我从网址获取JSON数据,并且想要在我的网站上显示数据。我成功显示了除JSON层次结构(JSON对象)数据以外的所有JSON数据。我能够访问JSONArray person
和error
数据。但是,我无法访问层次结构(JSON对象)updated
数据。
我要访问updated.time
。
import java.io.IOException;
import java.net.URL;
import org.apache.commons.io.IOUtils;
import org.json.simple.JSONArray;
import org.json.simple.JSONObject;
import org.json.simple.JSONValue;
import org.json.simple.parser.ParseException;
public class ParseJson1 {
public static void main(String[] args) {
String url = "http://freemusicarchive.org/api/get/genres.json?api_key=60BLHNQCAOUFPIBZ&limit=2";
/*
{
"person": [
{
"name": "John",
"city": "Mumbai"
},
{
"name": "Rahul",
"city": "Delhi"
},
{
"name": "Sanjana",
"city": "Amritsar"
},
{
"name": "Anjali",
"city": "Hyderabad"
},
{
"name": "Mukund",
"city": "Bangalore"
},
{
"name": "Raunak",
"city": "Patna"
}
],
"updated": {
"time": "14:17:48",
"date": "2016-04-10"
},
"error": "2353"
}
*/
try {
String genreJson = IOUtils.toString(new URL(url));
JSONObject genreJsonObject = (JSONObject) JSONValue.parseWithException(genreJson);
// get the error
System.out.println(genreJsonObject.get("error"));
//Get Array Values
JSONArray genreArray = (JSONArray) genreJsonObject.get("person");
// get the first genre
JSONObject firstGenre = (JSONObject) genreArray.get(0);
System.out.println(firstGenre.get("name"));
// get the Second
JSONObject firstGenre = (JSONObject) genreArray.get(1);
System.out.println(firstGenre.get("name"));
// get the third
JSONObject firstGenre = (JSONObject) genreArray.get(2);
System.out.println(firstGenre.get("city"));
} catch (IOException | ParseException e) {
e.printStackTrace();
}
}
}
最佳答案
此json结果显示您从api
URL获得的结果。对 ??
{
"person":[
{
"name":"John",
"city":"Mumbai"
},
{
"name":"Rahul",
"city":"Delhi"
},
{
"name":"Sanjana",
"city":"Amritsar"
},
{
"name":"Anjali",
"city":"Hyderabad"
},
{
"name":"Mukund",
"city":"Bangalore"
},
{
"name":"Raunak",
"city":"Patna"
}
],
"updated":{
"time":"14:17:48",
"date":"2016-04-10"
},
"error":"2353"
}
现在,这是一个代码,该代码如何迭代或解析您的
json
对象。我假设以上结果json
根据您的代码存储在String
变量String genreJson
中。在这里,我写了一个
method
来解决您的问题。您可以参考它,也可以尝试自己的代码。public void testYourJSON(String genreJson){
JSONParser parser=new JSONParser(); //parser used to parse String to Correct Json format.
JSONObject obj_ComplexData = (JSONObject) parser.parse(genreJson); // Now Your String Converted to a JSONObject Type.
//person tag Array Data is fetched and Stored into a JSONArray Object.
JSONArray obj_arrayPersonData = (JSONArray) parser.parse(obj_ComplexData.get("person").toString());
for (Object person : obj_arrayPersonData ) { //Iterate through all Person Array.
System.out.println(person.get("name"));
System.out.println(person.get("city"));
}
//Select "updated" Tag Json Data.
JSONObject obj_Updated = (JSONObject) parser.parse(obj_ComplexData.get("updated").toString());
System.out.println(obj_Updated.get("time")); //display time tag.
System.out.println(obj_Updated.get("date")); //display date tag.
System.out.println(obj_Updated.get("error")); //display Your Error.
}