spring如何实现两个xml配置文件间的互调
这篇文章主要介绍了spring如何实现两个xml配置文件间的互调,文中通过示例代码介绍的非常详细,对大家的学习或者工作具有一定的参考学习价值,需要的朋友可以参考下
首先建两个测试类
package soundsystem; public class Dog { private String Cry; private Cat Cat; public void setCry(String cry) { Cry = cry; } public void setCat(soundsystem.Cat cat) { Cat = cat; } public void DogCry(){ System.out.println("狗叫:"+Cry); Cat.CatCry(); } }
package soundsystem; public class Cat { private String Cry; public Cat(String cry){ this.Cry=cry; } public void CatCry(){ System.out.println("猫叫:"+Cry); } }
然后针对两类建两个xml配置文件
Bean_DogXML.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"> <import resource="Bean_CatXML.xml"></import> <bean id="Dog" class="soundsystem.Dog"> <property name="Cry" value="汪汪汪~"></property> <property name="cat" ref="Cat"></property> </bean> </beans>
Bean_CatXML.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"> <bean id="Cat" class="soundsystem.Cat"> <constructor-arg value="喵喵~"></constructor-arg> </bean> </beans>
现在开始测试:
package Test; import org.junit.runner.RunWith; import org.springframework.context.ApplicationContext; import org.springframework.context.support.ClassPathXmlApplicationContext; import org.springframework.test.context.junit4.SpringJUnit4ClassRunner; import soundsystem.Cat; import soundsystem.Dog; @RunWith(SpringJUnit4ClassRunner.class) public class Test { @org.junit.Test public static void main(String[] args) { ApplicationContext ap=new ClassPathXmlApplicationContext("config/Bean_DogXML.xml"); Dog dog=(Dog)ap.getBean("Dog"); dog.DogCry(); } }
输出结果是:
以上就是本文的全部内容,希望对大家的学习有所帮助,也希望大家多多支持我们。
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