Question

我是Spring MVC的新手，试图实现Spring Security功能。当我修改web.xml文件时，我在Catalina日志中遇到了这种错误：

27.11.2012 9:49:21 org.apache.catalina.core.StandardContext startInternal
SEVERE: Error listenerStart

web.xml文件是：

<?xml version="1.0" encoding="UTF-8"?>

<web-app version="3.0" 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_3_0.xsd">
    <!-- The definition of the Root Spring Container shared by all Servlets and Filters -->
    <context-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>/WEB-INF/spring/root-context.xml /WEB-INF/spring/appServlet/security-context.xml
        </param-value>
    </context-param>
    <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>
    <!-- Creates the Spring Container shared by all Servlets and Filters -->
    <listener>
        <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>

    </listener>

    <!-- Processes application requests -->
    <servlet>
        <servlet-name>app</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
        <init-param>
            <param-name>contextConfigLocation</param-name>
            <param-value>/WEB-INF/spring/appServlet/app-servlet.xml</param-value>
        </init-param>
        <load-on-startup>1</load-on-startup>
    </servlet>

    <servlet-mapping>
        <servlet-name>app</servlet-name>
        <url-pattern>/</url-pattern>
    </servlet-mapping>


    <servlet-mapping>
        <servlet-name>default</servlet-name>
        <url-pattern>*.gif</url-pattern>
        <url-pattern>*.png</url-pattern>
        <url-pattern>*.js</url-pattern>
        <url-pattern>*.css</url-pattern>
    </servlet-mapping>

    <!--CharsetFilter start-->
    <filter>
        <filter-name>CharsetFilter</filter-name>
        <filter-class>kz.bimash.FoodSec.filters.CharsetFilter</filter-class>
        <init-param>
            <param-name>requestEncoding</param-name>
            <param-value>UTF-8</param-value>
        </init-param>
    </filter>

    <filter-mapping>
        <filter-name>CharsetFilter</filter-name>
        <url-pattern>/*</url-pattern>
    </filter-mapping>
    <welcome-file-list>
        <welcome-file>redirect.jsp</welcome-file>
    </welcome-file-list>
</web-app>

此代码给出了一个错误，但是当我尝试删除Spring Security配置部分时，它开始正常工作。你能纠正web.xml中的错误吗？

这是Apache日志文件：

27.11.2012 10:57:37 org.apache.catalina.util.LifecycleBase stop
INFO: The stop() method was called on component    [StandardEngine[Catalina].StandardHost[localhost].StandardContext[/foodsec]] after stop() had already been called. The second call will be ignored.
27.11.2012 10:57:38 org.apache.catalina.startup.HostConfig deleteRedeployResources
INFO: Undeploying context [/foodsec]
27.11.2012 10:57:38 org.apache.catalina.startup.HostConfig deployDescriptor
INFO: Deploying configuration descriptor C:\Program Files\Apache Software  Foundation\tomcat7\conf\Catalina\localhost\foodsec.xml
27.11.2012 10:57:45 org.apache.catalina.core.StandardContext startInternal
SEVERE: Error listenerStart
27.11.2012 10:57:45 org.apache.catalina.core.StandardContext startInternal
SEVERE: Context [/foodsec] startup failed due to previous errors

Answer 1

您的多重配置可能会出现问题：

    <param-value>/WEB-INF/spring/root-context.xml /WEB-INF/spring/appServlet/security-context.xml

请尝试使用逗号：

    <param-value>/WEB-INF/spring/root-context.xml,/WEB-INF/spring/appServlet/security-context.xml

最佳做法是使用通配符，如下所示：

    <param-value>classpath*:META-INF/spring/applicationContext*.xml</param-value>

然后将您的上下文文件命名为：

src/main/resources/META-INF/spring/applicationContext.xml
src/main/resources/META-INF/spring/applicationContext-security.xml

Answer 2

您需要开始使用一些日志框架。（将日志框架jar放在类路径上）没有部署日志，没有人会帮助你（我们不想猜）。

看一下教程：http://fauzimh.wordpress.com/2011/01/26/spring-3-framework-hello-world-using-eclipse-slf4j-and-logback/

现在在项目中启用SLF4J和Logback。

启用它之后，您应该在Tomcat日志中获取有关Spring XML配置错误的信息。使用该信息来解决您的问题。是的，我认为问题不在你的web.xml中，而在你的Spring上下文文件中。

修改web.xml时出现安全性错误

2 个答案: