Question

我有这段代码，我试图从这个字符串中获取项目

但它失败了

我使用过eclipse IDE

我从远程主机解析json字符串

package selectDB;

import java.io.*;
import java.net.URL;
import java.net.URLConnection;
import java.sql.*;
import java.text.DateFormat;
import java.text.SimpleDateFormat;
import java.util.Date;
import java.util.Timer;
import java.util.TimerTask;
import org.json.simple.JSONArray;
import org.json.simple.JSONObject;
import org.json.simple.parser.JSONParser;
import org.json.simple.parser.ParseException;

public class selectDB
{


  public static void main(String[] args) throws IOException, ParseException
  {
      String line;
      String s = "";

      URL u = new URL("http://192.168.3.1/android/select.php");
      URLConnection c = u.openConnection();
      InputStream r = c.getInputStream();
      BufferedReader reader = new BufferedReader(new InputStreamReader(r));
      for(; (line = reader.readLine()) != null;)
          {


            s+=line;
          }
      System.out.println(s);

     try{
      JSONObject jObject  = new JSONObject(s);
      String projecname=(String) jObject.get("name");
      System.out.print(projecname);
     }catch(Exception e)
     {}




  }

}

结果Json字符串就像这样

{"result" : "true" , "messages" : [{"id":"866343023633578","latitute":"27","longitude":"31","number_phone":"01113171374"},{"id":"352168066354050","latitute":"27","longitude":"31","number_phone":"202222"},{"id":"50","latitute":"50","longitude":"100","number_phone":"50"},{"id":"110","latitute":"50","longitude":"50","number_phone":"110"},{"id":"120","latitute":"27","longitude":"31","number_phone":"120"},{"id":"130","latitute":"28","longitude":"29","number_phone":"120"},{"id":"140","latitute":"30","longitude":"40","number_phone":"140"},{"id":"800","latitute":"60","longitude":"30","number_phone":"800"},{"id":"353629054230064","latitute":"70","longitude":"80","number_phone":"120"}]}

请帮帮我

谢谢

Answer 1

您必须使用json string

将JSONObject解析为JSONParser

您不能简单地使用字符串对JSONObject进行参数设置。

所以你必须这样做：

JSONParser parser = new JSONParser();
JSONObject jObject=(JSONObject)parser.parse(jsonstr);
String projecname=(String) jObject.get("name");

不能将json字符串拆分为jsonarray中的参数

1 个答案: