我嘗試了Tiles2模板在我的新的春天MVC應用程序中的所有網頁。 我創建了一個名爲Login.jsp的登錄頁面,它使用j_spring_security_check進行表單發佈操作。Tiles2彈簧安全,登錄錯誤消息沒有得到顯示
我成功地使用AuthenticationProvider登錄並驗證身份驗證。 但是,我堅持如何顯示錯誤消息的情況下,不成功的登錄嘗試。我創建了一個控制器來檢查是否有錯誤的查詢參數。登錄控制器永遠不會在正常的登錄嘗試或不成功的登錄嘗試中被調用。控制器類位於上下文中:ApplicationContext.xml中的組件掃描。
我試着在控制器上添加斷點,但它永遠不會被調用。 我用於訪問登錄頁面的端點。
在我看來,像Controller沒有被使用和login.jsp顯示,我怎樣才能確保它通過控制器?
你能告訴我我在這裏錯過了什麼嗎?我看了很多示例配置和答案,我找不到我做錯了什麼。
春季版本:3.1.0 瓷磚版本:2.2.2
控制器:
@Controller
public class LoginController {
@RequestMapping(value = "/login.company", 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 && !StringUtils.isBlank(error)){
model.addObject("error", "Incorrect username or password.");
}else if(logout !=null && !StringUtils.isBlank(logout)){
model.addObject("msg", "You have been logged out.");
}
model.setViewName("login");
return model;
}
}
Tiles.xml
<tiles-definitions>
<definition name="mainLayout" template="/WEB-INF/tiles/mainLayout.jsp">
<put-attribute name="includes" value="/WEB-INF/tiles/includes.jsp"/>
<put-attribute name="content" value="/WEB-INF/tiles/blank.jsp"/>
<put-attribute name="footer" value="/WEB-INF/tiles/footer.jsp"/>
</definition>
<definition name="*" extends="mainLayout">
<put-attribute name="content" value="/WEB-INF/views/{1}.jsp"/>
</definition>
</tiles-definitions>
的security.xml
<beans:bean id="loginService" class="com.LoginService"/>
<authentication-manager>
<authentication-provider ref="loginService">
</authentication-provider>
</authentication-manager>
<http use-expressions="true">
<intercept-url pattern="/resources/**" access="permitAll"/>
<intercept-url pattern="/login.company" access="permitAll"/>
<intercept-url pattern="/**" access="hasRole('ROLE_USER')" />
<form-login login-page="/login.company" authentication-failure-url="/login.company?error=1" />
<logout logout-success-url="/login.company?logout" />
</http>
applicationContext.xml的
<bean id="viewResolver" class="org.springframework.web.servlet.view.UrlBasedViewResolver">
<property name="viewClass" value="org.springframework.web.servlet.view.tiles2.TilesView" />
</bean>
<bean id="tilesConfigurer" class="org.springframework.web.servlet.view.tiles2.TilesConfigurer"/>
的web.xml
<?xml version="1.0" encoding="UTF-8"?>
<web-app metadata-complete="false" 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/applicationContext.xml
/WEB-INF/spring/security.xml
</param-value>
</context-param>
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<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>
<servlet>
<servlet-name>mvc-dispatcher</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>
/WEB-INF/applicationContext.xml
</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>mvc-dispatcher</servlet-name>
<url-pattern>*.company</url-pattern>
</servlet-mapping>
<servlet-mapping>
<servlet-name>mvc-dispatcher</servlet-name>
<url-pattern>/index.html</url-pattern>
</servlet-mapping>
<welcome-file-list>
<welcome-file>index.html</welcome-file>
</welcome-file-list>
</web-app>