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我右鍵new 沒有web.xml 這個(gè)選擇,請問怎么新建一個(gè)xml?
2018-05-23
源自:Shiro安全框架入門 4-1
正在回答
右鍵創(chuàng)建不了,我試過~
等一束花開3990875 提問者
file—>Project Structure?
如圖所示
但是你得自己寫內(nèi)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?監(jiān)聽器--> ????<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> ????<!--?攔截器?攔截所有?--> ????<servlet-mapping> ????????<servlet-name>springDispatcherServlet</servlet-name> ????????<url-pattern>/</url-pattern> ????</servlet-mapping> </web-app>
舉報(bào)
從零入門Shiro安全框架
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2018-05-24
右鍵創(chuàng)建不了,我試過~
2018-05-24
file—>Project Structure?
如圖所示
但是你得自己寫內(nèi)容