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scala - 没有变量的Scala轮询循环

问题描述

轮询和 API 的 Scala 最佳实践是什么？

我正在尝试编写一个轮询 API 的 Scala 方法，检查它是否达到"SUCCESS". 在轮询时，它也可能达到像"FAILED"或"TIMEOUT“。

在 Java 中，我会写如下内容：

public String pollEndpoint() {
  boolean isPolling = true;
  String result = "NA";
  while (isPolling) {
    Response response = getResponse("http://myAPI.com/ready?id=1234");
    if (response.status == "FAILED") { throw new FailedException(response.reason);}
    else ... //Some other bad conditions
    else if (response.status == "SUCCESS") {
      isPolling = false;
      result = response.result; 
    }
    
    System.out.println("Current state is " + response.status); // When running, will be "RUNNING"
    Thread.sleep(1000);
  }
}

在 Scala 中，我可以这样做：

def pollEndpoint():String = {
  var isPolling = true
  var result = "NA"
  while (isPolling) {
    val response = getResponse("http://myAPI.com/ready?id=1234")
    if (response.status == "FAILED") { throw new FailedException(response.reason)}
    else ... //Some other bad conditions
    else if (response.status == "SUCCESS") {
      isPolling = false
      result = response.result 
    }
    println("Current state is " + response.status); // When running, will be "RUNNING"
    Thread.sleep(1000)  
  }
}

但是这个解决方案使用vars。

有什么好的方法可以做到这一点，只使用vals 吗？

标签： scalapollingidioms

解决方案

正如 Luis 在评论中提到的，您可以为此编写一个递归：

def pollEndpoint():String = {
  val response = getResponse("http://myAPI.com/ready?id=1234")
  println("Current state is " + response.status); // When running, will be "RUNNING"
  if (response.status == "FAILED") { ??? }
  else if (response.status == "SUCCESS") {
    response.result
  } else {
    Thread.sleep(1000)
    pollEndpoint()
  }
}

以下将模拟getResponse：

var i = 5
def getResponse(str: String): Response = {
  if (i < 0) {
    Response("SUCCESS", "Great success")
  } else {
    i = i - 1
    Response("Wait", "Not done yet.")
  }
}

和电话：

pollEndpoint()

将产生：

Current state is Wait
Current state is Wait
Current state is Wait
Current state is Wait
Current state is Wait
Current state is Wait
Current state is SUCCESS

说了这么多，与其getResponse按原样定义要好得多，我们可能会这样做：

def getResponse(str: String): Future[Response]

然后简单地说：

getResponse("http://myAPI.com/ready?id=1234").map { response =>
  // Handle response
}

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