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我一直在试用各种教程来让Spring Security工作。我正在自学一个学校项目。我有一个教程的基本登录功能，现在正在尝试实现sec taglib，因此我可以使用authorize标签根据用户角色选择性地向用户显示内容。无论我尝试的教程/官方文档的建议有什么组合，我都无法实现它（我确定它来自我对XML配置的乏味理解）。将不胜感激任何帮助。与Spring Security的基础知识苦苦挣扎 - 需要启用sec taglib
这里是我的pom.xml：
<project xmlns="http://maven.apache.org/POM/4.0.0"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://maven.apache.org/POM/4.0.0
http://maven.apache.org/xsd/maven-4.0.0.xsd">
<modelVersion>4.0.0</modelVersion>
<groupId>com.angels</groupId>
<artifactId>ccollier</artifactId>
<version>0.0.1-SNAPSHOT</version>
<packaging>war</packaging>
<dependencies>
<dependency>
<groupId>javax</groupId>
<artifactId>javaee-web-api</artifactId>
<version>6.0</version>
<scope>provided</scope>
</dependency>
<dependency>
<groupId>com.fasterxml.jackson.core</groupId>
<artifactId>jackson-databind</artifactId>
<version>2.5.3</version>
</dependency>
<dependency>
<groupId>org.springframework</groupId>
<artifactId>spring-webmvc</artifactId>
<version>4.2.2.RELEASE</version>
</dependency>
<dependency>
<groupId>org.hibernate</groupId>
<artifactId>hibernate-validator</artifactId>
<version>5.0.2.Final</version>
</dependency>
<dependency>
<groupId>org.webjars</groupId>
<artifactId>bootstrap</artifactId>
<version>3.3.6</version>
</dependency>
<dependency>
<groupId>org.webjars</groupId>
<artifactId>jquery</artifactId>
<version>1.11.1</version>
</dependency>
<dependency>
<groupId>log4j</groupId>
<artifactId>log4j</artifactId>
<version>1.2.17</version>
</dependency>
<dependency>
<groupId>javax.servlet</groupId>
<artifactId>jstl</artifactId>
<version>1.2</version>
</dependency>
<dependency>
<groupId>org.springframework.security</groupId>
<artifactId>spring-security-web</artifactId>
<version>4.0.1.RELEASE</version>
</dependency>
<dependency>
<groupId>org.springframework.security</groupId>
<artifactId>spring-security-config</artifactId>
<version>4.0.1.RELEASE</version>
</dependency>
<dependency>
<groupId>org.springframework.security</groupId>
<artifactId>spring-security-taglibs</artifactId>
<version>3.1.3.RELEASE</version>
</dependency>
</dependencies>
<build>
<pluginManagement>
<plugins>
<plugin>
<groupId>org.apache.maven.plugins</groupId>
<artifactId>maven-compiler-plugin</artifactId>
<version>3.2</version>
<configuration>
<verbose>true</verbose>
<source>1.8</source>
<target>1.8</target>
<showWarnings>true</showWarnings>
</configuration>
</plugin>
<plugin>
<groupId>org.apache.tomcat.maven</groupId>
<artifactId>tomcat7-maven-plugin</artifactId>
<version>2.2</version>
<configuration>
<path>/</path>
<contextReloadable>true</contextReloadable>
</configuration>
</plugin>
</plugins>
</pluginManagement>
</build>
我的web.xml：
<!-- webapp/WEB-INF/web.xml -->
<web-app 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"
version="3.0">
<display-name>To do List</display-name>
<welcome-file-list>
<welcome-file>login.do</welcome-file>
</welcome-file-list>
<servlet>
<servlet-name>dispatcher</servlet-name>
<servlet-class>
org.springframework.web.servlet.DispatcherServlet
</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/todo-servlet.xml</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>dispatcher</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
<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>
我的应用程序的命名空间（待办事项-servlet.xml中）：
<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:context="http://www.springframework.org/schema/context"
xmlns:security="http://www.springframework.org/schema/security"
xmlns:mvc="http://www.springframework.org/schema/mvc"
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/mvc http://www.springframework.org/schema/mvc/spring-mvc-4.0.xsd
http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context-4.0.xsd
http://www.springframework.org/schema/security http://www.springframework.org/schema/security/spring-security-3.0.3.xsd">
<context:component-scan base-package="com.angels" />
<bean
class="org.springframework.web.servlet.view.InternalResourceViewResolver">
<property name="prefix">
<value>/WEB-INF/views/</value>
</property>
<property name="suffix">
<value>.jsp</value>
</property>
</bean>
<mvc:resources mapping="/resources/**" location="/resources/"
cache-period="31556926"/>
<mvc:resources mapping="/webjars/**" location="/webjars/"/>
<mvc:annotation-driven />
<http auto-config="true">
<intercept-url pattern="/**" access="ROLE_USER" />
</http>
</beans>
现在造成什么问题是应用程序名称空间中的这一行。我直接从Spring文档http://docs.spring.io/spring-security/site/docs/3.1.x/reference/ns-config.html中了解到这一点。它表示，这些行需要启用网络安全。
<http auto-config="true">
<intercept-url pattern="/**" access="ROLE_USER" />
</http>
当我在它悬停在Eclipse中，我得到：
cvc-complex-type.2.4.a: Invalid content was found starting with element 'http'. One of '{"http://
www.springframework.org/schema/beans":import, "http://www.springframework.org/schema/beans":alias, "http://
www.springframework.org/schema/beans":bean, WC[##other:"http://www.springframework.org/schema/beans"], "http://
www.springframework.org/schema/beans":beans}' is expected.
我再尝试运行我的应用程序，这导致它不能启动时得到了类似的错误。感谢任何帮助。