问题描述
在网上看到这样一句话：“把DAO实现类注入到service实现类中，把service的接口(注意不要是service的实现类)注入到action中”，确实如此，比如UserServiceImpl实现了接口UserService，当我在UserAction中写private UserServiceImpl userService;@Resourcepublic void setUserService(UserServiceImpl userService) {this.userService = userService;}会报错。Unable to instantiate Action, com.action.UserAction, defined for 'user' in namespace '/'Failed to convert property value of type [$Proxy25 implementing com.service.UserService,org.springframework.aop.SpringProxy,org.springframework.aop.framework.Advised] to required type [com.service.UserServiceImpl] for property 'userService'; nested exception is java.lang.IllegalArgumentException: Cannot convert value of type [$Proxy25 implementing com.service.UserService,org.springframework.aop.SpringProxy,org.springframework.aop.framework.Advised] to required type [com.service.UserServiceImpl] for property 'userService': no matching editors or conversion strategy found把上面的类型改成UserService就ok了，经过多次测试，我采用的是注解方式配置，Service层一旦实现了接口，在action中注入接口正确，但是注入实现类不行，能不能讲讲为什么啊 问题补充：zyn010101 写道
解决方案
因为代理的原因，注入的是一个代理类，不能转型成UserServiceImpl，所以set会报错。而接口肯定不会报错
解决方案二：
UserAction 里面注入 UserService在配置文件中，(不管是注释还是手写)都会有这么一行<bean id="UserAction" class="UserAction"><property name="userService" ref="userService" /></bean><bean id="userService" class="UserServiceImpl"/>因此 注入方式是property name，要与setXxx)： private UserService userService; public void setUserService(UserService userService) { this.userService = userService; } 因此配置文件是这样也是可以的：<bean id="UserAction" class="UserAction"><property name="service" ref="userService" /></bean> 就需要写成 private UserService service; public void setService(UserService userService) { this.service = userService; } 而你的方式，setUserService(UserServiceImpl userService)配置文件应该如下，让spring去寻找 bean为 userServiceImpl的id<bean id="UserAction" class="UserAction"><property name="service" ref="userService" /></bean><bean id="userServiceImpl" class="UserServiceImpl"/>
解决方案三：
private UserService userService; @Resource public void setUserService(UserService userService) { this.userService = userService; } public UserService getUserService(){ return this.userService}在spring中这么配置<bean id="userService" class="UserServiceImpl"/>