解析Java中的JSON字符串

Question

我想解析java中的JSON字符串以分别打印各个值。但同时使程序运行我得到以下错误 -

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

我的类看起来喜欢 -

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);     
    } 
} 

让我知道什么是我缺少的，为什么我得到这个错误的原因每次运行应用程序。任何意见将不胜感激。

Answer 1

见我comment。 你需要包含完整的org.json library当作为运行时android.jar只包含需要编译的存根。

此外，您必须在longitude之后删除JSON数据中额外的}这两个实例。

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其实不是一个JSONObject但JSONArray。

这里是全工作并经过测试纠正代码：

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")); 
    } 
    } 
} 

下面是输出：

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

纠正我，如果我错了，但是JSON是只是文字的分隔“：”，所以只需使用使用st.nextToken（），直到你的数据

String line = ""; //stores the text to parse. 

StringTokenizer st = new StringTokenizer(line, ":"); 
String input1 = st.nextToken(); 

保持。确保使用“st.hasNextToken（）”，这样你就不会得到空的异常。

Answer 3

看起来像你的两个对象（在数组内），你在“经度”之后有一个额外的大括号。

Answer 4

你有一个额外中的每个对象 “}”， 您可以写信JSON字符串是这样的：

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\""    
    + "    }" 
    + "  }" 
    + " ]" 
    + "}"; 
} 

Answer 5

首先我们有充分的array object后，一个额外的}。

其次“地理数据”是JSONArray。因此，而不是JSONObject geoObject = jObject.getJSONObject("geodata");，你必须把它作为JSONArray geoObject = jObject.getJSONArray("geodata");

一旦你的JSONArray您可以获取使用geoObject.get(<index>)在JSONArray每个条目。我正在使用org.codehaus.jettison.json。

Answer 6

下面是一个对象的例子，对于你的情况你必须使用JSONArray。

public static final String JSON_STRING="{\"employee\":{\"name\":\"Sachin\",\"salary\":56000}}"; 
try{ 
    JSONObject emp=(new JSONObject(JSON_STRING)).getJSONObject("employee"); 
    String empname=emp.getString("name"); 
    int empsalary=emp.getInt("salary"); 

    String str="Employee Name:"+empname+"\n"+"Employee Salary:"+empsalary; 
    textView1.setText(str); 

}catch (Exception e) {e.printStackTrace();} 
    //Do when JSON has problem. 
} 

我没有时间，但试图给出一个想法。如果你仍然做不到，那我会帮忙。