Jackson 2.11.4：为缺少的 JSON 字段设置默认值

Question

我有以下 JSON

"Choices": [
        {
        "choiceId": "1"
        
    },
    {
        "choiceId": "2",
        "choiceType": null
    }
    ]

下面是 POJO，我需要两个构造函数，因为如果 json 中没有 choiceType 我需要将其默认为 Yes，如果 json 中的 choiceType 存在 null 值那么它应该不默认为Yes

@Getter
@ToString
@Setter
@NoArgsConstructor
public class Choices {
    @JsonProperty("choiceId")
    @NonNull
    private String choiceId;

    @JsonProperty("choiceType")
    private String choiceType;

    @JsonCreator
    @JsonIgnoreProperties(ignoreUnknown = true)
    public Choices(@JsonProperty("choiceId") String choiceId) {
        this.choiceType = choiceType !=null ? choiceType : "Yes";
        this.choiceId = choiceId;
    }

    public Choices(@JsonProperty("choiceId") String choiceId, @JsonProperty("choiceType") String choiceType) {
        this.choiceType = choiceType;
        this.choiceId = choiceId;
    }
}

我的目标是当上述 Json 被反序列化并且我有以下测试用例时，有一个选择列表

@Test
    public void testChoices(){
        ObjectMapper objectMapper = new ObjectMapper();
        String json = "[ { \"choiceId\": \"1\" }, { \"choiceId\": \"2\", \"choiceType\": null } ]";
        
        List<Choices> choices = objectMapper.convertValue(json, new TypeReference<List<Choices>>() {
        });
        assertTrue(choices.get(0).getChoiceId().equals("1"));
        assertTrue(choices.get(0).getChoiceType().equals("Yes"));
        assertTrue(choices.get(1).getChoiceType().equals("2"));
        assertNull(choices.get(1).getChoiceType());
    }

当我尝试反序列化下面的 json。我尝试了很多解决方案，但仍然没有运气，有人可以帮我解决这个问题吗？

Caused by: com.fasterxml.jackson.databind.exc.InvalidDefinitionException: Invalid definition for property `choiceType` (of type `com.onetrust.ds.request.dto.rest.Choices`): Could not find creator property with name 'choiceType' (known Creator properties: [choiceId])
 at [Source: UNKNOWN; line: -1, column: -1]

Answer 1

您可以摆脱 args 构造函数并将值默认为“是”。如果 value 显式设置为 null 或一个值，它将被分配。如果缺少值，它将被默认。

@Getter
@ToString
@Setter
@NoArgsConstructor
public class Choices {
    @JsonProperty("choiceId")
    @NonNull
    private String choiceId;

    @JsonProperty("choiceType")
    private String choiceType = "Yes";

}

参考 - Jackson: What happens if a property is missing?

Answer 2

看起来你的问题是你应该用@JsonCreator标记both构造函数。而且您只需要其中一个，Jackson 将默认将未设置的值填充为null（无需为此创建单独的构造函数）。

import com.fasterxml.jackson.annotation.JsonCreator;
import com.fasterxml.jackson.annotation.JsonProperty;
import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.databind.ObjectReader;
import lombok.Getter;

public static void main(String[] args) throws IOException {
    String json = "[ { \"choiceId\": \"1\" }, { \"choiceId\": \"2\", \"choiceType\": null } ]";
    ObjectMapper mapper = new ObjectMapper();
    ObjectReader reader = mapper.readerFor(Choices.class);
    List<Choices> choices = reader.<Choices>readValues(json).readAll();
}

@Getter
public static class Choices {

    private String choiceId;
    private String choiceType;

    @JsonCreator
    public Choices(@JsonProperty("choiceId") String choiceId,
                   @JsonProperty("choiceType") String choiceType) {
        this.choiceId = choiceId;
        this.choiceType = choiceType == null ? "Yes" : choiceType;
    }

}

我找到了这个例子here