Question

我在Apache HttpClient中使用scala，并通过示例进行操作。我收到以下错误：

/Users/name/IdeaProjects/JakartaCapOne/src/JakExamp.scala
   Error:Error:line (16)error: overloaded method value execute with alternatives 
(org.apache.http.HttpHost,org.apache.http.HttpRequest)org.apache.http.HttpResponse 
<and> 
(org.apache.http.client.methods.HttpUriRequest,org.apache.http.protocol.HttpContext)org.apache.http.HttpResponse
 cannot be applied to 
(org.apache.http.client.methods.HttpGet,org.apache.http.client.ResponseHandler[String])
val responseBody = httpclient.execute(httpget, responseHandler)

以下是突出显示错误和行的代码：

import org.apache.http.client.ResponseHandler
import org.apache.http.client.HttpClient
import org.apache.http.client.methods.HttpGet
import org.apache.http.impl.client.BasicResponseHandler
import org.apache.http.impl.client.DefaultHttpClient


object JakExamp {
 def main(args : Array[String]) : Unit = {
   val httpclient: HttpClient = new DefaultHttpClient
   val httpget: HttpGet = new HttpGet("www.google.com")

   println("executing request..." + httpget.getURI)
   val responseHandler: ResponseHandler[String] = new BasicResponseHandler
   val responseBody = httpclient.execute(httpget, responseHandler)
   // ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
   println(responseBody)

   client.getConnectionManager.shutdown

 }
}

我可以在java中成功运行该示例...

Answer 1

我也必须处理这个问题。尝试以下内容：

val handler:ResponseHandler[String] = new BasicResponseHandler
val request = new HttpGet("...")
val response = client execute request
val body = handler handleResponse response

这在2.7.7中适合我。它只有一条额外的线，所以也不会太糟糕。

Answer 2

对于Scala 2.7，据报道（2016），但收效甚微。也许有人可以重新打开它，给出一个更容易重现的案例。

您可以使用反射（通过Scala的结构类型）来解决此问题，如下所示：

import org.apache.http.client._
import org.apache.http.client.methods._
import org.apache.http.impl.client._
val httpclient = new DefaultHttpClient
val httpget = new HttpGet("www.google.com")
val brh = new BasicResponseHandler[String]
//httpclient.execute (httpget, brh)
httpclient.asInstanceOf[{
  def execute (request: HttpUriRequest,
               responseHandler: ResponseHandler[String]): String
}].execute (httpget, brh)

使用Scala 2.8，我发现以下代码有效：

import org.apache.http.client._
import org.apache.http.client.methods._
import org.apache.http.impl.client._
val httpclient = new DefaultHttpClient
val httpget = new HttpGet("www.google.com")
val brh = new BasicResponseHandler
httpclient.execute (httpget, brh)

因此我认为它已修复为2.8。

Scala和HttpClient：如何解决此错误？

2 个答案: