为了账号安全,请及时绑定邮箱和手机立即绑定

右键new web.xml,怎么新建?

我右键new 没有web.xml 这个选择,请问怎么新建一个xml?https://img1.sycdn.imooc.com//5b051f130001f2d505410495.jpg

正在回答

2 回答

右键创建不了,我试过~


0 回复 有任何疑惑可以回复我~
#1

等一束花開3990875 提问者

非常感谢!
2018-05-24 回复 有任何疑惑可以回复我~

https://img1.sycdn.imooc.com//5b061ed2000177ae13540929.jpg

file—>Project Structure 

如图所示

但是你得自己写内容

<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
         xmlns="http://java.sun.com/xml/ns/javaee"
         xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd"
         id="WebApp_ID" version="3.0">

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

    <!--spring 监听器-->
    <listener>
        <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
    </listener>

    <listener>
        <listener-class>org.springframework.web.context.request.RequestContextListener</listener-class>
    </listener>

    <!--shiro Filter-->
    <filter>
        <filter-name>shiroFilter</filter-name>
        <filter-class>org.springframework.web.filter.DelegatingFilterProxy</filter-class>
    </filter>

    <filter-mapping>
        <filter-name>shiroFilter</filter-name>
        <url-pattern>/*</url-pattern>
    </filter-mapping>

    <!-- 解析springMvc -->
    <servlet>
        <servlet-name>springDispatcherServlet</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
        <init-param>
            <param-name>contextConfigLocation</param-name>
            <param-value>classpath*:spring-mvc.xml</param-value>
        </init-param>
        <load-on-startup>1</load-on-startup>
    </servlet>

    <!-- 拦截器 拦截所有 -->
    <servlet-mapping>
        <servlet-name>springDispatcherServlet</servlet-name>
        <url-pattern>/</url-pattern>
    </servlet-mapping>

</web-app>


6 回复 有任何疑惑可以回复我~

举报

0/150
提交
取消

右键new web.xml,怎么新建?

我要回答 关注问题
意见反馈 帮助中心 APP下载
官方微信