Question

运行项目时，出现此错误。我可以正确地编写所有内容，但是有一天我无法解决我编写不正确的错误。项目根本无法启动，并且总是会出现此错误。我使用netbeans和tomcat。请帮我 enter image description here

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_3_0.xsd"
         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">

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

    <error-page>

        <location>/errors</location>
    </error-page>


    <filter>
        <filter-name>sitemesh</filter-name>
        <filter-class>org.sitemesh.config.ConfigurableSiteMeshFilter</filter-class>
    </filter>
    <filter-mapping>
        <filter-name>sitemesh</filter-name>
        <url-pattern>/*</url-pattern>
    </filter-mapping>






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




        <!-- Loads Spring Security config file -->
    <context-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>
            /WEB-INF/spring-security.xml
        </param-value>
    </context-param>

    <!-- Spring Security -->
    <filter>
        <filter-name>springSecurityFilterChain</filter-name>
        <filter-class>org.springframework.web.filter.DelegatingFilterProxy
        </filter-class>
    </filter>

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






</web-app>

Spring-Security.XML

<?xml version="1.0" encoding="UTF-8"?>
<beans:beans
    xmlns="http://www.springframework.org/schema/security"
    xmlns:beans="http://www.springframework.org/schema/beans"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xsi:schemaLocation="http://www.springframework.org/schema/beans
                    http://www.springframework.org/schema/beans/spring-beans-4.0.xsd
                    http://www.springframework.org/schema/security
                     http://www.springframework.org/schema/security/spring-security-4.2.xsd">

    <http auto-config="true">
        <intercept-url pattern="/admin**" access="hasRole('ROLE_ADMIN')" />

        <!-- user-defined login form redirection -->
        <form-login login-page="/login" default-target-url="/" authentication-failure-url="/login?error" />

        <!-- logout url -->
        <logout logout-success-url="/login?logout" />

        <!-- csrf disabled -->
        <csrf disabled="true" />
    </http>

    <authentication-manager>
        <authentication-provider>
            <user-service>
                <user name="admin" password="1234" authorities="ROLE_ADMIN" />                
            </user-service>
        </authentication-provider>
    </authentication-manager>
</beans:beans>

MVC-Dispatcher-Servlet.XML

<?xml version="1.0" encoding="UTF-8"?>
<beans
    xmlns="http://www.springframework.org/schema/beans"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xmlns:p="http://www.springframework.org/schema/p"
    xmlns:context="http://www.springframework.org/schema/context"
    xsi:schemaLocation="http://www.springframework.org/schema/beans  
    http://www.springframework.org/schema/beans/spring-beans-3.0.xsd  
    http://www.springframework.org/schema/context  
    http://www.springframework.org/schema/context/spring-context-3.0.xsd">

    <context:component-scan base-package="adil.java.schoolmaven" />

    <!-- Resolves Views Selected For Rendering by @Controllers to *.jsp Resources in the /WEB-INF/ Folder -->
    <bean
        class="org.springframework.web.servlet.view.InternalResourceViewResolver">
        <property name="prefix" value="/WEB-INF/views/" />
        <property name="suffix" value=".jsp" />
    </bean>
</beans>

Answer 1

您需要在ContextLoaderListener中指定web.xml：

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

Answer 2

欢迎使用堆栈溢出。：）

似乎您的web.xml中缺少ContextLoaderListener。

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

此外，我认为您将必须为您的Spring应用程序上下文 mvc-dispatcher-servlet.xml 和 spring-security.xml 加载多个文件。

有多种方法可以实现这一目标。您可以在web.xml配置中指定多个路径，也可以将spring-security导入mvc-despatcher-servlet.xml。有关更多信息，请参考以下链接：

https://www.baeldung.com/spring-web-contexts

Splitting applicationContext to multiple files

如果这对您没有帮助，请尝试在stackoverflow中搜索“ No bean springSecurityFilterChain”，您可能会找到许多解决方案。

错误-无法使用名为'springSecurityFilterChain'的bean

2 个答案: