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

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。