播放框架2.5移动Action.async逻辑从控制器到服务

Question

我有一个控制器，有很多业务逻辑，我想移动Action.async代码块。此代码的工作，我怎么能移动到另一个类（服务）是

def tweetsnew(query: String) = Action.async { 
// Move From Here... 
credentials.map { 
    case (consumerKey, requestToken) => 
    ws.url("https://api.twitter.com/1.1/search/tweets.json") 
     .sign(OAuthCalculator(consumerKey, requestToken)) 
     .withQueryString("q" -> query) 
     .withQueryString("max_id" -> "833342796736167936") 
     .get().map { twitterResponse => 
     if (twitterResponse.status == 200) { 
     // Here There Are More Complex Logic 
     Ok("That is fine: "+twitterResponse.body) 
     } else { 
     throw new Exception(s"Could not retrieve tweets for $query query term") 
     } 
    } 
}.getOrElse { 
    Future.failed(new Exception("You did not correctly configure the Twitter credentials")) 
} 
//....To Here. To Another Class 
} 

我已经检查了docummentation的Action.async？：里面的代码，一些涉及到create a Future[Result]但我不能使该函数返回Action.async期望的相同类型。

Answer 1

Action.async预计Future[Result]返回类型，所以你需要创建一个返回Future[Result]

第一步的功能，提取您的代码的函数：

object TwitterService { 
    def search(query: String, consumerKey: ConsumerKey, requestToken: RequestToken)(implicit ws: WSClient, ec: ExecutionContext): Future[Result] = { 
    // your code that make the ws call that returns Ok("...") 
    } 
} 

然后在控制器中调用函数：

def tweetsnew(query: String) = Action.async { 
    credentials.map { 
    case (consumerKey, requestToken) => TwitterService.search(query, consumerKey, requestToken) 
    }.getOrElse { 
    // Better to send a bad request or a redirect instead of an Exception 
    Future.successful(BadRequest("Credentials not set")) 
    } 
}