Spring Security 3.2更新web.xml

时间：2017-04-22 03:57:31

标签： java spring spring-mvc spring-security

我是Spring的新手，我正在使用spring security创建一个简单的登录过程。我正在学习本教程http://www.mkyong.com/spring-security/spring-security-form-login-example/，但对如何更新web.xml感到困惑。

以下是使用spring mvc项目的eclipse sts中的项目设置和代码：

servlet的context.xml中

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

<annotation-driven />

static resources in the ${webappRoot}/resources directory -->
<resources mapping="/resources/**" location="/resources/" />

<beans:bean class="org.springframework.web.servlet.view.InternalResourceViewResolver">
    <beans:property name="prefix" value="/WEB-INF/views/" />
    <beans:property name="suffix" value=".jsp" />
</beans:bean>

<context:component-scan base-package="com.at.ccts" />
</beans:beans>

弹簧security.xml文件

<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-3.0.xsd
    http://www.springframework.org/schema/security
    http://www.springframework.org/schema/security/spring-security-3.2.xsd">

<http auto-config="true">
    <intercept-url pattern="/admin**" access="ROLE_USER" />

    <form-login 
        login-page="/login" 
        login-processing-url="/j_spring_security_check" 
        authentication-failure-url="/login?error" 
        username-parameter="username"
        password-parameter="password" />
    <logout logout-success-url="/login?logout"  />
    <!-- enable csrf protection -->
    <csrf/>
</http>

<authentication-manager>
    <authentication-provider>
        <user-service>
            <user name="admin" password="admin" authorities="ROLE_USER" />
        </user-service>
    </authentication-provider>
</authentication-manager>

</beans:beans>

HomeController.java 

package com.at.ccts.controller;

import org.springframework.stereotype.Controller;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.bind.annotation.RequestMethod;
import org.springframework.web.bind.annotation.RequestParam;
import org.springframework.web.servlet.ModelAndView;

@Controller
public class HomeController {

@RequestMapping(value = "/admin**", method = RequestMethod.GET)
public ModelAndView adminPage() {

    ModelAndView model = new ModelAndView();
    model.addObject("title", "Spring Security Custom Login Form");
    model.addObject("message", "This is protected page!");
    model.setViewName("admin");

    return model;
}

@RequestMapping(value = "/login", method = RequestMethod.GET)
public ModelAndView login(@RequestParam(value = "error", required = false) String error,
        @RequestParam(value = "logout", required = false) String logout) {

    ModelAndView model = new ModelAndView();
    if (error != null) {
        model.addObject("error", "Invalid username and password!");
    }

    if (logout != null) {
        model.addObject("msg", "You've been logged out successfully.");
    }
    model.setViewName("login");

    return model;
}
}

的login.jsp 

<%@ taglib prefix="c" uri="http://java.sun.com/jsp/jstl/core"%>
<%@page session="true"%>
<html>
<head>
<title>Login Page</title>
</head>
<body onload='document.loginForm.username.focus();'>

<h1>Spring Security Custom Login Form (XML)</h1>

<div id="login-box">

    <h3>Login with Username and Password</h3>

    <c:if test="${not empty error}">
        <div class="error">${error}</div>
    </c:if>
    <c:if test="${not empty msg}">
        <div class="msg">${msg}</div>
    </c:if>

    <form name='loginForm'
        action="<c:url value='/j_spring_security_check' />" method='POST'>

        <table>
            <tr>
                <td>User:</td>
                <td><input type='text' name='username'></td>
            </tr>
            <tr>
                <td>Password:</td>
                <td><input type='password' name='password' /></td>
            </tr>
            <tr>
                <td colspan='2'><input name="submit" type="submit"
                    value="submit" /></td>
            </tr>
        </table>

        <input type="hidden" name="${_csrf.parameterName}"
            value="${_csrf.token}" />

    </form>
</div>

</body>
</html>

和我的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/spring/root-context.xml</param-value>
</context-param>

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

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

</web-app>

我可以直接在我的web.xml上添加这段代码，还是需要修改我当前的web.xml？

<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>

root.xml但这里没有声明：

<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="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.xsd">

<!-- Root Context: defines shared resources visible to all other web components -->

</beans>

1 个答案:

答案 0 :(得分：0)

我刚刚修改了我的web.xml，删除了root xml并添加了spring security

的web.xml 

<servlet>
    <servlet-name>SpringController</servlet-name>
    <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
    <init-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>/WEB-INF/spring-servlet.xml</param-value>
    </init-param>
    <load-on-startup>1</load-on-startup>
</servlet>

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

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

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

<!-- Spring Security  Filter -->
<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>
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