3

嘿，我的基于 springsecurity 的登录有点乱

我不断收到错误“凭据错误”

这是我的用户表：

![用户表][1]

这是来自applicationContext的我的数据源：

<!-- database driver/location -->
<bean id="dataSource" class="org.springframework.jdbc.datasource.DriverManagerDataSource">
    <property name="driverClassName" value="com.mysql.jdbc.Driver" />
    <property name="url" value="jdbc:mysql://localhost:3306/ams" />
    <property name="username" value="root" />
    <property name="password" value="root" />
</bean>

和我的securityContext：

<?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"
    xmlns:tx="http://www.springframework.org/schema/tx"
    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-3.0.xsd">

    <!-- <security:http auto-config="true" access-decision-manager-ref="accessDecisionManager"> -->
    <security:http auto-config="true">
        <security:intercept-url pattern="/login/login.do" access="IS_AUTHENTICATED_ANONYMOUSLY" />
        <security:intercept-url pattern="/login/doLogin.do" access="IS_AUTHENTICATED_ANONYMOUSLY" />
        <security:intercept-url pattern="/lib/**" access="IS_AUTHENTICATED_ANONYMOUSLY" />
        <security:intercept-url pattern="/css/**" access="IS_AUTHENTICATED_ANONYMOUSLY" />
        <security:intercept-url pattern="/images/**" access="IS_AUTHENTICATED_ANONYMOUSLY" />
        <security:intercept-url pattern="/resources/**" access="IS_AUTHENTICATED_ANONYMOUSLY" />
        <security:intercept-url pattern="/**" access="IS_AUTHENTICATED_REMEMBERED" />
        <security:form-login login-page="/login/login.do" authentication-failure-url="/login/login.do?login_error=true" default-target-url="/test/showTest.do"/>
        <security:logout logout-success-url="/login/login.do" invalidate-session="true" />
        <security:remember-me key="rememberMe"/>
    </security:http>    


    <security:authentication-manager>
        <security:authentication-provider>
            <security:jdbc-user-service data-source-ref="dataSource" 
            users-by-username-query="select USERNAME as username, PASSWORD as password, DELETED as deleted from ams.user where USERNAME=?"
            authorities-by-username-query="
                select distinct user.USERNAME as username, permission.NAME as authority 
            from scu.user, scu.user_role, scu.role, scu.role_permission, scu.permission
            where user.ID=user_role.USER_ID AND user_role.ROLE_ID=role_permission.ROLE_ID AND role_permission.PERMISSION_ID=permission.ID AND user.USERNAME=?"/>
            <!-- security:password-encoder ref="passwordEncoder" /> -->
        </security:authentication-provider>
    </security:authentication-manager>

    <bean id="passwordEncoder"
        class="org.springframework.security.authentication.encoding.ShaPasswordEncoder">
        <constructor-arg value="256" />
    </bean>
</beans>

当我尝试登录时：admin 和 init01

它给了我错误的错误凭据... =（

任何建议表示赞赏！

4

2 回答 2

4

您的password-encoder引用authentication-provider已被注释掉。如果您使用散列密码（应该如此），则需要密码编码器。还要检查这个答案，特别是关于编写测试以确保您使用的密码编码器与您存储在数据库中的密码编码器相匹配的第 2 点。

您可能还想查看有关使用 bcrypt作为普通 SHA 哈希的更安全替代方案的答案。

于 2013-02-20T19:36:12.967 回答
0

您的密码正在被散列。如果您添加密码“init01”，实际上意味着原始密码的哈希是“init01”，因为 Spring 对提供的密码进行哈希处理并与您输入的密码匹配。所以 SHA('init01') 不是 'init01'

于 2013-02-20T18:53:09.180 回答