Spring中如何获取泛型信息的技巧

/**
 * 通过反射, 获得Class定义中声明的父类的泛型参数的类型.
 * 
 * 注意泛型必须定义在父类处. 这是唯一可以通过反射从泛型获得Class实例的地方.
 * 
 * 如无法找到, 返回Object.class.
 * 
 * 如public UserDao extends HibernateDao<User,Long>
 * 
 * @param clazz clazz The class to introspect
 * @param index the Index of the generic declaration, start from 0.
 * @return the index generic declaration, or Object.class if cannot be determined
 */
public static Class getClassGenericType(final Class clazz, final int index) {

    Type genType = clazz.getGenericSuperclass();

    if (!(genType instanceof ParameterizedType)) {
        logger.warn(clazz.getSimpleName() + "'s superclass not ParameterizedType");
        return Object.class;
    }

    Type[] params = ((ParameterizedType) genType).getActualTypeArguments();

    if ((index >= params.length) || (index < 0)) {
        logger.warn("Index: " + index + ", Size of " + clazz.getSimpleName() + "'s Parameterized Type: "
                + params.length);
        return Object.class;
    }
    if (!(params[index] instanceof Class)) {
        logger.warn(clazz.getSimpleName() + " not set the actual class on superclass generic parameter");
        return Object.class;
    }

    return (Class) params[index];
}

private HashMap<Integer, List<String>> myMap;

public void example() {
    ResolvableType t = ResolvableType.forField(getClass().getDeclaredField("myMap"));
    t.getSuperType(); // AbstractMap<Integer, List<String>>
    t.asMap(); // Map<Integer, List<String>>
    t.getGeneric(0).resolve(); // Integer
    t.getGeneric(1).resolve(); // List
    t.getGeneric(1); // List<String>
    t.resolveGeneric(1, 0); // String
}

ResolvableType.forField(Field)

ResolvableType.forMethodParameter(Method, int)

ResolvableType.forMethodReturnType(Method)

ResolvableType.forConstructorParameter(Constructor, int)

ResolvableType.forClass(Class)

ResolvableType.forType(Type)

ResolvableType.forInstance(Object)