Question

当我使用spring mvc时。我想得到json string.But遇到了问题。

我有一个班级用户：

public class User {
        private String username;
        private String password;
        private boolean accountNonExpired = true;
        private boolean accountNonLocked = true;
        private boolean credentialsNonExpired = true;
        private boolean enabled = true;
        private DateTime registeTime;
        private String ip;
        private DateTime loginTime;
        private DateTime logoutTime;
        private String loginIp;
        private DateTime lastLogoutTime;
        private int passwordExpiredDays = -1;
        private DateTime passwordChangeTime;
        private boolean loginAtSameTime = false;
        private int loginAttempt = 0;
        private int status = 0;
        private String problem;
        private String solution;
}

我想要像json一样：

{"username":"a","ip":"127.0.0.1"}

现在，我使用：

@JsonIgnoreProperties(value = {"password","accountNonExpired", 
"accountNonLocked","credentialsNonExpired", "enabled", "registeTime", 
"loginTime", "logoutTime", "passwordExpiredDays", "passwordChangeTime", 
"loginAtSameTime", "loginAttempt", "status", "problem", "solution"})
public class User {
        ...
}

但是!!!这非常复杂。我能找到解决方法吗？例如：

@JsonInclude("Annotation")
public class User {
    @JsonInclude
    private String username;
    ...
    @JsonInclude
    private String ip;
    ...
}

或

@JsonIncludeProperties(value={"username", "ip"})
public class User {
        ...
}

Answer 1

我认为你可以在字段级别使用@JsonIgnore注释来避免在序列化过程中使用它。

Answer 2

我相信你需要做的就是用@JsonProperty标记你想要的字段。例如：

public class User {
    @JsonProperty
    private String username;
    ...
    @JsonProperty
    private String ip;
    ...
}

Jackson默认如何序列化？

2 个答案: