How to execute on start code in scala Play! framework application?

Question

I need to execute a code allowing the launch of scheduled jobs on start of the application, how can I do this? Thanks.

Bertsche

Answer 1

Use the Global object which - if used - must be defined in the default package:

object Global extends play.api.GlobalSettings {

  override def onStart(app: play.api.Application) {
    ...
  }

}

Remember that in development mode, the app only loads on the first request, so you must trigger a request to start the process.

---

Since Play Framework 2.6x

The correct way to do this is to use a custom module with eager binding:

import scala.concurrent.Future
import javax.inject._
import play.api.inject.ApplicationLifecycle

// This creates an `ApplicationStart` object once at start-up and registers hook for shut-down.
@Singleton
class ApplicationStart @Inject() (lifecycle: ApplicationLifecycle) {

  // Start up code here

  // Shut-down hook
  lifecycle.addStopHook { () =>
    Future.successful(())
  }
  //...
}

import com.google.inject.AbstractModule

class StartModule extends AbstractModule {
  override def configure() = {
    bind(classOf[ApplicationStart]).asEagerSingleton()
  }
}

See https://www.playframework.com/documentation/2.6.x/ScalaDependencyInjection#Eager-bindings

Canteen

I have added a note at the end because this bit me, I hope you don't mind. – Empirin

Where is the default package in Play 2? – Melise

@poliu2s: i think its the app/ folder. – Sherry

For anyone looking at this post, the new way to do this is to use a module and eager bidings. See playframework.com/documentation/2.6.x/… – Hesperidin

Answer 2

I was getting a similar error. Like @Leo said, create Global object in app/ directory.

Only thing I had to make sure was to change "app: Application" to "app: play.api.Application".

app: Application referred to class Application in controllers package.

Sherry