ssh，怎样不用applicationContext.getBean得到类实例

Question

applicationContext是ApplicaionContext实例
如何配置可以在服务器开始时就自动加载spring配置文件，并实例里面的bean 
而不需要 ApplicionContext applicationContext = new ClasspathApplicionContext（“..”） 
            applicationContext.getBean（".."）;

Answer 1

在web.xml里面配置
<listener>
   <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
  </listener>
  <context-param>
   <param-name>contextConfigLocation</param-name>
   <param-value>/WEB-INF/applicationContext.xml</param-value>
  </context-param>
不知道是不是你说的那样

Answer 2

第一个Bean
package com.bean;public class User {
private String name;
private String pass;
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public String getPass() {
return pass;
}
public void setPass(String pass) {
this.pass = pass;
}
public void print(){
System.out.print("av");
}
}第二个Bean public class LoginAction extends Action {
private User user;
public void setUser(User user) {
this.user = user;
}

public ActionForward execute(ActionMapping mapping, ActionForm form,
HttpServletRequest request, HttpServletResponse response) {
user.print();
return null;
}
}
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-2.0.xsd">

<bean id="user"
class="com.bean.User">
<property name="name"
value="123">
</property>
<property name="pass"  value="123">
</property>
</bean>
<bean id ="loginAction" class="com.yourcompany.struts.action.LoginAction">
  <property name="user" ref="user"></property>
</bean>

</beans>
web.xml:
<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns="http://java.sun.com/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" version="2.5" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee   http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd">
  <servlet>
    <servlet-name>action</servlet-name>
    <servlet-class>org.apache.struts.action.ActionServlet</servlet-class>
    <init-param>
      <param-name>config</param-name>
      <param-value>/WEB-INF/struts-config.xml</param-value>
    </init-param>
    <init-param>
      <param-name>debug</param-name>
      <param-value>3</param-value>
    </init-param>
    <init-param>
      <param-name>detail</param-name>
      <param-value>3</param-value>
    </init-param>
    <load-on-startup>0</load-on-startup>
  </servlet>
  
  <servlet-mapping>
    <servlet-name>action</servlet-name>
    <url-pattern>*.do</url-pattern>
  </servlet-mapping>
  
  <listener>  
    <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>  
  </listener> 
  
   <context-param>
   <param-name>contextConfigLocation</param-name>
   <param-value>classpath*:applicationContext.xml</param-value>
  </context-param>
  
   
 
  
  
  <welcome-file-list>
    <welcome-file>index.jsp</welcome-file>
  </welcome-file-list>
</web-app>

Answer 3

上面是大致的框架，会出现null pointer错误啊

Answer 4

<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>classpath:applicationContext*.xml</param-value>
</context-param>

Answer 5

这种都是必须的啊，谁做web开发不在 web.xml里面配？？楼主，你既然要用 spring，就必须按照他的BeanFactory方式参数Bean。他只是帮你new出来而已，和你自己new没有任何区别。不过他能帮你管理。如果你不用 spring，那把类定义成，除非你把类定义成 静态类，这样加载时就不用初始化。不过静态类一般都是内部类。

Answer 6

可以自己写个基类 然后子类继承它
   protected static Object getBean(ServletContext servletContext,String beanName){
   WebApplicationContext wac = WebApplicationContextUtils.getWebApplicationContext(servletContext);
   return wac.getBean(beanName);
   }
servletContext这个你因该在你的web项目中能得到的 这样就很方便的直接返回你要的bean了 HOHO~