解析Java中的JSON字符串

Question

我想解析java中的JSON字符串以分別打印各個值。但同時使程序運行我得到以下錯誤 -

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

我的類看起來喜歡 -

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);     
    } 
} 

讓我知道什麼是我缺少的，爲什麼我得到這個錯誤的原因每次運行應用程序。任何意見將不勝感激。

Answer 1

見我comment。 你需要包含完整的org.json library當作爲運行時android.jar只包含需要編譯的存根。

此外，您必須在longitude之後刪除JSON數據中額外的}這兩個實例。

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其實不是一個JSONObject但JSONArray。

這裏是全工作並經過測試糾正代碼：

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")); 
    } 
    } 
} 

下面是輸出：

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

糾正我，如果我錯了，但是JSON是隻是文字的分隔「：」，所以只需使用使用st.nextToken（），直到你的數據

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

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

保持。確保使用「st.hasNextToken（）」，這樣你就不會得到空的異常。

Answer 3

看起來像你的兩個對象（在數組內），你在「經度」之後有一個額外的大括號。

Answer 4

你有一個額外中的每個對象 「}」， 您可以寫信JSON字符串是這樣的：

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

首先我們有充分的array object後，一個額外的}。

其次「地理數據」是JSONArray。因此，而不是JSONObject geoObject = jObject.getJSONObject("geodata");，你必須把它作爲JSONArray geoObject = jObject.getJSONArray("geodata");

一旦你的JSONArray您可以獲取使用geoObject.get(<index>)在JSONArray每個條目。我正在使用org.codehaus.jettison.json。

Answer 6

下面是一個對象的例子，對於你的情況你必須使用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. 
} 

我沒有時間，但試圖給出一個想法。如果你仍然做不到，那我會幫忙。