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idea提示 org.springframework.web.context.ContextLoaderListener没找到
 he***ba  分类:Java代码  人气:1291  回帖:1  发布于6年前 收藏

web.xml

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
         xmlns="http://java.sun.com/xml/ns/javaee"
         xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"
         xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"
         id="WebApp_ID" version="2.5">
    <display-name>SalamanderBBS</display-name>

    <!-- 起始欢迎界面 -->
    <welcome-file-list>
        <welcome-file />
    </welcome-file-list>

    <!-- 读取spring配置文件 -->
    <context-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>classpath:spring-mybatis.xml</param-value>
    </context-param>

    <!-- 设计路径变量值 -->
    <context-param>
        <param-name>webAppRootKey</param-name>
        <param-value>springmvc.root</param-value>
    </context-param>


    <!-- Spring字符集过滤器 -->
    <filter>
        <filter-name>SpringEncodingFilter</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>
        <init-param>
            <param-name>forceEncoding</param-name>
            <param-value>true</param-value>
        </init-param>
    </filter>
    <filter-mapping>
        <filter-name>SpringEncodingFilter</filter-name>
        <url-pattern>/*</url-pattern>
    </filter-mapping>

    <listener>
        <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
    </listener>

    <!-- springMVC核心配置 -->
    <servlet>
        <servlet-name>dispatcherServlet</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
        <init-param>
            <param-name>contextConfigLocation</param-name>
            <!--spingMVC的配置路径 -->
            <param-value>classpath:spring-mvc.xml</param-value>
        </init-param>
        <load-on-startup>1</load-on-startup>
    </servlet>
    <!-- 拦截设置 -->
    <servlet-mapping>
        <servlet-name>dispatcherServlet</servlet-name>
        <url-pattern>/</url-pattern>
    </servlet-mapping>


    <!-- 错误跳转页面 -->
    <error-page>
        <!-- 路径不正确 -->
        <error-code>404</error-code>
        <location>/WEB-INF/view/errpage/404.jsp</location>
    </error-page>
    <error-page>
        <!-- 没有访问权限,访问被禁止 -->
        <error-code>405</error-code>
        <location>/WEB-INF/view/errpage/405.jsp</location>
    </error-page>
    <error-page>
        <!-- 内部错误 -->
        <error-code>500</error-code>
        <location>/WEB-INF/view/errpage/500.jsp</location>
    </error-page>
</web-app>

项目结构:

讨论这个帖子(1)垃圾回帖将一律封号处理……

Lv1 新人
十***刻 产品经理 6年前#1

解决了:
File > Project Structure > Artifacts > 在右侧Output Layout右击项目名,选择Put into Output Root,见下面截图:

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