在Java中解析JSON数据 [英] Parsing JSON data in Java

问题描述

我想分析该网页上的一些数据：
http://www.bbc.co.uk /radio1/programmes/schedules/england/2013/03/1.json

I want to parse the some data from this page: http://www.bbc.co.uk/radio1/programmes/schedules/england/2013/03/1.json

我想分析的数据是不过标题我不确定我怎么能提取数据。这是我迄今所做的：

The data I want to parse is the titles however I am unsure how I can extract the data. This is what I have done so far:

 import java.io.BufferedReader;
 import java.io.InputStream;
 import java.io.InputStreamReader;
 import java.net.HttpURLConnection;
 import java.net.URL;
 import org.json.simple.JSONObject;
 import org.json.simple.parser.JSONParser;

 public class Test
 {
    public Test() { }

    public static void main(String[] args)
    {
            URL url;
            HttpURLConnection connection = null;
            InputStream is = null;
            JSONParser parser = new JSONParser();

            try
            {
                    url = new URL("http://www.bbc.co.uk/radio1/programmes/schedules/england/2013/03/1.json");
                    connection = (HttpURLConnection) url.openConnection();
                    connection.setRequestMethod("GET");
                    connection.connect();
                    is = connection.getInputStream();
                    BufferedReader theReader = new BufferedReader(new InputStreamReader(is, "UTF-8"));
                    String reply;
                    while ((reply = theReader.readLine()) != null)
                    {
                            System.out.println(reply);
                            Object obj = parser.parse(reply);
                            JSONObject jsonObject = (JSONObject) obj;
                            String title = (String) jsonObject.get("time");
                            System.out.println(title);
                    }
            }
            catch (Exception e) {
                    e.printStackTrace();
            }
    }

}

这只是返回null。谁能告诉我什么，我需要改变吗？谢谢你。

This just returns null. Can anybody tell me what I need to change? Thanks.

推荐答案

如果你读的JSONObject＃GET（字符串）的Javadoc，这实际上是 HashMap.get（字符串），它规定

If you read the javadoc of JSONObject#get(String) which is actually HashMap.get(String), it states

返回：值以指定键映射，或无效如果
  此映射包含的键的映射关系。

Returns: the value to which the specified key is mapped, or null if this map contains no mapping for the key

您JSON不包含该键的映射时间。

Your JSON does not contain a mapping for the key time.

编辑：

如果你的意思是标题而不是时间，以JSON的这种提取物

If you meant title instead of time, take this extract of the JSON

{"schedule":{"service":{"type":"radio","key":"radio1","title":"BBC Radio 1",...

您需要先获得进度为的JSONObject ，那么服务为的JSONObject ，然后标题作为一个正常的字符串值。应用此不同，这取决于JSON值的类型。

You need to first get schedule as a JSONObject, then service as a JSONObject, and then title as a normal String value. Apply this differently depending on the type of JSON value.

这篇关于在Java中解析JSON数据的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！