How to wait for one job to complete before starting the next one?

[henrypratt]

I am using an AsyncScheduledJob to queue up 4 jobs, but I need them to be run one after another instead of in parallel. I can't find any information online about how to do this.

[MahdiBM]

You can create a parent job that executes those child jobs one by one.

[henrypratt]

That is what I am trying to do, but can't seem to figure out how.

[0xTim]

In the run of one job, when you get to the end, queue the next job up

[henrypratt]

But jobs are dispatched asynchronously and as far as I know there is no way to do that, unless you mean dispatching the next job from within the previous job? Here is a sample of what I am doing:

// Dispatch job 1
try await context
    .queue
    .dispatch(
        AsyncJob1.self,
        .init(""),
        maxRetryCount: 3
    )

// Dispatch job 2
try await context
    .queue
    .dispatch(
        AsyncJob2.self,
        .init(""),
        maxRetryCount: 3
    )

// Dispatch job 3
try await context
    .queue
    .dispatch(
        AsyncJob3.self,
        .init(""),
        maxRetryCount: 3
    )

// Dispatch job 4
try await context
    .queue
    .dispatch(
        AsyncJob4.self,
        .init(""),
        maxRetryCount: 3
    )

// At this point all jobs are dispatched but none have been completed.

Even though I am using await, the next job gets queued up immediately. Shouldn't it wait for the completion of the job? This is all from within an AsyncScheduledJob by the way.

[0xTim]

unless you mean dispatching the next job from within the previous job

That's exactly what I mean. Jobs are not designed to be sequential or one at a time, a worker will process them as it's available to

[henrypratt]

Ok thanks. So I think at this point it just makes more sense to not run the work as jobs.

[0xTim]

Or just run everything in one job - either works

[henrypratt]

Yeah I am going to run it all in a scheduled job. Thanks!