it is not possible to test a step with JobLauncherTestUtils.launchStep(..) which itself is a local slave step for a partition step.

I tagged this issue as minor, i am not sure if a slave step should be testable like this, but it would sure be nice.

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Affects: 2.2.6

[spring-projects-issues]

Michael Minella commented

Do you have a specific example of why it isn't? The slave step of a partitioned step is a regular step so I'd be surprised if there was an issue...

[spring-projects-issues]

Michael Pralow commented

while using JavaConfig and some debugging, it seems to be tied to the SimpleJob, SimpleJob.steps does not contain the slave step (as expected it can be found inside the partition step)

[spring-projects-issues]

Michael Minella commented

Ah. I see. Actually, it's not even PartitionStep that holds the child step. If it was, PartitionStep could implement StepLocator and it would just work. However, the issue with using JobLauncherTestUtils#launchStep(String stepName) to launch the slave step is that it assumes the step is local, which it may not be (in the remote partitioning case). I'm open to suggestions or a pull request, but since the slave step really has no direct relation to the job, I'm not sure there's an easy fix...

[spring-projects-issues]

Michael Pralow commented

I updated the title and description to be more specific to "local" slave steps and i reproduced it with XmlConfig as well.
I do not see this as mission critical, just let it stay tagged as minor improvement and wait for more similar requests.