I have written this web service in play framework.
controller
def getByGenre(genre: String) = Action {
val result = Await.result(Movies.getByGenre(genre), 5 seconds)
Ok(toJson(result))
}
routes
GET /movies/genre/:genre controllers.MoviesController.getByGenre(genre: String)
However a user may select multiple Genre. Therefore I need to convert the genre parameter to a List[String]
I also need to know how to pass that Array parameter to the web service using CURL.
If you can pass the genres parameter as part of the query string, just repeat the parameter with different values and then retrieve it like this:
def getByGenre() = Action.async { implicit request =>
val genres = request.queryString.get("genres")
Movies.getByGenre(genres).map { movies =>
Ok(toJson(movies))
}
}
Your route will be:
GET /movies/genre controllers.MoviesController.getByGenre()
Also, notice that you will need to change the Movies.getByGenre signature to:
def getByGenre(genres: Option[Seq[String]]): Seq[Movies]
An final url will be something like @mfirry showed:
myhost.com/movies/genre?genre=action&genre=drama
Finally, as you may have noticed, I've removed the blocking code from you action. Using Await at your controller means that you action would be blocking for at least 5 seconds at the worst case scenario. I suggest you to take a look at the following page of Play docs:
https://www.playframework.com/documentation/2.5.x/ScalaAsync
