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(yīng)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(duì)應(yīng)的上下文配置,?默認(rèn)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(qǐng)求--> ????<url-pattern>/</url-pattern> </servlet-mapping> ?? </web-app>
上面的listener-class和servlet-class的值,我們?cè)趧偱渲脮r(shí)時(shí)通過(guò)什么途徑獲取的呢?
2017-05-05
謝謝dtd文件
2017-02-13
通過(guò)什么途徑獲得?這個(gè)問(wèn)的尷尬了。首先你必須非常熟悉你這個(gè)項(xiàng)目,因?yàn)檫@些class值都是你導(dǎo)入要用到各種框架jar包的地址,各個(gè)包的功能,一般導(dǎo)入的都是必須的,網(wǎng)上教程也一大堆