[英]How to repeat request in Play! framework 2.1?
有沒有簡單的方法來重復請求,直到獲得成功結果Play2.1(scala)? 以及如何限制嘗試次數?
我想做這樣的事情:
WS.url("some.url").get().map{ response =>
val strval = someFunction(response)
strval match {
case "success" => println("do something after successful request")
case "error" => println("repeat same request until success - and repeat maximum N times!")
}
}
提前致謝!
未經測試
你可以這樣做:
import scala.concurrent._
import play.api.libs.concurrent.Execution.Implicits._
def withRetry[T](retries:Int = 5)(f: => Future[T]) =
f.recoverWith {
case t:Throwable if (retries > 0) => withRetry(retries - 1)(f)
}
然后在您自己的代碼中,您可以像這樣使用它:
withRetry(retries = 2) {
WS.url("some.url").get
.map { response =>
require(someFunction(response) != "error", "Please retry")
response
}
}
如果你願意將someFunction
重寫為Response => Boolean
你可以像這樣使用它:
def someFunction(r: Response): Boolean = ???
withRetry(retries = 2) {
WS.url("some.url").get
.filter(someFunction)
}
試試這個:
def wSCall = WS.url("http://foo/bar").get()
def ƒ(response: Response, n: Int): Result = {
val strval = someFunction(response)
strval match {
case "success" => Ok("Ok!")
case "error" => {
if (n > 0)
Async { wSCall.map(response => ƒ(response, n - 1)) }
else
BadRequest("Fail :(")
}
}
}
Async { wSCall.map(response => ƒ(response, 10)) }
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.