SqlSesion怎么获取一个Mapper?
一个Mapper接口没有一个实现类怎么能够实例化?
public <T> T getMapper(Class<T> type) { // 通过 configuration 的getMapper方法获取Mapper对象 return configuration.<T>getMapper(type, this); } public <T> T getMapper(Class<T> type, SqlSession sqlSession) { // 通过MapperRegistry对象获取 Mapper的对象 return mapperRegistry.getMapper(type, sqlSession); } public <T> T getMapper(Class<T> type, SqlSession sqlSession) { // 通过注册的 mapperProxy 获取mapperProxyFactory 这里的knownMappers 在解析 mapper 的时候添加进入的 final MapperProxyFactory<T> mapperProxyFactory = (MapperProxyFactory<T>) knownMappers.get(type); if (mapperProxyFactory == null) throw new BindingException("Type " + type + " is not known to the MapperRegistry."); try { // 通过mapperProxyFactory 实例化一个对象 return mapperProxyFactory.newInstance(sqlSession); } catch (Exception e) { throw new BindingException("Error getting mapper instance. Cause: " + e, e); } } public T newInstance(SqlSession sqlSession) { //MapperProxy 继承了 InvocationHandler final MapperProxy<T> mapperProxy = new MapperProxy<T>(sqlSession, mapperInterface, methodCache); return newInstance(mapperProxy); } protected T newInstance(MapperProxy<T> mapperProxy) { // 动态代理的对像 return (T) Proxy.newProxyInstance(mapperInterface.getClassLoader(), new Class[] { mapperInterface }, mapperProxy); }
答案:动态代理
原文地址:https://www.cnblogs.com/jas0/p/9545504.html
时间: 2024-11-07 22:44:13