SerializationFeature.WRAP_ROOT_VALUE作为jackson json中的注释

Pat*_*ick 9 java serialization json jackson

有没有办法SerializationFeature.WRAP_ROOT_VALUE在根元素上配置注释而不是使用ObjectMapper

例如,我有:

@JsonRootName(value = "user")
public class UserWithRoot {
    public int id;
    public String name;
}
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使用ObjectMapper:

@Test
public void whenSerializingUsingJsonRootName_thenCorrect()
  throws JsonProcessingException {
    UserWithRoot user = new User(1, "John");

    ObjectMapper mapper = new ObjectMapper();
    mapper.enable(SerializationFeature.WRAP_ROOT_VALUE);
    String result = mapper.writeValueAsString(user);

    assertThat(result, containsString("John"));
    assertThat(result, containsString("user"));
}
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结果:

{
    "user":{
        "id":1,
        "name":"John"
    }
}
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有没有办法将此SerializationFeature作为注释而不是作为配置objectMapper

使用依赖:

<dependency>
     <groupId>com.fasterxml.jackson.core</groupId>
     <artifactId>jackson-databind</artifactId>
     <version>2.7.2</version>
</dependency>
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Isa*_*ank 21

import com.fasterxml.jackson.annotation.JsonTypeInfo;
import com.fasterxml.jackson.annotation.JsonTypeName;
import com.fasterxml.jackson.core.JsonProcessingException;
import com.fasterxml.jackson.databind.ObjectMapper;

public class Test2 {
    public static void main(String[] args) throws JsonProcessingException {
        UserWithRoot user = new UserWithRoot(1, "John");

        ObjectMapper objectMapper = new ObjectMapper();

        String userJson = objectMapper.writerWithDefaultPrettyPrinter().writeValueAsString(user);

        System.out.println(userJson);
    }

    @JsonTypeName(value = "user")
    @JsonTypeInfo(include = JsonTypeInfo.As.WRAPPER_OBJECT, use = JsonTypeInfo.Id.NAME)
    private static class UserWithRoot {
        public int id;
        public String name;
    }
}
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@JsonTypeName@JsonTypeInfo使它成为可能.

结果:

{
  "user" : {
    "id" : 1,
    "name" : "John"
  }
}
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  • 谢谢您,我不敢相信仅包装响应就变得多么复杂。 (2认同)
  • @Sean - 这是真的,它很复杂,需要时间习惯.他们可以拿出像`@ JsonWrapper`或`@ JsonRootValue`将带包装的名称. (2认同)