Spring中优雅的获取泛型信息的方法

/**
 * 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
 * 
 * @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> myMap;

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