Question

如果ElementType是一个静态已知类型，那么它很容易＆＃34;创建表示Type s：

集合的ElementType对象

Type listType = new TypeToken<ObservableList<ElementType>>(){}.getType();

但这不可能，因为我的elementType是一个动态值，只在运行时知道：

@Override
public ListProperty<?> deserialize(JsonElement json, Type typeOfT, JsonDeserializationContext context) throws
        JsonParseException {
    Type elementType = ((ParameterizedType) typeOfT).getActualTypeArguments()[0];
    Type listType = // ??? create the type representing a ObservableList<elementType>
    ObservableList<?> list = context.deserialize(json, listType);
    return new SimpleListProperty<>(list);
}

现在我希望listType代表ObservableList<>哪个类型参数应该是elementType的值。有办法吗？

Answer 1

如果你有番石榴，你可以使用他们提供你需要的TypeToken版本：

{{1}}

Answer 2

根据this post的建议，我最终创建了自己的ParametrizedType实现：

private static class CustomParameterizedType implements ParameterizedType {

    private Type rawType;
    private Type ownerType;
    private Type[] typeArguments;

    private CustomParameterizedType(Type rawType, Type ownerType, Type... typeArguments) {
        this.rawType = rawType;
        this.ownerType = ownerType;
        this.typeArguments = typeArguments;
    }

    @Override
    public Type[] getActualTypeArguments() {
        return typeArguments;
    }

    @Override
    public Type getRawType() {
        return rawType;
    }

    @Override
    public Type getOwnerType() {
        return ownerType;
    }

    @Override
    public boolean equals(Object o) {
        if (this == o) {
            return true;
        }
        if (o == null || getClass() != o.getClass()) {
            return false;
        }
        CustomParameterizedType that = (CustomParameterizedType) o;
        return Objects.equals(rawType, that.rawType) &&
                Objects.equals(ownerType, that.ownerType) &&
                Arrays.equals(typeArguments, that.typeArguments);
    }

    @Override
    public int hashCode() {
        return Objects.hash(rawType, ownerType, typeArguments);
    }
}

然后我可以这样使用它：

@Override
public ListProperty<?> deserialize(JsonElement json, Type typeOfT, JsonDeserializationContext context) throws
        JsonParseException {
    Type elementType = ((ParameterizedType) typeOfT).getActualTypeArguments()[0];
    Type listType = new CustomParametrizedType(ObservableList.class, null, elementType);
    ObservableList<?> list = context.deserialize(json, listType);
    return new SimpleListProperty<>(list);
}

从动态元素类型中获取ParametrizedType对象

2 个答案: