Question

为了学习目的，我简单的Spring-hibernate应用程序。它具有一对多数据库关联，您可以在其中添加仓库，并且可以为每个仓库添加项目。现在我想添加spring-security进行身份验证，但是当我尝试访问该站点时，我收到了HTTP ERROR 404.

这是web.xml中的spring-security par：

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

这是spring-security.xml：

<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 use-expressions="true">
        <intercept-url pattern="/record/**" access="permitAll" />
        <form-login />
    </http>

    <authentication-manager>
        <authentication-provider>
            <user-service>
                <user name="admin" password="password" authorities="supervisor, teller, user" />
                <user name="dianne" password="emu" authorities="teller, user" />
                <user name="scott" password="wombat" authorities="user" />
                <user name="peter" password="opal" authorities="user" />
            </user-service>
        </authentication-provider>
    </authentication-manager>
</beans:beans>

如果我将access="permitAll"更改为`access =“hasRole（'supervisor'）”当我尝试访问该应用程序时，系统会要求我输入用户名和密码，但是当我输入它们时，我也会得到404错误。没有spring-security，当我访问“http：// localhost：8080 / testapp / record / list”时应用程序正常工作...... 我使用这个网站来了解春季安全 - http://static.springsource.org/spring-security/site/tutorial.html

这是我的MainController：

@Controller
@RequestMapping("/record")
public class MainController {

    protected static Logger logger = Logger.getLogger("controller");

    @Resource(name="warehouseService")
     private WarehouseService warehouseService;

     @Resource(name="itemService")
     private ItemService itemService;

        @RequestMapping(value = "/list", method = RequestMethod.GET)
        public String getRecords(Model model) {
         logger.debug("Received request to show records page");

         // Retrieve all persons
         List<Warehouse> warehouses = warehouseService.getAll();

         // Prepare model object
         List<WarehouseDTO> warehousesDTO = new ArrayList<WarehouseDTO>();

         for (Warehouse warehouse: warehouses) {
          // Create new data transfer object
             WarehouseDTO dto = new WarehouseDTO();

       dto.setId(warehouse.getId());
       dto.setName(warehouse.getName());
       dto.setAddress(warehouse.getAddress());
       dto.setManager(warehouse.getManager());
       dto.setItems(itemService.getAll(warehouse.getId()));

       warehousesDTO.add(dto);
         }

         model.addAttribute("warehouses", warehousesDTO);

      return "records";
     }

有什么想法吗？

Answer 1

尝试在form-login标记中指定default-target-url。我猜测你的身份验证和Spring安全性是默认将请求路由到应用程序的根目录，而不是映射导致404。

<form-login default-target-url="/record/list"/>

如果失败尝试（不确定如何映射视图）：

<form-login default-target-url="/record/list/"/>

Answer 2

这可能听起来很愚蠢，但您是否尝试将pattern =“/ record / ”更改为pattern =“/ testapp / record / ”？

通常安全访问问题会产生403错误（访问被拒绝），而不是404（找不到资源）。

spring security http 404错误

2 个答案: