如何解析JSONArray中的JSON对象? [英] How do I parse JSON objects from a JSONArray?

问题描述

我有一个非常大的JSON文件，格式如下:

I have a very large JSON file in the following format:

[{"fullname": "name1", "id": "123"}, {"fullname": "name2", "id": "245"}, {"fullname": "name3", "id": "256"}]

它看起来像一个JSONArray.所有记录都写在同一行中.

It looks like a JSONArray. All the records are written in the same line.

您能帮我如何使用Java解析此文件.我想读取每个JSON对象并显示所有全名和ID.以下是我的尝试，但是我的代码无法正常工作:

Can you help me how can I parse this file using Java. I want to read each JSON object and display all the fullname and ids. Below is my attempt, but my code is not working:

import org.apache.commons.lang3.StringEscapeUtils;
import org.json.simple.JSONArray;
import org.json.simple.JSONObject;
import org.json.simple.parser.JSONParser;
import org.json.simple.parser.ParseException;

public class JSONFileReaderDriver {

public static void main(String[] args) throws FileNotFoundException, 
IOException, ParseException 
{
 String filename="Aarau";
 String src="C:\\"+filename+".json";
 JSONParser parser = new JSONParser();
 JSONObject obj;
 try
    {
        BufferedReader br=new BufferedReader (new FileReader(src));  
        obj = (JSONObject) new JSONParser().parse(row);
        String fullname=obj.get("fullname");
        String id=obj.get("id");
        System.out.println ("fullname: "+fullname+" id: "+id);
    }catch(Exception e)
     {e.printStackTrace();}
   br.close();
  }
 }

推荐答案

好吧...刚解决了我的问题.我正在发布解决方案，以防有人再次遇到相同的问题，可以使用我的解决方案.我的解决方案部分是由Rahul Rabhadiya推动的.谢谢，伙计.

Okay folks...just solved my problem. I am posting the solution in case someone runs into the same issue again, can use my solution. My solution is partly motivated by Rahul Rabhadiya. Thanks, dude.

try{
 row=br.readLine();
            JSONArray root = (JSONArray) JSONValue.parseWithException(row);

            for (int i=0;i<root.size();i++)
            {

            JSONObject rootObj = (JSONObject) root.get(i);
            String fullname=(String) rootObj.get("fullname");

            System.out.println (fullname);
            }
    }catch(Exception e)
         {
        e.printStackTrace();
        }

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