Jackson序列化：忽略空值（或null）

Question

我目前正在使用杰克逊2.1.4，并且在将对象转换为JSON字符串时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;
        ErrorCode errorCode;
        String expiry;
        int coins;
        String email;
        String birthday;
        String pictureUrl;
        ArrayList <Performer> performer;
    }
}
}

这是我如何转换它：

ObjectMapper mapper = new ObjectMapper();
mapper.setVisibility(PropertyAccessor.FIELD, Visibility.ANY);

JsonOperation subscribe = new JsonOperation();

subscribe.request.requestType = "login";

subscribe.request.data.username = "Vincent";
subscribe.request.data.password = "test";


Writer strWriter = new StringWriter();
try {
    mapper.writeValue(strWriter, subscribe.request);
} catch (JsonGenerationException e) {
    // TODO Auto-generated catch block
    e.printStackTrace();
} catch (JsonMappingException e) {
    // TODO Auto-generated catch block
    e.printStackTrace();
} catch (IOException e) {
    // TODO Auto-generated catch block
    e.printStackTrace();
}

Log.d("JSON", strWriter.toString())

这是输出：

{"data":{"birthday":null,"coins":null,"coinsPackage":null,"email":null,"username":"Vincent","password":"test","transactionId":null,"isLoggedIn":false},"requestType":"login"}

如何避免这些空值？我只想获取“订阅”目的所需的信息！

这正是我要查找的输出：

{"data":{"username":"Vincent","password":"test"},"requestType":"login"}

我还尝试了@JsonInclude（Include.NON_NULL）并将所有变量都设置为null，但是它也不起作用！感谢您的帮助！

Answer 1

您将注解放置在错误的位置-注解必须在类上，而不是在字段上。即：

@JsonInclude(Include.NON_NULL) //or Include.NON_EMPTY, if that fits your use case 
public static class Request {
  // ...
}

如注释中所述，在2.x以下的版本中，此注释的语法为：

@JsonSerialize(include = JsonSerialize.Inclusion.NON_NULL) // or JsonSerialize.Inclusion.NON_EMPTY

另一种选择是ObjectMapper直接配置，只需调用
mapper.setSerializationInclusion(Include.NON_NULL);

（为了便于记录，我认为此答案的流行程度表明该注释 应 在逐个字段的基础上应用@fasterxml）