Spring配置文件location的几种设置方法

默认会去加载WEB-INF下的applicationContext.xml文件，如果该文件不存在，则会抛出以下的异常。

org.springframework.beans.factory.BeanDefinitionStoreException: IOException parsing XML document from ServletContext resource [/WEB-INF/applicationContext.xml]; nested exception is java.io.FileNotFoundException: Could not open ServletContext resource [/WEB-INF/applicationContext.xml]

2. web.xml中通过servlet name自定义

通过以下的定义，会去加载WEB-INF下面的test-servlet.xml作为spring的配置文件

<servlet>
	<servlet-name>test</servlet-name>		
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
	<load-on-startup>1</load-on-startup>
</servlet>

<servlet-mapping>
	<servlet-name>test</servlet-name>
	<url-pattern>*.do</url-pattern>
</servlet-mapping>

3.web.xml中通过DispatcherServlet的init-param自定义

通过以下的定义，会去加载src/config文件夹下的test-servlet.xml作为spring的配置文件

<servlet>  
    <servlet-name>test</servlet-name>  
    <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>  
    <init-param>  
        <param-name>contextConfigLocation</param-name>  
        <param-value>classpath:config/test-servlet.xml</param-value>  
    </init-param>  
</servlet>  
<servlet-mapping>  
    <servlet-name>test</servlet-name>  
    <url-pattern>*.do</url-pattern>  
</servlet-mapping>  

4.web.xml中通过ContextLoaderListener自定义

通过以下的定义，会去加载WEB-INF文件夹下的test-servlet.xml作为spring的配置文件

<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<context-param>
	<param-name>contextConfigLocation</param-name>
	<param-value>/WEB-INF/test-servlet.xml</param-value>
</context-param>

注：其他的配置文件也可以用这种方法设定，比如要去加载src/config/文件夹下的test-context.xml

<listener>  
    <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>  
</listener>  
<context-param>  
    <param-name>contextConfigLocation</param-name>  
    <param-value>classpath:config/test-context.xml</param-value>  
</context-param>

注意点

如果就是想将/WEB-INF/applicationContext.xml作为配置文件呢。

单单使用<listener>和<context-param>是不行的。因为你要让Servlet去处理某种请求（URL，比如*.do），就必须得定义<servlet>,这种情况下，完整的web.xml如下

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
	xmlns="http://java.sun.com/xml/ns/javaee" xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"
	xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"
	id="WebApp_ID" version="2.5">
	
	<display-name>SpringFormTagTest</display-name>
	
	<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>
	
	<servlet>
		<servlet-name>test</servlet-name>
		<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
		<init-param>
		<param-name>contextConfigLocation</param-name>
		<param-value>/WEB-INF/applicationContext.xml</param-value> 
		</init-param>
		<load-on-startup>1</load-on-startup>
	</servlet>
	
	<servlet-mapping>
		<servlet-name>test</servlet-name>
		<url-pattern>*.*</url-pattern>
	</servlet-mapping>
	
	<welcome-file-list>
		<welcome-file>jsp/input.jsp</welcome-file>
	</welcome-file-list>
</web-app>