出于学习目的，我简单的 Spring-hibernate 应用程序。它具有一对多的数据库关联，您可以在其中添加一个仓库，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，我得到了 HTTP ERROR 404。
这是 web.xml 中的 spring-security 标准：
<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>
如果我将其更改access="permitAll"为 `access="hasRole('supervisor')" 当我尝试访问应用程序时，系统会要求我输入用户名和密码，但当我输入它们时，我也会收到 404 错误。没有spring-security，当我访问“http://localhost:8080/testapp/record/list”时，应用程序可以正常工作......我使用这个网站来了解spring security - http://static.springsource.org/spring -security/site/tutorial.html
这是我的主控制器：
@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";
}
有任何想法吗？