Question

这是我的JSON解析应用程序的代码。一旦我在我的模拟器上运行它就会强制关闭。我是Android编程新手，我自学成才所以请耐心等待。谢谢！

JSONfunctions.java：

package com.pxr.tutorial.json;

import java.io.BufferedReader;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.HashMap;

import org.apache.http.HttpEntity;
import org.apache.http.HttpResponse;
import org.apache.http.client.HttpClient;
import org.apache.http.client.entity.UrlEncodedFormEntity;
import org.apache.http.client.methods.HttpPost;
import org.apache.http.impl.client.DefaultHttpClient;
import org.json.JSONArray;
import org.json.JSONException;
import org.json.JSONObject;

import android.util.Log;

public class JSONfunctions {

public static JSONObject getJSONfromURL(String url){
    InputStream is = null;
    String result = "";
    JSONObject jArray = null;

    //http post
    try{
            HttpClient httpclient = new DefaultHttpClient();
            HttpPost httppost = new HttpPost(url);
            HttpResponse response = httpclient.execute(httppost);
            HttpEntity entity = response.getEntity();
            is = entity.getContent();

    }catch(Exception e){
            Log.e("log_tag", "Error in http connection "+e.toString());
    }

  //convert response to string
    try{
            BufferedReader reader = new BufferedReader(new            
                InputStreamReader(is,"iso-8859-1"),8);
            StringBuilder sb = new StringBuilder();
            String line = null;
            while ((line = reader.readLine()) != null) {
                    sb.append(line + "\n");
            }
            is.close();
            result=sb.toString();
    }catch(Exception e){
            Log.e("log_tag", "Error converting result "+e.toString());
    }

    try{

        jArray = new JSONObject(result);            
    }catch(JSONException e){
            Log.e("log_tag", "Error parsing data "+e.toString());
    }

    return jArray;
}
}

Main.java：

package com.pxr.tutorial.json;

import java.util.ArrayList;
import java.util.HashMap;

import org.json.JSONArray;
import org.json.JSONException;
import org.json.JSONObject;
import org.w3c.dom.Document;
import org.w3c.dom.Element;
import org.w3c.dom.NodeList;

import com.pxr.tutorial.xmltest.R;

import android.app.ListActivity;
import android.content.Intent;
import android.os.Bundle;
import android.util.Log;
import android.view.View;
import android.widget.AdapterView;
import android.widget.AdapterView.OnItemClickListener;
import android.widget.ListAdapter;
import android.widget.ListView;
import android.widget.SimpleAdapter;
import android.widget.Toast;

public class Main extends ListActivity {
/** Called when the activity is first created. */
@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.listplaceholder);

    ArrayList<HashMap<String, String>> mylist = new ArrayList<HashMap<String, 
    String>> ();


    JSONObject json = JSONfunctions.getJSONfromURL
("http://www.tastekid.com/ask/ws?q=mosdef&f=musifin2125&k=mjjlnzkyzwuz&format=JSON");                
    try{

        JSONArray  similar = json.getJSONArray("similar");

        for(int i=0;i<similar.length();i++){                        
            HashMap<String, String> map = new HashMap<String, String>();    
            JSONObject e = similar.getJSONObject(i);

            map.put("id",  String.valueOf(i));
            map.put("name", "Name:" + e.getString("name"));
            map.put("type", "Type: " +  e.getString("type"));
            mylist.add(map);            
        }       
    }catch(JSONException e)        {
         Log.e("log_tag", "Error parsing data "+e.toString());
    }

    ListAdapter adapter = new SimpleAdapter(this, mylist , R.layout.main, 
                    new String[] { "name", "type" }, 
                    new int[] { R.id.item_title, R.id.item_subtitle });

    setListAdapter(adapter);

    final ListView lv = getListView();
    lv.setTextFilterEnabled(true);  
    lv.setOnItemClickListener(new OnItemClickListener() {
        public void onItemClick(AdapterView<?> parent, View view, int position,
            long id) {                  
            @SuppressWarnings("unchecked")
            HashMap<String, String> o = (HashMap<String, String>)
lv.getItemAtPosition(position);                 
            Toast.makeText(Main.this, "ID '" + o.get("id") + "' was clicked.",   
Toast.LENGTH_SHORT).show(); 
        }
    });
}
}

编辑 - 修复了未处理的异常，但是当我尝试解析URL时，它现在显示“无数据”。

再次感谢！

Answer 1

以下是您的错误：

JSONArray similar = json.getJSONArray("similar");

在解析JSON时你必须注意它是区分大小写，我看到你的json对象是Similiar但是你试图把它作为similiar。
Similiar是一个JSON对象，但您尝试将其作为JSONArray

以下是您的代码的修复程序，我为您提供了在Info对象中获取Similiar的解决方法：

        JSONObject earthquakes = json.getJSONObject("Similar");
        JSONArray info = earthquakes.getJSONArray("Info");

        for (int i = 0; i < info.length(); i++) {
            HashMap<String, String> map = new HashMap<String, String>();
            JSONObject e = info.getJSONObject(i);

            map.put("id", String.valueOf(i));
            map.put("name", "Name:" + e.getString("Name"));
            map.put("type", "Type: " + e.getString("Type"));
            mylist.add(map);
        }

在try之后替换JSONObject json = JSONFun.getJSONfromURL("http://www.tastekid.com/ask/ws? q=mosdef&f=musifin2125&k=mjjlnzkyzwuz&format=JSON");内的代码块上面的代码。

Answer 2

您是否在清单文件中添加了允许使用互联网的权限标记

JSON解析应用程序“没有数据”？

2 个答案: