How do I deserialize a generic type within a JSON object using Jackson in Java?

Jackson and Generic Type Reference

In Java, generics are used to represent types that can be parameterized with other types. This allows for code reuse and greater flexibility. However, when working with JSON using Jackson, generic types can pose certain challenges.

One common issue arises when accessing generic properties within a JSON object. Consider the following Java code:

<code class="java">public MyRequest<T> tester() {
    TypeReference<MyWrapper<T>> typeRef = new TypeReference<MyWrapper<T>>();
    MyWrapper<T> requestWrapper = (MyWrapper<T>) JsonConverter.fromJson(jsonRequest, typeRef);
    return requestWrapper.getRequest();
}

public class MyWrapper<T> {
    private MyRequest<T> request;
    public MyRequest<T> getRequest() { return request; }
    public void setRequest(MyRequest<T> request) { this.request = request; }
}

public class MyRequest<T> {
    private List<T> myobjects;
    public void setMyObjects(List<T> ets) { this.myobjects = ets; }
    @NotNull
    @JsonIgnore
    public T getMyObject() { return myobjects.get(0); }
}</code>

When calling getMyObject(), which is within the MyRequest object, Jackson returns the nested custom object as a LinkedHashMap. This is because Jackson cannot determine the actual type of T based on the generic declaration alone.

To resolve this issue and specify the type of T, an explicit class argument is required. This can be done by constructing a JavaType object, as follows:

<code class="java">TypeReference<MyWrapper<T>> typeRef = new TypeReference<MyWrapper<T>>() {};
JavaType type = mapper.getTypeFactory().constructCollectionType(List.class, Foo.class);
MyWrapper<Foo> requestWrapper = mapper.readValue(jsonRequest, type);</code>

In this example, the Foo class represents the actual type of T. By constructing a JavaType with the desired type, Jackson can properly deserialize the JSON object and return the correct type of T.

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