如何使用ObjectMapper将JSON属性转换为对象

Question

我使用以下代码将JSON转换为Object:

JSON:

{
"color": "blue"
}

代码:

somethingDto = objectMapper.readValue(myJson, SomethingDTO.class);

但是，我们假设JSON的格式如下:

{
"something": {
"color": "blue"
}
}

哪些代码可以达到相同的结果?

编辑:"某某"旁边可能有更多属性

Answer 1

您应该为它创建一个根类，例如

public class RootDto {
SomethingDTO somthing;
public SomethingDTO getSomething() {
return somthing;
}


public void setSomething(SomethingDTO somthing) {
this.somthing = somthing;
}
}

然后将JSON映射到它

RootDto rootDto = objectMapper.readValue(myJson, RootDto.class);

或者你可以使用JsonNode

JsonNode rootNode = objectMapper.readTree(myJson);
JsonNode somethingNode = rootNode.get("something");
SomethingDTO somethingDto = objectMapper.convertValue(somethingNode, SomethingDTO.class);

同样，你可以这样处理它们

JsonNode rootNode = objectMapper.readTree(myJson);
JsonNode somethingNode = rootNode.get("something");
SomethingDTO somethingDto;
if (somethingNode != null) {
somethingDto = objectMapper.convertValue(somethingNode, SomethingDTO.class);
} else {
somethingDto = objectMapper.convertValue(rootNode, SomethingDTO.class);
}

Answer 2

有一个名为@JsonRootName的注释，它与@XmlRootElement类似。该注释指示"根"的名称。包装。

@JsonRootName("something")
public class SomethingDto {
String color;
public String getColor() {
return color;
}
public SomethingDto setColor(String color) {
this.color = color;
return this;
}
}

之后，必须启用该功能:

ObjectMapper om = new ObjectMapper()
.enable(SerializationFeature.WRAP_ROOT_VALUE)
.enable(DeserializationFeature.UNWRAP_ROOT_VALUE);

现在，它将绕/解绕对象到所需的根元素。

class WrapTest {
private static final ObjectMapper om = new ObjectMapper()
.enable(SerializationFeature.WRAP_ROOT_VALUE)
.enable(DeserializationFeature.UNWRAP_ROOT_VALUE);
//language=json
private static final String input = """
{
"something": {
"color" : "blue"
}
}
""";
@Test
void wrappedTest() throws JsonProcessingException {
var dto = om.readValue(input, SomethingDto.class);
Assertions.assertAll(
() -> assertNotNull(dto),
() -> assertEquals("blue", dto.getColor(), "color field mismatch")
);
}
}