用Java解析JSON字符串

Z时代
2024-01-10
分类：问答

我正在尝试解析java中的JSON字符串，以单独打印各个值。但是，在使程序运行时，出现以下错误-

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);                   
   }
}

让我知道我丢失了什么，或者为什么每次运行该应用程序时都会得到该错误。任何意见，将不胜感激。

回答：

看我的评论。当以 android.jar身份运行时，您需要包含完整的org.json库。仅包含要针对其进行编译的存根。

此外，您还必须}在JSON数据中删除额外的两个实例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a JSONObject而是a JSONArray。

这是经过全面测试并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"));
    }
  }
}