Android Eclipse：HTTP无法解析为类型

Question

有人可以解释为什么以及如何解决此错误消息，我在编程或Android时在Eclipse中获取？

HTTP不能被解析为一个类型

代码示例如下。

try { 
       // Call class "HTTP" passing an URL using Async "execute" 
       // Save returned string in "result" 
       String result = new HTTP().execute("http://domain.com/file.php?ID=1&R=T").get(); 

       Log.i("EmonLog", "Result: "+result); 

       result = result.replaceAll("\"",""); 
       TextView powerval = (TextView) findViewById(R.id.information_message); 
       powerval.setText("Result: " + result); 
      } catch (Exception e) { 
       Log.i("EmonLog", "Error: "+e); 
      } 

的程序编译和完美的作品，但我得到这个错误，在HTTP下的红线，我想知道为什么，以及如何解决它。

编辑...添加了一些信息。这是execute调用的线程。

class HTTP extends AsyncTask<String, Void, String> 
{ 
    @Override 
    protected String doInBackground(String... params) { 
     String result = ""; 
     try { 
      // Get the first parameter of the "params" which has been related to "String" 
      String urlstring = params[0]; 
      // Log the string that contains the url 
      Log.i("EmonLog", "HTTP Connecting: "+urlstring); 
      // Convert the String into an actual URL 
      URL url = new URL(urlstring); 
      // Make the HTTP connection to the Internet 
      HttpURLConnection urlConnection = (HttpURLConnection) url.openConnection(); 
      // Once the connection is done, try to store the information received into "reader" 
      try { 
       InputStream reader = new BufferedInputStream(urlConnection.getInputStream()); 
       // Create a new String which will extract and contain the information received into the Buffered Input Stream 
       String text = ""; 
       int i = 0; 
       while((i=reader.read())!=-1) 
       { 
        text += (char)i; 
       } 
       Log.i("EmonLog", "HTTP Response: "+text); 
       result = text; 

      } catch (Exception e) { 
       Log.i("EmonLog", "HTTP Exception: "+e); 
      } 
      finally { 
       Log.i("EmonLog", "HTTP Disconnecting"); 
       urlConnection.disconnect(); 
      } 

     } catch (Exception e) { 
      e.printStackTrace(); 
      Log.i("EmonLog", "HTTP Exception: "+e); 
     } 

     return result; 
    } 
} 

我进口：

import android.support.v7.app.ActionBarActivity; 
import android.os.Bundle; 
import android.os.Handler; 
import android.util.Log; 
import android.view.Menu; 
import android.view.View; 
import android.widget.Button; 
import android.widget.TextView; 

非常感谢您的帮助。

Answer 1

您不能将HTTP用作AsyncTask的名称。

正如你已经有

org.apache.http.protocol.HTTP 

为HTTP。

因此，请将您的AsyncTask重命名为其他内容。

可能，

class HTTPTask extends AsyncTask<String, Void, String> 

就不够好。