j_spring_security_check HTTP状态404（自定义登录） [英] j_spring_security_check HTTP Status 404 (Custom Login)
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问题描述
我使用Spring 4 + Hibernate 4.我检查了每篇文章。但我找不到问题所在：
Web.xml
I am using Spring 4 + Hibernate 4. I checked every articles. but I could not find out what is the problem:
Web.xml
<?xml version="1.0" encoding="UTF-8"?>
<web-app 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">
<!-- The definition of the Root Spring Container shared by all Servlets
and Filters -->
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/spring/root-context.xml, /WEB-INF/spring/appSecurity/spring-security.xml</param-value>
</context-param>
<!-- Creates the Spring Container shared by all Servlets and Filters -->
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<!-- Processes application requests -->
<servlet>
<servlet-name>appServlet</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/spring/appServlet/servlet-context.xml</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>appServlet</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
<!-- 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>
spring-security.xml
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.xsd">
<!-- enable use-expressions -->
<http auto-config="true" use-expressions="true">
<intercept-url pattern="/panel/**"
access="hasAnyRole('ROLE_USER', 'ROLE_ADMIN')" />
<intercept-url pattern="/admin/**" access="hasRole('ROLE_ADMIN')" />
<!-- access denied page -->
<access-denied-handler error-page="/403" />
<form-login login-page="/login" default-target-url="/"
authentication-failure-url="/login?error" username-parameter="username"
password-parameter="password" />
<logout logout-success-url="/login?logout" />
<!-- enable csrf protection -->
<csrf />
</http>
<authentication-manager>
<authentication-provider>
<jdbc-user-service data-source-ref="dataSource"
users-by-username-query="select username,password,enabled from users where username=?"
authorities-by-username-query="select u1.username, u2.role from users u1, user_roles u2 where u1.role_id = u2.role_id and u1.username =?" />
</authentication-provider>
</authentication-manager>
</beans:beans>
login.jsp
login.jsp
<%@ page language="java" contentType="text/html; charset=ISO-8859-1"
pageEncoding="ISO-8859-1"%>
<%@ taglib uri="http://java.sun.com/jsp/jstl/core" prefix="c"%>
<!DOCTYPE html>
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1">
<title>Login</title>
</head>
<body>
<form action="<c:url value='/j_spring_security_check' />" method='POST'>
<table>
<tr>
<td>User:</td>
<td><input type='text' name='username' value=''></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>
<input type="hidden" name="${_csrf.parameterName}"
value="${_csrf.token}" />
</form>
</body>
</html>
当我想登录的时候，url是 localhost：8080 / App / j_spring_security_check 和我得到404错误。
有人可以解释是什么问题？
When I want to login the url is localhost:8080/App/j_spring_security_checkand I get 404 error.
Could somebody explain that what is the problem?
推荐答案
在 spring-security.xml 中进行以下更改。提及 login-processing-url =/ j_spring_security_check
Make the below changes in your spring-security.xml. Mention the login-processing-url="/j_spring_security_check"
<form-login login-processing-url="/j_spring_security_check" login-page="/login" default-target-url="/"
authentication-failure-url="/login?error" username-parameter="username"
password-parameter="password" />
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