web.xml中配置的class老師是如何獲得的呢?
<?xml?version="1.0"?encoding="UTF-8"?> <web-app?version="2.4"?xmlns="http://java.sun.com/xml/ns/j2ee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee? http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd"> ??<display-name>Spring?MVC?Study</display-name> ?? ????<!--?Spring應用上下文,?理解層次化的ApplicationContext?--> ??<context-param> ? <param-name>contextConfigLocation</param-name> <param-value>/WEB-INF/configs/spring/applicationContext*.xml</param-value> ??</context-param> ?? ??<listener> <listener-class> org.springframework.web.context.ContextLoaderListener </listener-class> ??</listener> ?? ????<!--?DispatcherServlet,?Spring?MVC的核心?--> <servlet> ????<servlet-name>mvc-dispatcher</servlet-name> ????<servlet-class>?org.springframework.web.servlet.DispatcherServlet</servlet-class> ????<!--?DispatcherServlet對應的上下文配置,?默認為/WEB-INF/$servlet-name$-servlet.xml?--> <init-param> ????<param-name>contextConfigLocation</param-name> ????<param-value>/WEB-INF/configs/spring/mvc-dispatcher-servlet.xml</param-value> </init-param> <load-on-startup>1</load-on-startup> </servlet> <servlet-mapping> ????<servlet-name>mvc-dispatcher</servlet-name> ????<!--?mvc-dispatcher攔截所有的請求--> ????<url-pattern>/</url-pattern> </servlet-mapping> ?? </web-app>
上面的listener-class和servlet-class的值,我們在剛配置時時通過什么途徑獲取的呢?
2017-05-05
謝謝dtd文件
2017-02-13
通過什么途徑獲得?這個問的尷尬了。首先你必須非常熟悉你這個項目,因為這些class值都是你導入要用到各種框架jar包的地址,各個包的功能,一般導入的都是必須的,網上教程也一大堆