Spring学习笔记(一)第一个Spring程序

开始Spring之旅,编写第一个Spring程序, 只是简单地从Xml配置文件中读取Bean, 实现依赖注入DI

UserInfoService接口:

package com.skymr.spring.test;

public interface UserInfoService {

public abstract void save();
}

UserInfoServiceBean:

package com.skymr.spring.test;

public class UserInfoServiceBean implements UserInfoService{

public void save() {
System.out.println("第一个Spring程序");
}

}

Spring配置文件beans.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-2.0.xsd"> 
<bean id = "userInfoService" class="com.skymr.spring.test.UserInfoServiceBean"></bean>
</beans>

最好放在类路径下

测试类：

package com.skymr.spring.test;

import org.junit.BeforeClass;
import org.springframework.context.ApplicationContext;
import org.springframework.context.support.ClassPathXmlApplicationContext;

public class Test {
@BeforeClass
public static void setUpBeforeClass() throws Exception{

}
@org.junit.Test
public void instanceSpring(){
ApplicationContext ctx = new ClassPathXmlApplicationContext("beans.xml");
UserInfoService userInfoService = (UserInfoService) ctx.getBean("userInfoService");
userInfoService.save();
}
}

在junit中运行

打印出: 第一个Spring程序

两种实现化Spring容器方法:

1.在类路径下搜索配置文件:

ApplicationContext ctx = new ClassPathXmlApplicationContext("beans.xml");

2.在系统路径下搜索:

ctx = new FileSystemXmlApplicationContext("d:/a.xml");