java.lang.IllegalStateException:找不到WebApplicationContext:没有注册ContextLoaderListener？

Question

这是WEB-INF中的文件web.xml

<?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">

    <context-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>/WEB-INF/applicationContext.xml</param-value>
    </context-param>

    <filter>
        <filter-name>LoginFilter</filter-name>
        <filter-class>glpi.filter.LoginFilter</filter-class>
    </filter>
    <filter-mapping>
        <filter-name>LoginFilter</filter-name>
        <url-pattern>/index.jsp</url-pattern>
    </filter-mapping>

     <servlet>
        <servlet-name>context</servlet-name>
        <servlet-class>org.springframework.web.context.ContextLoaderServlet</servlet-class>
        <load-on-startup>1</load-on-startup>
     </servlet>

     <servlet>
        <servlet-name>dispatcher</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
        <load-on-startup>1</load-on-startup>
     </servlet>

         <servlet>
        <servlet-name>action</servlet-name>
        <servlet-class>org.apache.struts.action.ActionServlet</servlet-class>
        <init-param>
            <param-name>config</param-name>
            <param-value>/WEB-INF/struts-config.xml</param-value>
        </init-param>
        <init-param>
            <param-name>debug</param-name>
            <param-value>2</param-value>
        </init-param>
        <init-param>
            <param-name>detail</param-name>
            <param-value>2</param-value>
        </init-param>
        <load-on-startup>2</load-on-startup>
    </servlet>

     <servlet-mapping>
        <servlet-name>dispatcher</servlet-name>
        <url-pattern>*.htm</url-pattern>
    </servlet-mapping>
    <servlet-mapping>
        <servlet-name>action</servlet-name>
        <url-pattern>*.do</url-pattern>
    </servlet-mapping>
    <session-config>
        <session-timeout>
            30
        </session-timeout>
    </session-config>
    <welcome-file-list>
        <welcome-file>/login.jsp</welcome-file>
    </welcome-file-list>

</web-app>

Answer 1

我认为你缺少上下文加载器监听器(选择你的弹簧上下文文件).

将其添加到您的web.xml

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

您还可以查看初始Web配置部分@ http://static.springsource.org/spring/docs/2.0.x/reference/beans.html

Answer 2

您将ContextLoaderServlet和DispatcherServlet都设置为load-on-startup = 1.这意味着其中任何一个都可以先启动,并且您需要首先启动ContextLoaderServlet,因为这是创建错误所述的根WebApplicationContext的原因.因此,将ContextLoaderServlet的load-on-startup保留为1,并将DispatcherServlet更改为2或更高.

实际上,最好使用ContextLoaderListener而不是Servlet,除非你在一个看起来不能正常工作的旧容器上.

Answer 3

在web.xml文件中添加以下代码,bcs它查找要加载的contex,因此我们必须最初声明它.

    <context-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>/WEB-INF/HelloWeb-servlet.xml</param-value>
    </context-param>

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