fastJson在bean中加入@JsonProperty转换出的对象中依然有字段为n

发布时间：2020-12-16 18:49:03 所属栏目：百科来源：网络整理

导读：1.转换的代码如下 String json = "{"id":1059827483,"idstr":"1059827483","class":1,"screen_name":"DancingToDeath"}"; System.out.println(json); U u = JSONObject.parSEObject(json,U.class); System.out.println(u.getId()); System.out.p

1.转换的代码如下

String json = "{"id":1059827483,"idstr":"1059827483","class":1,"screen_name":"DancingToDeath"}";
        System.out.println(json);
        U u = JSONObject.parSEObject(json,U.class);
        System.out.println(u.getId());
        System.out.println(u.getIdstr());
        System.out.println(u.getMyClass());
        System.out.println(u.getScreen());

2.U实体定义如下：

public class U {

    private long id;
    private String idstr;
    @JSONField(name = "class")
    private int myClass;
    @JsonProperty("screen_name")
    private String screen;

    public long getId() {
        return id;
    }

    public void setId(long id) {
        this.id = id;
    }

    public String getIdstr() {
        return idstr;
    }

    public void setIdstr(String idstr) {
        this.idstr = idstr;
    }

    public int getMyClass() {
        return myClass;
    }

    public void setMyClass(int myClass) {
        this.myClass = myClass;
    }

    public String getScreen() {
        return screen;
    }

    public void setScreen(String screenName) {
        this.screen = screenName;
    }

}

3.转换结果：

{"id":1059827483,"idstr":"1059827483","class":1,"screen_name":"DancingToDeath"}
1059827483
1059827483
1
null

可见，screen_name并未转换出来，原因在于，对于普通的json解析类，用JsonProperty可以，但是在用fastJson进行解析时，需要用@JSONField(name = "screen_name")来进行标注。

4.改变U

private long id;
    private String idstr;
    @JSONField(name = "class")
    private int myClass;
    @JSONField(name = "screen_name")
    private String screen;

5.转换结果如下：

1059827483
1059827483
1
DancingToDeath

以此记录，希望能帮助到遇到同样问题的人。

（编辑：李大同）

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