使用 <context:component-scan> 特性，可以自动扫描 base-package 下类名有注解 @Component、@Service 或者 @Controller 的类，为其在 Spring 容易里创建一个对象。
@Service、@Controller 和 @Component 在语法上作用是一样的，区别是各自有自己的语义，例如 SpringMVC 里用 @Controller 表明类是 Controller。
spring-beans.xml
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| <?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:context="http://www.springframework.org/schema/context"
xsi:schemaLocation="
http://www.springframework.org/schema/beans
http://www.springframework.org/schema/beans/spring-beans.xsd
http://www.springframework.org/schema/context
http://www.springframework.org/schema/context/spring-context.xsd">
<context:component-scan base-package="com.xtuer.beans"/>
</beans>
|
House
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| package com.xtuer.beans;
import org.springframework.stereotype.Component;
@Component
public class House {
private String description;
public String getDescription() {
return description;
}
public void setDescription(String description) {
this.description = description;
}
}
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测试
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| import com.xtuer.beans.House;
import com.xtuer.util.CommonUtils;
import org.junit.Assert;
import org.junit.BeforeClass;
import org.junit.Test;
import org.springframework.context.ApplicationContext;
import org.springframework.context.support.ClassPathXmlApplicationContext;
public class ComponentScanTest {
private static ApplicationContext context;
@BeforeClass
public static void setup() {
context = new ClassPathXmlApplicationContext("spring-beans.xml");
}
@Test
public void test() {
House house1 = context.getBean(House.class);
CommonUtils.output(house1);
House house2 = context.getBean("house", House.class);
CommonUtils.output(house2);
Assert.assertEquals(house1, house2);
}
}
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输出
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| {
"description" : null
}
{
"description" : null
}
|
