Java 为什么 j_spring_security_check 404?

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时间:2020-08-13 02:55:44  来源:igfitidea点击:

Why j_spring_security_check 404?

javaspringspring-security

提问by xwhyz

I'm trying to implement form based authentication with Spring security. Redirection works fine: my main page works correctly and for http://localhost:8080/master/adminI'm redirected to login page (http://localhost:8080/master/login/):

我正在尝试使用 Spring 安全性实现基于表单的身份验证。重定向工作正常:我的主页工作正常,并且 http://localhost:8080/master/admin我被重定向到登录页面 ( http://localhost:8080/master/login/):

    <form action="j_spring_security_check" method="POST">
        <label for="username">User Name:</label>
        <input id="username" name="j_username" type="text"/>
        <label for="password">Password:</label>
        <input id="password" name="j_password" type="password"/>
        <input type="submit" value="Log In"/>
      </form>

but when I submit I get 404 on address: `http://localhost:8080/master/login/j_spring_security_check`

Here's my configuration web.xml:

这是我的配置 web.xml:

    <context-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>/WEB-INF/spring/root-context.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>

    <!-- 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>

And my root-context.xml (i made redirection to google so that I know page exists in context)

还有我的 root-context.xml(我重定向到 google,以便我知道页面存在于上下文中)

<sec:http auto-config="true">
    <sec:intercept-url pattern="/admin/**" access="ROLE_USER" />
    <sec:form-login login-page="/login/"
        authentication-failure-url="http://www.google.com" default-target-url="http://www.google.com" />
    <sec:logout logout-success-url="/logout" />
</sec:http>

<sec:authentication-manager>
    <sec:authentication-provider>
        <sec:user-service>
            <sec:user name="test" password="test" authorities="ROLE_USER, ROLE_ADMIN" />
            <sec:user name="testuser" password="testuserpassword"
                authorities="ROLE_USER" />
        </sec:user-service>
    </sec:authentication-provider>
</sec:authentication-manager> 

I've lost quite a lot of time already, trying different combinations but no luck. Any help is appreciated!

我已经失去了很多时间,尝试不同的组合但没有运气。任何帮助表示赞赏!

采纳答案by Joshua Moore

I think it may be the fact your security form should point to

我认为这可能是您的安全表格应该指出的事实

/master/j_spring_security_check