fastjson 的簡單使用，fastjson使用

fastjson 的簡單使用，fastjson使用

public static void main(String[] args) {/*普通對象與json相互轉換*/User u = new User("miquan", "000");//{"@type":"testjava.User","password":"000","userName":"miquan"}String jsonUser = JSON.toJSONString(u, SerializerFeature.WriteClassName);System.out.println(jsonUser);User user = JSON.parseObject(jsonUser, User.class);/*帶集合的對象與json字串相互轉換*/List<String> list = new ArrayList<String>();list.add("miquan");list.add("qiantu");Download d = new Download();d.setEmojiSetName("guoqiao");d.setEmojiPathList(list);//{"@type":"testjava.Download","emojiPathList":["miquan","qiantu"],"emojiSetName":"guoqiao"}String jsonList = JSON.toJSONString(d, SerializerFeature.WriteClassName);System.out.println(jsonList);Download download = JSON.parseObject(jsonList, Download.class);/*json與List對象相互轉換*/List<User> listUser = new ArrayList<User>();listUser.add(new User("liangguoqiao", "iloveyou"));listUser.add(new User("liangguoqiao", "iloveyou"));//[{"password":"iloveyou","userName":"liangguoqiao"},//{"password":"iloveyou","userName":"liangguoqiao"}]String json = JSON.toJSONString(listUser);System.out.println(json);List<User> users = JSON.parseArray(json, User.class);}

注意：1、

    2、User和Download等實體類必須添加空的建構函式

            3、轉為JSON字串的時候加上SerializerFeature.WriteClassName

            4、坐等下班。。。

package com.yangshidesign.weixinface.bean;import java.util.List;public class Download {private String emojiSetName;/** 表情圖片名字 */private List<String> emojiNameList;public Download() {super();}public String getEmojiSetName() {return emojiSetName;}public void setEmojiSetName(String emojiSetName) {this.emojiSetName = emojiSetName;}public List<String> getEmojiNameList() {return emojiNameList;}public void setEmojiNameList(List<String> emojiNameList) {this.emojiNameList = emojiNameList;}}

package testjava;public class User {private String userName;private String password;public User() {super();}public User(String userName, String password) {super();this.userName = userName;this.password = password;}public String getUserName() {return userName;}public void setUserName(String userName) {this.userName = userName;}public String getPassword() {return password;}public void setPassword(String password) {this.password = password;}}

json對象在java中怎麼修改value的值簡單使用fastjson來處理
//轉成Map
Map<String,String> value = JSON.parseObject(json,Map.class);
value.put("zzmm","newValue");//改變zzmm值
JSON.toJSONString(value)；//重新轉成json字串｛name:"張三","age":"20","xb":男,"zzmm":"newValue"｝

value值已經改變了
怎樣在java代碼裡擷取一個json對象，然後對其進行排序，排好了之後，再返回給json？有多種方式實現使用java比較簡單使用json類庫直接通過類庫json轉換java bean對象對javabean對象排序再通過類庫直接轉換成json字串輸出即人比較常用GSON類庫或也用fastJSON等百度搜尋下多簡單用哪了

使用json想必應該要用javascript其實也直接用javascript進行排序操作