스프링 보안이 어떻게 작동 하는지를 배우려고합니다. 따라서 샘플 프로젝트를 다운로드 한 다음 그 솔루션을 프로젝트에 구현하려고했습니다. 하지만 로그인을 시도하면 404
오류가 발생하고 주소 표시 줄에 http://localhost:8080/fit/j_spring_security_check
이 있습니다. 나는 비슷한 질문을 여기에서 보려고했지만, 나는 그것을 실현할 수 없었다. 어떻게 그것을 나의 프로젝트에 적용 할 수 있는가. 더 많은 경험을 가진 누군가 나를 도울 수 있다면 정말 감사 할 것입니다.스프링 3 보안 j_spring_security_check
내 응용 프로그램 구조는 다음과 같습니다
applicationContext.xml :
<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:context="http://www.springframework.org/schema/context"
xmlns:security="http://www.springframework.org/schema/security"
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
http://www.springframework.org/schema/security http://www.springframework.org/schema/security/spring-security-3.1.xsd">
<context:annotation-config/>
<context:component-scan base-package="cz.cvut.fit"/>
<import resource="classpath:applicationContext-security.xml"/>
</beans>
의 ApplicationContext-web.xml의 :
<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:mvc="http://www.springframework.org/schema/mvc"
xmlns:context="http://www.springframework.org/schema/context"
xmlns:security="http://www.springframework.org/schema/security"
xsi:schemaLocation="
http://www.springframework.org/schema/mvc http://www.springframework.org/schema/mvc/spring-mvc-3.0.xsd
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
http://www.springframework.org/schema/security http://www.springframework.org/schema/security/spring-security-3.1.xsd">
<context:annotation-config/>
<context:component-scan base-package="cz.cvut.fit" />
<mvc:annotation-driven />
<security:global-method-security jsr250-annotations="enabled"
proxy-target-class="true"/>
</beans>
의 ApplicationContext-security.xml :
<beans xmlns:security="http://www.springframework.org/schema/security"
xmlns="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">
<security:http pattern="/css/**" security="none"/>
<security:http pattern="/views/login.jsp*" security="none"/>
<security:http pattern="/views/denied.jsp" security="none"/>
<security:http auto-config="true" access-denied-page="/denied.jsp" servlet-api-provision="false">
<security:intercept-url pattern="/views/login.jsp*" access="IS_AUTHENTICATED_ANONYMOUSLY"/>
<security:intercept-url pattern="/views/edit/**" access="ROLE_EDIT"/>
<security:intercept-url pattern="/views/admin/**" access="ROLE_ADMIN"/>
<security:intercept-url pattern="/**" access="ROLE_USER"/>
<security:form-login login-page="/views/login.jsp" authentication-failure-url="/denied.jsp"
default-target-url="/home.jsp"/>
<security:logout/>
</security:http>
<security:authentication-manager>
<security:authentication-provider>
<security:user-service>
<security:user name="adam" password="adampassword" authorities="ROLE_USER"/>
<security:user name="jane" password="janepassword" authorities="ROLE_USER, ROLE_ADMIN"/>
<security:user name="sue" password="suepassword" authorities="ROLE_USER, ROLE_EDIT"/>
</security:user-service>
</security:authentication-provider>
</security:authentication-manager>
</beans>
는'j_spring_security_check' 실제 인증이 만든 서블릿이며,이 서블릿에 로그인 양식의 조치를 매핑해야합니다. 로그인 페이지에서 다음과 같이 하시겠습니까? -
'? – Lionweb.xml을 보여주십시오./j_spring_security_check URL은 springSecurityFilterChain 필터로 처리해야합니다. –
예, 저는 ...하지만 다음에해야 할 일을 잘 이해하기 위해 단서가 없습니다. : -/ – Dworza