use*_*286 2 java jackson deserialization fasterxml
比方说,我有一堂课
public class A{
private UUID typeId;
private B data;
}
public abstract class B{
private String a;
}
public class BChildOne extends B{
... some variables
}
public class BChildTwo extends B{
... some variables
}
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B类的type根据A的typeId在变化,所以如果A的typeId为“XXX”,则数据字段类型为BChildOne,如果A的typeId为“YYY”,则数据字段类型为BChildTwo。
我怎样才能做到这一点?
所以我试过了;
@JsonAutoDetect(fieldVisibility = JsonAutoDetect.Visibility.ANY, getterVisibility =
JsonAutoDetect.Visibility.NONE, setterVisibility =
JsonAutoDetect.Visibility.NONE)
@JsonIgnoreProperties(ignoreUnknown = true)
@JsonTypeInfo(use = JsonTypeInfo.Id.NAME, include =
JsonTypeInfo.As.EXTERNAL_PROPERTY , property = "typeId")
@JsonSubTypes({
@JsonSubTypes.Type(value = BChildOne.class, name = "40ad2fe6-e672-4f0e-
986e-
619c7a1a3223") }
)
public abstract class B{
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但我遇到了以下错误;
意外的标记 (END_OBJECT),预期的 FIELD_NAME:缺少包含类型 ID 的属性“typeId”(对于 B 类)
这是显而易见的,因为 typeId 字段在 A 类而不是 B 类中。
假设你的 JSON 文档是这样的:
{
"type": "foo",
"data": {
"someCommonProperty": "common property",
"fooProperty": "foo specific property"
}
}
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{
"type": "bar",
"data": {
"someCommonProperty": "common property",
"barProperty": "bar specific property"
}
}
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您可以使用:
public class Wrapper {
private String type;
@JsonTypeInfo(use = Id.NAME, property = "type", include = As.EXTERNAL_PROPERTY)
@JsonSubTypes(value = {
@JsonSubTypes.Type(value = Foo.class, name = "foo"),
@JsonSubTypes.Type(value = Bar.class, name = "bar")
})
private AbstractData data;
// Getters and setters
}
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public abstract class AbstractData {
private String someCommonProperty;
// Getters and setters
}
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public class Foo extends AbstractData {
private String fooProperty;
// Getters and setters
}
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public class Bar extends AbstractData {
private String barProperty;
// Getters and setters
}
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在这种方法中,@JsonTypeInfo被设置为type用作外部属性来确定映射该data属性的正确类。JSON 文档可以反序列化如下:
ObjectMapper mapper = new ObjectMapper();
Wrapper wrapper = mapper.readValue(json, Wrapper.class);
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