2012-12-21 61 views
5

我使用@JsonTypeInfo指示Jackson 2.1.0在'discriminator'属性中查找具体的类型信息。这很有效,但是在反序列化过程中,歧义属性并未设置为POJO。@JsonTypeInfo属性在POJO反序列化过程中被忽略

根据同时杰克逊的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 ""; 

这里是一个failling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 { 
} 

任何想法?

回答

14

使用 '可见' 属性,像这样:

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

然后将暴露类型属性;默认情况下它们不可见,因此不需要为此元数据添加显式属性。

+0

从杰克逊用户邮件列表复制/粘贴,但没关系。 –

+4

是的;主要是为了不在名单上的读者的利益。 – StaxMan

+2

杰克森1.9有没有办法做到这一点? – bananasplit

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