Spring Security / Hibernate：即使凭证正确，也有错误的凭证吗？

Question

嘿，我的基于springsecurity的登录有点混乱

我不断收到错误“凭据错误”

这是我的用户表：

！[Usertable] [1]

这是来自applicationContext的dataSource：

<!-- 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>

当我尝试使用admin和init01登录时

它给了我错误的错误凭据… =（

任何建议表示赞赏！

Answer 1

password-encoder您的参考authentication-provider已被注释掉。如果您使用的是哈希密码，则需要一个密码编码器（应该如此）。还要检查此答案，特别是关于编写测试的第2点，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匹配。

您可能还想检查有关使用bcrypt作为普通SHA哈希的更安全替代方法的答案。