用Jackson反序列化多态类型

问题描述

如果我有这样的类结构：

If I have a class structure like so:

public abstract class Parent {
    private Long id;
    ...
}

public class SubClassA extends Parent {
    private String stringA;
    private Integer intA;
    ...
}

public class SubClassB extends Parent {
    private String stringB;
    private Integer intB;
    ...
}

是否有其他方法可以反序列化不同的 @JsonTypeInfo ？在我的父类上使用此注释：

Is there an alternative way to deserialize different then @JsonTypeInfo? Using this annotation on my parent class:

@JsonTypeInfo(use = JsonTypeInfo.Id.NAME, property = "objectType")

我宁愿不必强迫我的API客户包含objectType ：SubClassA反序列化 Parent 子类。

I would rather not have to force clients of my API to include "objectType": "SubClassA" to deserialize a Parent subclass.

而不是使用 @JsonTypeInfo ，杰克逊是否提供了一种注释子类的方法，并通过一个独特的属性将其与其他子类区分开来？在上面的示例中，这将是如果JSON对象具有stringA：... 将其反序列化为 SubClassA ，如果它有stringB：... 将其反序列化为 SubClassB 。

Instead of using @JsonTypeInfo, does Jackson provide a way to annotate a subclass and distinguish it from other subclasses via a unique property? In my example above, this would be something like, "If a JSON object has "stringA": ... deserialize it as SubClassA, if it has "stringB": ... deserialize it as SubClassB".

推荐答案

这就像 @JsonTypeInfo 和 @JsonSubTypes 应该用于但是我已经选择了文档，并且没有任何可以提供的属性看起来与你所描述的相匹配。

This feels like something @JsonTypeInfo and @JsonSubTypes should be used for but I've picked through the docs and none of the properties that can be supplied quite seem to match what you're describing.

您可以编写一个自定义反序列化器，它以非标准方式使用 @JsonSubTypes 'name和value属性来完成您想要的任务。将在您的基类上提供反序列化器和 @JsonSubTypes ，并且反序列化器将使用name值来检查是否存在属性，如果存在，则反序列化将JSON放入value属性中提供的类中。你的类看起来像这样：

You could write a custom deserializer that uses @JsonSubTypes' "name" and "value" properties in a non-standard way to accomplish what you want. The deserializer and @JsonSubTypes would be supplied on your base class and the deserializer would use the "name" values to check for the presence of a property and if it exists, then deserialize the JSON into the class supplied in the "value" property. Your classes would then look something like this:

@JsonDeserialize(using = PropertyPresentDeserializer.class)
@JsonSubTypes({
        @Type(name = "stringA", value = SubClassA.class),
        @Type(name = "stringB", value = SubClassB.class)
})
public abstract class Parent {
    private Long id;
    ...
}

public class SubClassA extends Parent {
    private String stringA;
    private Integer intA;
    ...
}

public class SubClassB extends Parent {
    private String stringB;
    private Integer intB;
    ...
}

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