How to config spring security - forms

Question

I'm trying learning how to protect my web Application using Spring security I have some problems,
this is my web.xml
<welcome-file-list>
<welcome-file>loginPage.html</welcome-file>
<welcome-file>loginPage.htm</welcome-file>
<welcome-file>loginPage.jsp</welcome-file>
</welcome-file-list>
<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>*.html</url-pattern>
</servlet-mapping>
<!-- Spring context files to be loaded -->
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>
/WEB-INF/dispatcher-servlet.xml,
/WEB-INF/spring-security.xml
</param-value>
</context-param>
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<!-- filter declaration for 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>
</web-app>
and this my 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-3.1.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="/*" access="ROLE_USER" />
<form-login login-processing-url="/login" login-page="/loginPage"
username-parameter="username" password-parameter="password"
default-target-url="/secured/mypage" authentication-failure-url="/loginPage?auth=fail" />
<logout logout-url="/logout" logout-success-url="/logoutPage" />
</http>
<authentication-manager>
<authentication-provider>
<user-service>
<user name="srccodes" password="password" authorities="ROLE_USER" />
</user-service>
</authentication-provider>
</authentication-manager>
and finally my servlet-dispatcher.xml
<bean id="viewResolver" class="org.springframework.web.servlet.view.InternalResourceViewResolver">
<property name="prefix" value="/WEB-INF/pages/" />
<property name="suffix" value=".jsp" />
</bean>
My application run normally and I have my login page, when I type the right login et password I have error 404 not found!! with url
localhost:8084/LoginFormExampleSpring/secured/mypage
this is my form
<form action="${pageContext.request.contextPath}/login" method="post">
<table>
<tr>
<td>Username:</td>
<td><input type='text' name='username' /></td>
</tr>
<tr>
<td>Password:</td>
<td><input type='password' name='password'></td>
</tr>
<tr>
<td colspan='2'><input name="submit" type="submit" value="Submit"></td>
</tr>
</table>
</form>
anyone can help me please?

Answer 1

It sounds like you need to add a mapping for /secured/mypage To do so:
Create a file at /WEB-INF/pages/secured/mypage.jsp
Add a mvc:view-controller to dispatcher-servlet.xml
For example:
dispatcher-servlet.xml
<mvc:view-controller path="/secured/mypage"/>