Hey I am having a bit of a mess with my springsecurity based login
I'm keep getting the error "bad credentials"
Here's my user table:
![Usertable][1]
Here's my dataSource from the applicationContext:
<!-- 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>
and my 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>
When i try to login with: admin and init01
it gives me the error bad credentials... =(
ANY suggestions are appreciated!!!
The
password-encoder
reference in yourauthentication-provider
is commented out. You need a password encoder if you are using hashed passwords (as you should be). Also check this answer, particularly point 2 about writing a test to make sure the password encoder you are using matches what you have stored in the database.You might also want to check this answer on using bcrypt as a more secure alternative to plain SHA hashes.
Your passwords are getting hashed. If you add a password 'init01', it actually means the hash of the original password is 'init01' because Spring hashes the supplied password and matches with the one you enter. So SHA('init01') is something other than 'init01'