Spring 源码是个大宝库,咱们能遇到的大部分工具在源码里都能找到,因此笔者开源的 mica[1] 彻底基于 Spring 进行基础加强,不重复造轮子。今天我要分享的是在 Spring 中优雅的获取泛型。ide
咱们以前的处理方式,代码来源 vjtools(江南白衣)。工具
/** * 经过反射, 得到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. 这是惟一能够经过反射从泛型得到Class实例的地方. * * 如没法找到, 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];}
从 Spring 4.0 开始 Spring 中添加了 ResolvableType 工具,这个类能够更加方便的用来回去泛型信息。 首先咱们来看看官方示例:spa
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)