不能從GIven URL訪問JSON不知道爲什麼？

Question

`package com.tricks.readjsonfromurl;

import java.io.IOException; 

    import org.apache.http.HttpEntity; 
    import org.apache.http.HttpResponse; 
    import org.apache.http.client.ClientProtocolException; 
    import org.apache.http.client.HttpClient; 
    import org.apache.http.client.methods.HttpGet; 

    import org.apache.http.impl.client.DefaultHttpClient; 
    import org.apache.http.util.EntityUtils; 
    import org.json.JSONArray; 
    import org.json.JSONException; 
    import org.json.JSONObject; 

    import android.os.Bundle; 
    import android.os.StrictMode; 
    import android.app.Activity; 
    import android.widget.ScrollView; 
    import android.widget.TextView; 



    public class MainActivity extends Activity 
    { 
     @Override 
     protected void onCreate(Bundle savedInstanceState) 
     { 
      super.onCreate(savedInstanceState); 
      setContentView(R.layout.activity_main); 

      StrictMode.setThreadPolicy(new StrictMode.ThreadPolicy.Builder() 
        .detectDiskReads() 
        .detectDiskWrites() 
        .detectNetwork() // or .detectAll() for all detectable problems 
        .penaltyLog() 
        .build()); 

      TextView showJSON = (TextView) findViewById(R.id.JSONfromURL); 

      //ScrollView ScrollJSON = (ScrollView) findViewById(R.id.scrollView); 

      JSONObject json = null; 

      String str = ""; 
      HttpResponse response; 
      HttpClient myClient = new DefaultHttpClient(); 

      //This the url from which i am expecting to get JSON Data. 

      HttpGet myConnection = new HttpGet("http://www.clusterdev.com/flipkart-api-demo/category-example.html"); 

//Same Code but different URL i've commented below and this URL is //returning Data WTF?? 
      //HttpGet myConnection = new HttpGet("http://api.openweathermap.org/data/2.1/find/station?bbox=12,32,15,37,10&cluster=yes"); 



      try 
      { 
       response = myClient.execute(myConnection); 
       str = EntityUtils.toString(response.getEntity(), "UTF-8"); 

      } 
      catch (ClientProtocolException e) 
      { 
       e.printStackTrace(); 
      } 
      catch (IOException e) 
      { 
       e.printStackTrace(); 
      } 

      try 
      { 
//Getting nothing from the URL. 

       JSONObject jsonObject = new JSONObject(str);  

       String showJSONIntoTextView = jsonObject.toString(); 

       showJSON.setText(showJSONIntoTextView); 


      } 

      catch (JSONException e) 
      { 
       e.printStackTrace(); 
      } 
     } 
    }`I want to get `JSON` Data from the `URL`, `URL` contains `JSON` data only but giving nothing. 

http://www.clusterdev.com/flipkart-api-demo/category-example.html

所有我想要做的是從上面的網址 返回JSON數據並顯示在一個Android的TextView。

謝謝您提前。

Answer 1

 protected String doInBackground(String... params) { 
    try { 
     URL url=new URL(serviceURL); 
     HttpURLConnection con= (HttpURLConnection) url.openConnection(); 
     con.setRequestProperty("Accept-Language", "en-US,en;q=0.5"); 
     con.setRequestProperty("Content-Type", "application/json"); 

     String urlParameters = data; 

     switch (REQTYPE) { 
      case 0: 
       con.setRequestMethod("GET"); 
       break; 
      case 1: 
       con.setRequestMethod("POST"); 

       if(urlParameters.trim().length()>0) { 
        // Send post request 
        con.setDoOutput(true); 
        DataOutputStream wr = new DataOutputStream(con.getOutputStream()); 
        wr.writeBytes(urlParameters); 
        wr.flush(); 
        wr.close(); 
       } 


       break; 
      default: 
       break; 
     } 

     int responseCode = con.getResponseCode(); 
     System.out.println("\nSending 'POST' request to URL : " + serviceURL); 
     System.out.println("Post parameters : " + urlParameters); 
     System.out.println("Response Code : " + responseCode); 

     BufferedReader in = new BufferedReader(
       new InputStreamReader(con.getInputStream())); 

     String inputLine; 
     StringBuilder resp = new StringBuilder(); 

     while ((inputLine = in.readLine()) != null) { 
      resp.append(inputLine); 
     } 
     in.close(); 

     System.out.println(resp.toString()); 
     Response = resp.toString(); 


    } catch (Exception e) 
    { 
     e.printStackTrace(); 

    } 

    return Response; 
} 

創建延伸的AsyncTask做以下的東西一類...在背景