关于数据序列化（3），使用JSON的方式，以及几个注意的点,以及fastJson的坑

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JSON也是对象序列化的一种方式

public class FriendLikeVO {public String sn; public int score; public long time;public FriendLikeVO(JSONObject jo) {this.sn = jo.getString("sn");this.score = jo.getIntValue("score");this.time = jo.getLong("time");}public FriendLikeVO() {}public FriendLikeVO(String sn, int score, long time) {this.sn = sn;this.score = score;this.time = time;}/** * 把Json转换为List */public static List<FriendLikeVO> jsonToList(String json) {List<FriendLikeVO> result = new ArrayList<FriendLikeVO>();JSONArray ja = JSON.parseArray(json);if (ja == null) {return result;}for (int i = 0; i < ja.size(); i++) {FriendLikeVO vo = new FriendLikeVO(ja.getJSONObject(i));result.add(vo);}return result;}/** * 将List转换为Json */public static String listToJson(List<FriendLikeVO> likeList) {JSONArray ja = new JSONArray();likeList.stream().forEach(vo -> {JSONObject jo = new JSONObject();jo.put("sn", vo.sn);jo.put("score", vo.score);jo.put("time", vo.time);ja.add(jo);});return ja.toJSONString();}}

但是用fastjson的序列化更简单一些，如果不想序列化的字段就标注transient就可以了

public class FriendLike {public String sn; // 好友ID，目前为IPpublic int score; // 点赞的点数public long time;        public transient int cc;public FriendLike() {}public FriendLike(String sn, int score, long time) {this.sn = sn;this.score = score;this.time = time;}/** * 把Json转换为List */@SuppressWarnings("unchecked")public static List<FriendLike> jsonToList(String json) {if(Util.isEmpty(json)){return new ArrayList<FriendLike>();}return JSON.parseObject(json, List.class);}/** * 将List转换为Json */public static String listToJson(List<FriendLike> likeList) {if(null == likeList){return null;}return JSON.toJSONString(likeList);}

tips:

fastjson的循环引用问题，反序列化也是可以成功的

测试代码

FriendLike fk = new FriendLike("a", 2, System.currentTimeMillis());
FriendLike fk2 = new FriendLike("b", 2, System.currentTimeMillis());
List<FriendLike> list = new ArrayList<>();
list.add(fk);
list.add(fk2);
list.add(fk2);
String a = JSON.toJSONString(list);
List<FriendLike> list2 = JSON.parseObject(a, List.class);
System.out.println(a);
System.out.println(list);

输出结果

[{"score":2,"sn":"a","time":1470211141179},{"score":2,"sn":"b","time":1470211141179},{"$ref":"$[1]"}]
[FriendLike[sn=a,score=2,time=1470211141179], FriendLike[sn=b,score=2,time=1470211141179], FriendLike[sn=b,score=2,time=1470211141179]]

注意一个坑，如果要序列化List的时候应该使用JSON.parseArray这个方法，并且注意，被序列化的对象类型要有无参构造方法！

@SuppressWarnings("unchecked")public static List<FriendHelpVO> jsonToList(String json) {if (Util.isEmpty(json)) {return new ArrayList<FriendHelpVO>();}return JSON.parseArray(json, FriendHelpVO.class);}

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