在Java中解析JSON字符串

Question

我试图在java中解析一个JSON字符串,以单独打印单个值.但是在程序运行的同时我得到以下错误 -

Exception in thread "main" java.lang.RuntimeException: Stub!
       at org.json.JSONObject.<init>(JSONObject.java:7)
       at ShowActivity.main(ShowActivity.java:29)

我的班级看起来像 -

import org.json.JSONException;
import org.json.JSONObject;

public class ShowActivity {
   private final static String  jString = "{" 
   + "    \"geodata\": [" 
   + "        {" 
   + "                \"id\": \"1\"," 
   + "                \"name\": \"Julie Sherman\","                  
   + "                \"gender\" : \"female\"," 
   + "                \"latitude\" : \"37.33774833333334\"," 
   + "                \"longitude\" : \"-121.88670166666667\""            
   + "                }" 
   + "        }," 
   + "        {" 
   + "                \"id\": \"2\"," 
   + "                \"name\": \"Johnny Depp\","          
   + "                \"gender\" : \"male\"," 
   + "                \"latitude\" : \"37.336453\"," 
   + "                \"longitude\" : \"-121.884985\""            
   + "                }" 
   + "        }" 
   + "    ]" 
   + "}"; 
   private static JSONObject jObject = null;

   public static void main(String[] args) throws JSONException {
       jObject = new JSONObject(jString);
       JSONObject geoObject = jObject.getJSONObject("geodata");

       String geoId = geoObject.getString("id");
           System.out.println(geoId);

       String name = geoObject.getString("name");
       System.out.println(name);

       String gender=geoObject.getString("gender");
       System.out.println(gender);

       String lat=geoObject.getString("latitude");
       System.out.println(lat);

       String longit =geoObject.getString("longitude");
       System.out.println(longit);                   
   }
}

让我知道我错过了什么,或者每次运行应用程序时我都会遇到错误的原因.任何意见将不胜感激.

Answer 1

看我的评论.运行时需要包含完整的org.json库,因为android.jar只包含要编译的存根.

此外,您必须删除}以下JSON数据中的两个额外实例longitude.

   private final static String JSON_DATA =
     "{" 
   + "  \"geodata\": [" 
   + "    {" 
   + "      \"id\": \"1\"," 
   + "      \"name\": \"Julie Sherman\","                  
   + "      \"gender\" : \"female\"," 
   + "      \"latitude\" : \"37.33774833333334\"," 
   + "      \"longitude\" : \"-121.88670166666667\""
   + "    }," 
   + "    {" 
   + "      \"id\": \"2\"," 
   + "      \"name\": \"Johnny Depp\","          
   + "      \"gender\" : \"male\"," 
   + "      \"latitude\" : \"37.336453\"," 
   + "      \"longitude\" : \"-121.884985\""
   + "    }" 
   + "  ]" 
   + "}"; 

除此之外geodata,实际上不是一个JSONObject而是一个JSONArray.

以下是完全正常运行且经过测试的更正代码:

import org.json.JSONArray;
import org.json.JSONException;
import org.json.JSONObject;

public class ShowActivity {


  private final static String JSON_DATA =
     "{" 
   + "  \"geodata\": [" 
   + "    {" 
   + "      \"id\": \"1\"," 
   + "      \"name\": \"Julie Sherman\","                  
   + "      \"gender\" : \"female\"," 
   + "      \"latitude\" : \"37.33774833333334\"," 
   + "      \"longitude\" : \"-121.88670166666667\""
   + "    }," 
   + "    {" 
   + "      \"id\": \"2\"," 
   + "      \"name\": \"Johnny Depp\","          
   + "      \"gender\" : \"male\"," 
   + "      \"latitude\" : \"37.336453\"," 
   + "      \"longitude\" : \"-121.884985\""
   + "    }" 
   + "  ]" 
   + "}"; 

  public static void main(final String[] argv) throws JSONException {
    final JSONObject obj = new JSONObject(JSON_DATA);
    final JSONArray geodata = obj.getJSONArray("geodata");
    final int n = geodata.length();
    for (int i = 0; i < n; ++i) {
      final JSONObject person = geodata.getJSONObject(i);
      System.out.println(person.getInt("id"));
      System.out.println(person.getString("name"));
      System.out.println(person.getString("gender"));
      System.out.println(person.getDouble("latitude"));
      System.out.println(person.getDouble("longitude"));
    }
  }
}

这是输出:

C:\dev\scrap>java -cp json.jar;. ShowActivity
1
Julie Sherman
female
37.33774833333334
-121.88670166666667
2
Johnny Depp
male
37.336453
-121.884985

Answer 2

要将JSON 字符串转换为 hashmap，您可以使用以下命令：

HashMap<String, Object> hashMap = new HashMap<>(Utility.jsonToMap(response)) ;

使用此类:)（甚至处理列表、嵌套列表和 json）

public class Utility {

    public static Map<String, Object> jsonToMap(Object json) throws JSONException {

        if(json instanceof JSONObject)
            return _jsonToMap_((JSONObject)json) ;

        else if (json instanceof String)
        {
            JSONObject jsonObject = new JSONObject((String)json) ;
            return _jsonToMap_(jsonObject) ;
        }
        return null ;
    }


   private static Map<String, Object> _jsonToMap_(JSONObject json) throws JSONException {
        Map<String, Object> retMap = new HashMap<String, Object>();

        if(json != JSONObject.NULL) {
            retMap = toMap(json);
        }
        return retMap;
    }


    private static Map<String, Object> toMap(JSONObject object) throws JSONException {
        Map<String, Object> map = new HashMap<String, Object>();

        Iterator<String> keysItr = object.keys();
        while(keysItr.hasNext()) {
            String key = keysItr.next();
            Object value = object.get(key);

            if(value instanceof JSONArray) {
                value = toList((JSONArray) value);
            }

            else if(value instanceof JSONObject) {
                value = toMap((JSONObject) value);
            }
            map.put(key, value);
        }
        return map;
    }


    public static List<Object> toList(JSONArray array) throws JSONException {
        List<Object> list = new ArrayList<Object>();
        for(int i = 0; i < array.length(); i++) {
            Object value = array.get(i);
            if(value instanceof JSONArray) {
                value = toList((JSONArray) value);
            }

            else if(value instanceof JSONObject) {
                value = toMap((JSONObject) value);
            }
            list.add(value);
        }
        return list;
    }
}