Nicht bekommen Frühling Sicherheit automatisch generierte Login-Seite

Question

Hallo, ich bin neu in Spring Security Ich mache ein einfaches Sppring Security Program, aber nicht bekommen Frühling Sicherheit automatisch generierte Login-Seite.
Das ist mein Controller-

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

@Controller 
public class HelloWorldController { 

    String message = "Welcome to Spring MVC!"; 

    @RequestMapping("/hello") 
    public ModelAndView showMessage(
      @RequestParam(value = "name", required = false, defaultValue = "World") String name) { 
     System.out.println("in controller"); 

     ModelAndView mv = new ModelAndView("helloworld"); 
     mv.addObject("message", message); 
     mv.addObject("name", name); 
     return mv; 
    } 

    @RequestMapping("/") 
    public ModelAndView showIndex() { 
     System.out.println("in index controller"); 

     ModelAndView mv = new ModelAndView("index"); 
     return mv; 
    } 

    @RequestMapping("/admin") 
    public ModelAndView showMessageAdmin() { 
     System.out.println("in controller"); 

     ModelAndView mv = new ModelAndView("helloworld"); 
     return mv; 
    } 
} 

Dieser

mein Web.xml ist

<?xml version="1.0" encoding="UTF-8"?> 
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://java.sun.com/xml/ns/javaee" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd" id="WebApp_ID" version="3.0"> 


    <display-name>Archetype Created Web Application</display-name> 

    <servlet> 
     <servlet-name>dispatcher</servlet-name> 
     <servlet-class> 
      org.springframework.web.servlet.DispatcherServlet 
     </servlet-class> 
     <load-on-startup>1</load-on-startup> 
    </servlet> 
    <welcome-file-list> 
    <welcome-file>index.jsp</welcome-file> 

    </welcome-file-list> 
    <servlet-mapping> 
     <servlet-name>dispatcher</servlet-name> 
     <url-pattern>/</url-pattern> 
    </servlet-mapping> 

    <context-param> 
     <param-name>contextConfigLocation</param-name> 
     <param-value> 
     /WEB-INF/dispatcher-servlet.xml 
     /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> 

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

</web-app> 

Dies ist meine Klasse XML Spring Security

<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-4.0.xsd 
http://www.springframework.org/schema/security 
http://www.springframework.org/schema/security/spring-security-4.0.xsd"> 
    <http auto-config="true"> 
     <intercept-url pattern="/admin" access="ROLE_ADMIN" /> 
    </http> 
    <authentication-manager> 

     <authentication-provider> 
     <user-service> 
     <user name="Hello" password="Pass" authorities="ROLE_USER" /> 
     </user-service> 
     </authentication-provider> 
    </authentication-manager> 
</beans:beans> 

Das ist mein Dispatcher servlet.xml ist

<beans xmlns="http://www.springframework.org/schema/beans" 
    xmlns:context="http://www.springframework.org/schema/context" 
    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/context 
http://www.springframework.org/schema/context/spring-context-3.0.xsd"> 

    <context:component-scan base-package="com.ge.web.Controller" /> 

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

for/admin Ich möchte die Spring Security-Anmeldeseite zur Authentifizierung erhalten. Aber ich verstehe es nicht, es wird direkt zu sehen, wie kann ich es bekommen, wo ich falsch liege? Vielen Dank im Voraus ....

Answer 1

First Fix Sie web.xml: Sie brauchen nicht /WEB-INF/dispatcher-servlet.xml in Zusammenhang param angeben, da die Datei unter src/main/webapp/WEB-INF ist, die für Dispatcher Servlet Standardverzeichnis ist.

<?xml version="1.0" encoding="UTF-8"?> 
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://java.sun.com/xml/ns/javaee" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd" id="WebApp_ID" version="3.0"> 

    <display-name>Archetype Created Web Application</display-name> 

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

    <welcome-file-list> 
     <welcome-file>index.jsp</welcome-file> 
    </welcome-file-list> 

    <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> 

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

</web-app> 

Dann in Ihrer XML-pring Sicherheit: Hier fehlten Sie <form-login /><logout /> die Standard-Login-Formular erhalten erforderlich ist.

<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-4.0.xsd 
    http://www.springframework.org/schema/security 
    http://www.springframework.org/schema/security/spring-security-4.0.xsd"> 

    <http> 
     <intercept-url pattern="/admin" access="ROLE_ADMIN" /> 
     <form-login /> 
     <logout /> 
    </http> 

    <authentication-manager> 
     <authentication-provider> 
     <user-service> 
     <user name="Admin" password="AdminPassword" authorities="ROLE_USER, ROLE_ADMIN" /> 
     </user-service> 
     </authentication-provider> 
    </authentication-manager> 

</beans:beans>