1. 實體類或集合轉JSON串
String jsonString = JSONObject.toJSONString(實體類);
2.JSON串轉JSONObject
JSONObject jsonObject = JSONObject.parseObject(jsonString);
3.JSON串轉實體類
實體類 javaBean = JSON.parseObject(json, 實體類.class);
4.JSON串轉帶泛型的List的集合
List<實體類或其他泛型> list = JSON.parseObject(json, new TypeReference<List<實體類或其他泛型>>(){});原文地址:https://blog.csdn.net/aa494661239/article/details/80515420
1 <dependency> 2 <groupId>com.alibaba</groupId> 3 <artifactId>fastjson</artifactId> 4 <version>1.2.7</version> 5 </dependency>
首先引入相關jar包,
假設有一個實體類User
public class User{ private int id; private String name; }
然后通過fastJson進行類型轉換
public static void main(String[] args){ String userString = "{"id":1,"name","lz"}"; JSONObject userJson = JSONObject.parseObject(userString); User user = JSON.toJavaObject(userJson,User.class);}