2

我在使用 Springs 安全配置时遇到问题。我在下面分享我的代码。

我的问题是我想要/user*url 并且/admin*只有在用户登录到我的应用程序时才应该访问 url,我的应用程序有主要的 ajax 调用,所以我不希望用户在/user*没有登录的情况下访问 URL。但是当我尝试在其中键入 URL 时网络浏览器我什至没有被重定向到登录页面,而是进入了输入 URL 的页面。

所以任何人都可以帮我解决这个问题。

弹簧安全.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-3.0.xsd
     http://www.springframework.org/schema/security
     http://www.springframework.org/schema/security/spring-security-3.1.xsd">

<http auto-config="true">
    <intercept-url pattern="/user**" access="ROLE_USER" />
    <form-login login-page="/login" default-target-url="/user/home"
        authentication-failure-url="/loginfailed" />
    <logout invalidate-session="true" logout-success-url="/logout" />
</http>

<authentication-manager>
    <authentication-provider>
        <jdbc-user-service data-source-ref="dataSource"
            users-by-username-query="select USERNAME, PASSWORD from USER where USERNAME = ?"
            authorities-by-username-query="ROLE_USER"
        />
    </authentication-provider>
</authentication-manager>

web.xml

    <?xml version="1.0" encoding="utf-8" standalone="no"?><web-app 
      xmlns="http://java.sun.com/xml/ns/javaee"
      xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"
      xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" version="2.5"
      xsi:schemaLocation="http://java.sun.com/xml/ns/javaee
      http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd">

<servlet>
    <servlet-name>Admin</servlet-name>
    <servlet-class>
        org.springframework.web.servlet.DispatcherServlet
    </servlet-class>
    <load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
    <servlet-name>Admin</servlet-name>
    <url-pattern>/admin*</url-pattern>
</servlet-mapping>

<servlet>
    <servlet-name>User</servlet-name>
    <servlet-class>
        org.springframework.web.servlet.DispatcherServlet
    </servlet-class>
    <load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
    <servlet-name>User</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/User-servlet.xml,
        /WEB-INF/Spring-Datasource.xml,
        /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>

<welcome-file-list>
    <welcome-file>index</welcome-file>
</welcome-file-list>
    <servlet>
          <servlet-name>SystemServiceServlet</servlet-name>
          <servlet-class>com.google.api.server.spi.SystemServiceServlet</servlet-class>
    <init-param>
        <param-name>services</param-name>
        <param-value/>
    </init-param>
</servlet>
<servlet-mapping>
    <servlet-name>SystemServiceServlet</servlet-name>
    <url-pattern>/_ah/spi/*</url-pattern>
</servlet-mapping>
    </web-app>

ControllerServlet.java

    @Controller
    public class UserController {
    @RequestMapping(value = "/login", method = RequestMethod.GET)
    public String getUserLoginPage(){
        return "user/index";
    }

    @RequestMapping(value = "/loginfailed", method = RequestMethod.GET)
    public String showErrorLoginPage(ModelMap modelMap){
        modelMap.addAttribute("message", "Invalid Login Credentials");
        return "user/index";
    }

    @RequestMapping(value = "/user/home", method = RequestMethod.GET)
    public String getUserHomePage(ModelMap modelMap, Principal principal){
       //String name = principal.getName();
       //modelMap.addAttribute("name", name);
       return "user/home";  
    }

    @RequestMapping(value = "/logout", method = RequestMethod.GET)
    public String showLogoutPage(ModelMap modelMap){
       return "user/index";
    }
    }

数据库设计/代码:

    create table USER(ID BIGINT NOT NULL AUTO_INCREMENT, USERNAME varchar(20) NOT NULL UNIQUE, PASSWORD varchar(20) NOT NULL, FIRSTNAME varchar(25) NOT NULL, LASTNAME varchar(25) NOT NULL, UPDATED_ON varchar(25) NOT NULL, PRIMARY KEY (ID));

登录表格代码:

    <form class="form-horizontal" method="POST"
                action="<c:url value='/j_spring_security_check' />" id="loginForm">
                <div class="span4"></div>
                <div class="span5" style="background-color: #FBFBFC; border: solid 1px #CCC;padding: 30px 5px 30px 5px;">
                <div class="span1"></div>
                <fieldset>                      
                    <legend>Login Here</legend>
                    <div class="control-group">
                        <label for="username">Username</label>
                        <input type="text" name="j_username" id="username"
                                placeholder="Username"
                                title="Please enter your username" data-placement="right" />
                        <div class="clear"></div>
                        <span id="errorSpan"></span>
                    </div>

                    <div class="control-group">
                        <label for="password">Password</label>
                        <input type="password" name="j_password" id="password"
                                placeholder="Password" title="Please enter your password"
                                data-placement="right" />
                        <div class="clear"></div>
                        <span id="errorSpan"></span>
                    </div>


                    <div class="control-group">
                        <label>&nbsp;</label>
                            <input type="submit" class="btn btn-primary" value="Sign In" />
                    </div>
                    <div class="span1"></div>
                </fieldset>
                </div>
                <div class="span3"></div>
            </form>
4

3 回答 3

2

你的模式是错误的

pattern="/user/**"

应该是对的。
为了更好地理解图案中的星星:

  • xx/**意味着下面的完整树结构xx是安全的。
  • xx/*意味着只有其中的数据xx是安全的。
  • xx/*.rar在这种情况下*是文件的通配符,因此所有 .rar 文件都是安全的。

希望这可以帮助。

于 2012-12-23T11:42:30.140 回答
2

有两个问题:

首先,您必须更正地址模式:例如user/**

其次,如果您对受保护的资源进行了安静的调用,则很可能会发生这种情况。
检查萤火虫。您会看到一个302 redirect作为对您请求的响应。在这种情况下,您最好不要使用重定向,而是给出403 access denied响应并在您的 ajax 框架中手动处理它。

于 2012-12-23T12:27:09.810 回答
1

您还应该像对 users-by-username-query所做的那样,为authority-by-username-query指定一个查询

<jdbc-user-service data-source-ref="dataSource"
     users-by-username-query="select USERNAME, PASSWORD from USER where USERNAME = ?"
     authorities-by-username-query="ROLE_USER"
/>

例如

authorities-by-username-query="select u.username, ur.authority from user u, user_roles ur 
              where u.user_id = ur.user_id and u.username =?"

还可以考虑在您的配置中添加密码编码器,这样您的数据库中就没有纯文本密码

于 2012-12-23T12:15:09.577 回答