8

我正在使用 @JsonTypeInfo 来指示 Jackson 2.1.0 在“鉴别器”属性中查找具体类型信息。这很好用，但是在反序列化期间没有将鉴别器属性设置到 POJO 中。

根据 Jackon 的 Javadoc (com.fasterxml.jackson.annotation.JsonTypeInfo.Id)，它应该：

/**
 * Property names used when type inclusion method ({@link As#PROPERTY}) is used
 * (or possibly when using type metadata of type {@link Id#CUSTOM}).
 * If POJO itself has a property with same name, value of property
 * will be set with type id metadata: if no such property exists, type id
 * is only used for determining actual type.
 *<p>
 * Default property name used if this property is not explicitly defined
 * (or is set to empty String) is based on
 * type metadata type ({@link #use}) used.
 */
public String property() default "";

这是一个失败的测试

 @Test
public void shouldDeserializeDiscriminator() throws IOException {

    ObjectMapper mapper = new ObjectMapper();
    Dog dog = mapper.reader(Dog.class).readValue("{ \"name\":\"hunter\", \"discriminator\":\"B\"}");

    assertThat(dog).isInstanceOf(Beagle.class);
    assertThat(dog.name).isEqualTo("hunter");
    assertThat(dog.discriminator).isEqualTo("B"); //FAILS
}

@JsonTypeInfo(
        use = JsonTypeInfo.Id.NAME,
        include = JsonTypeInfo.As.PROPERTY,
        property = "discriminator")
@JsonSubTypes({
        @JsonSubTypes.Type(value = Beagle.class, name = "B"),
        @JsonSubTypes.Type(value = Loulou.class, name = "L")
})
private static abstract class Dog {
    @JsonProperty("name")
    String name;
    @JsonProperty("discriminator")
    String discriminator;
}

private static class Beagle extends Dog {
}

private static class Loulou extends Dog {
}

有任何想法吗 ？

4

1 回答 1

23

像这样使用“可见”属性：

@JsonTypeInfo(use = JsonTypeInfo.Id.NAME,
    include = JsonTypeInfo.As.PROPERTY,
    property = "discriminator", visible=true)

然后将暴露类型属性；默认情况下，它们是不可见的，因此无需为此元数据添加显式属性。

于 2013-01-03T20:05:25.143 回答