java web 44 : Spring

src/main/resources  applicationContext.xml

<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xsi:schemaLocation="http://www.springframework.org/schema/beans
        http://www.springframework.org/schema/beans/spring-beans.xsd">
    
    <!-- 将EmpService接口的实现类作为bean注册到spring容器中,
        即让spring容器创建该类的实例 。
        如果注册的类有父接口,id值通常是接口名(首字母小写)
        如果注册的类没有父接口,id值通常为类名(首字母小写)-->
    <bean id="empService" class="com.tedu.service.EmpServiceImpl"></bean>
</beans>

com.tedu.service/EmpService.java(接口)

package com.tedu.service;

/** 模拟员工模块的service层接口 **/
public interface EmpService {
    //新增员工信息
    public void add();
}

com.tedu.service/EmpServiceImpl.java

package com.tedu.service;

/** 模拟员工模块的servic接口的实现类(子类)*/
public class EmpServiceImpl implements EmpService {

    @Override
    public void add() {
        System.out.println("EmpServiceImpl...add()...");
    }

}

com.tedu/TestService.java

package com.tedu;

import org.junit.Test;
import org.springframework.context.support.ClassPathXmlApplicationContext;

import com.tedu.service.EmpService;
import com.tedu.service.EmpServiceImpl;

public class TestService {
    @Test
    public void testService() {
        //创建Spring的容器对象
        ClassPathXmlApplicationContext ac = new ClassPathXmlApplicationContext("applicationContext.xml");
        //通过spring容器获取EmpService接口的子类实例
        EmpService service = (EmpService) ac.getBean("empService");
        service.add();
    }
}

Run As Junit Test:

 

posted @ 2020-08-22 19:55  Saturn5  阅读(15)  评论(0)    收藏  举报