如何在myeclipse中搭建ssm框架 _电脑

最简单的ssm框架搭建方式

需要这些哦
电脑
myeclipse8.5
方式/
1起首呢打开eclipse点击左上角的File-->New-->Report Web Project

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2然后在弹出的窗口输入项目名称点击Finish

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3项目布局

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4接下来呢我们直接生当作Spring 在项目上右击选择 Myeclipse-->Add Spring Cababilities 之后会弹出窗口(如图) 然后勾选JDBC(它会主动勾选三个)之后再勾选一个web3.0,再点击finish

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5此刻的项目布局

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6之后我们再设置装备摆设web.xml:
【如何在myeclipse中搭建ssm框架】<?xml version="1.0" encoding="UTF-8"?><web-app version="2.5" xmlns="http://java.sun.com/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"> <welcome-file-list> <welcome-file>index.jsp</welcome-file> </welcome-file-list> <filter> <filter-name>encoding</filter-name> <filter-class>org.springframework.web.filter.CharacterEncodingFilter</filter-class> <init-param> <param-name>encoding</param-name> <param-value>UTF-8</param-value> </init-param> </filter> <filter-mapping> <filter-name>encoding</filter-name> <url-pattern>/*</url-pattern> </filter-mapping> <servlet> <servlet-name>springDispatcher</servlet-name> <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class> <init-param> <param-name>contextConfigLocation</param-name> <param-value>classpath:applicationContext.xml</param-value> </init-param> <load-on-startup>0</load-on-startup> </servlet> <servlet-mapping> <servlet-name>springDispatcher</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> </web-app>