相关疑难解决方法(0)

杰克逊序列化:忽略空值(或null)

我目前正在使用jackson 2.1.4,当我将对象转换为JSON字符串时,我在忽略字段时遇到了一些麻烦.

这是我的类,它充当要转换的对象:

public class JsonOperation {

public static class Request {
    @JsonInclude(Include.NON_EMPTY)
    String requestType;
    Data data = new Data();

    public static class Data {
        @JsonInclude(Include.NON_EMPTY)
        String username;
        String email;
        String password;
        String birthday;
        String coinsPackage;
        String coins;
        String transactionId;
        boolean isLoggedIn;
    }
}

public static class Response {
    @JsonInclude(Include.NON_EMPTY)
    String requestType = null;
    Data data = new Data();

    public static class Data {
        @JsonInclude(Include.NON_EMPTY)
        enum ErrorCode { ERROR_INVALID_LOGIN, ERROR_USERNAME_ALREADY_TAKEN, ERROR_EMAIL_ALREADY_TAKEN };
        enum Status { ok, error };

        Status status; …
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java json jackson

131
推荐指数
6
解决办法
27万
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Jackson JSON 序列化:如何在嵌套对象的所有字段都为空时忽略它?

我正在使用 Jackson 并且我有一些 JSON 模式对象设置如下:

@JsonInclude(JsonInclude.Include.NON_EMPTY)
public class Person {

    String name;
    Child child = new Child();
    Sibling sibling = new Sibling();

    public String getName() {
        return name;
    }
    public void setName(String name) {
        this.name = name;
    }

    public Child getChild() {
        return child;
    }
    public void setChild(Child child) {
        this.child = child;
    }

    public Sibling getSibling() {
        return sibling;
    }
    public void setSibling(Sibling sibling) {
        this.sibling = sibling;
    }
}

@JsonInclude(JsonInclude.Include.NON_EMPTY)
public class Child {

    String name;

    public String …
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java json jackson

6
推荐指数
1
解决办法
3116
查看次数

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