Question

我尝试使用Spring Security UserDetailService身份验证使用Email而不是UserName登录，但我无法登录并获取

org.springframework.security.authentication.InternalAuthenticationServiceException。

当我调试代码时，调试器不会追踪

用户user = userDao.findByUserName（useremail）;来自UserServiceImpl.java并发生错误。

可能在某个地方需要更多步骤。配置和Java代码和错误日志如下： -

的applicationContext.xml

<security:http auto-config="true">
        <security:intercept-url pattern="/admin/**" access="ROLE_ADMIN" />
        <security:intercept-url pattern="/customer/**" access="ROLE_USER" />
        <security:form-login login-page="/login"
            default-target-url="/product/productList/all" authentication-failure-url="/login?error"
            username-parameter="username" password-parameter="password" />
        <security:logout logout-success-url="/login?logout" />
    </security:http>

    <security:authentication-manager>
        <security:authentication-provider user-service-ref="userServiceImpl">
                <security:password-encoder ref="bcryptEncoder" />
        </security:authentication-provider>
    </security:authentication-manager>

    <bean id="userServiceImpl" class="com.emusicstore.service.impl.UserServiceImpl" />

    <bean id="bcryptEncoder" class="org.springframework.security.crypto.bcrypt.BCryptPasswordEncoder" >
    <constructor-arg name="strength" value="11" />
    </bean>

的Login.jsp

<form name="loginForm" action="<c:url value="/j_spring_security_check" />" method="post">

                <c:if test="${not empty error}">
                    <div class="error" style="color: #ff0000;">${error}</div>
                </c:if>

                <div class="form-group">
                    <label for="username">Email : </label>
                    <input type="text" id="username" name="username" class="form-control" />
                </div>
                <div class="form-group">
                    <label for="password">Passwrod:</label>
                    <input type="password" id="password" name="password" class="form-control" />
                </div>
                <input type="submit" value="Submit" class="btn btn-default">
                <input type="hidden" name="${_csrf.parameterName}" value="${_csrf.token}" />
            </form>

UserServiceImpl.java

import java.util.ArrayList;
import java.util.Collection;
import java.util.List;

import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.beans.factory.annotation.Qualifier;
import org.springframework.security.authentication.jaas.AuthorityGranter;
import org.springframework.security.core.GrantedAuthority;
import org.springframework.security.core.authority.SimpleGrantedAuthority;
import org.springframework.security.core.userdetails.UserDetails;
import org.springframework.security.core.userdetails.UserDetailsService;
import org.springframework.security.core.userdetails.UsernameNotFoundException;
import org.springframework.stereotype.Component;
import org.springframework.stereotype.Service;
import org.springframework.transaction.annotation.Transactional;

import com.emusicstore.dao.UserDao;
import com.emusicstore.model.Users;
import com.emusicstore.service.UserService;
@Service
@Transactional
@Qualifier("userServiceImpl")
public class UserServiceImpl implements UserDetailsService{

    @Autowired
    private UserDao userDao;

    @Override
    public UserDetails loadUserByUsername(String useremail)
            throws UsernameNotFoundException {
        Users user = userDao.findByUserEmail(useremail);
        if(user == null){
            throw new UsernameNotFoundException("UserName or Password Invalid.");
        }
        return new org.springframework.security.core.userdetails.User(user.getUsername(), user.getPassword(), user.getEnabled(), true, true, true, getGrantedAuthorities(userDao.getUserRole(user.getUsersId())));
    }

    private final List<GrantedAuthority> getGrantedAuthorities(final List<String> roleList) {
        final List<GrantedAuthority> authorities = new ArrayList<GrantedAuthority>();
        for (final String role : roleList) {
            authorities.add(new SimpleGrantedAuthority(role));
        }
        return authorities;
}
}

控制台错误如下：

错误： org.springframework.security.web.authentication.UsernamePasswordAuthenticationFilter - 尝试验证用户时发生内部错误。 org.springframework.security.authentication.InternalAuthenticationServiceException 在 org.springframework.security.authentication.dao.DaoAuthenticationProvider.retrieveUser（DaoAuthenticationProvider.java:110）在 org.springframework.security.authentication.dao.AbstractUserDetailsAuthenticationProvider.authenticate（AbstractUserDetailsAuthenticationProvider.java:132）在 org.springframework.security.authentication.ProviderManager.authenticate（ProviderManager.java:156）在 org.springframework.security.authentication.ProviderManager.authenticate（ProviderManager.java:177）在 org.springframework.security.web.authentication.UsernamePasswordAuthenticationFilter.attemptAuthentication（UsernamePasswordAuthenticationFilter.java:92）在 org.springframework.security.web.authentication.AbstractAuthenticationProcessingFilter.doFilter（AbstractAuthenticationProcessingFilter.java:211）在 org.springframework.security.web.FilterChainProxy $ VirtualFilterChain.doFilter（FilterChainProxy.java:342）在 org.springframework.security.web.authentication.logout.LogoutFilter.doFilter（LogoutFilter.java:110）在 org.springframework.security.web.FilterChainProxy $ VirtualFilterChain.doFilter（FilterChainProxy.java:342）在 org.springframework.security.web.context.request.async.WebAsyncManagerIntegrationFilter.doFilterInternal（WebAsyncManagerIntegrationFilter.java:50）在 org.springframework.web.filter.OncePerRequestFilter.doFilter（OncePerRequestFilter.java:107）在 org.springframework.security.web.FilterChainProxy $ VirtualFilterChain.doFilter（FilterChainProxy.java:342）在 org.springframework.security.web.context.SecurityContextPersistenceFilter.doFilter（SecurityContextPersistenceFilter.java:87）在 org.springframework.security.web.FilterChainProxy $ VirtualFilterChain.doFilter（FilterChainProxy.java:342）在 org.springframework.security.web.FilterChainProxy.doFilterInternal（FilterChainProxy.java:192）在 org.springframework.security.web.FilterChainProxy.doFilter（FilterChainProxy.java:160）在 org.springframework.web.filter.DelegatingFilterProxy.invokeDelegate（DelegatingFilterProxy.java:346）在 org.springframework.web.filter.DelegatingFilterProxy.doFilter（DelegatingFilterProxy.java:262）在 org.apache.catalina.core.ApplicationFilterChain.internalDoFilter（ApplicationFilterChain.java:239）在 org.apache.catalina.core.ApplicationFilterChain.doFilter（ApplicationFilterChain.java:206）在 org.apache.catalina.core.StandardWrapperValve.invoke（StandardWrapperValve.java:212）在 org.apache.catalina.core.StandardContextValve.invoke（StandardContextValve.java:106）在 org.apache.catalina.authenticator.AuthenticatorBase.invoke（AuthenticatorBase.java:502）在 org.apache.catalina.core.StandardHostValve.invoke（StandardHostValve.java:141）在 org.apache.catalina.valves.ErrorReportValve.invoke（ErrorReportValve.java:79）在 org.apache.catalina.valves.AbstractAccessLogValve.invoke（AbstractAccessLogValve.java:616）在 org.apache.catalina.core.StandardEngineValve.invoke（StandardEngineValve.java:88）在 org.apache.catalina.connector.CoyoteAdapter.service（CoyoteAdapter.java:518）在 org.apache.coyote.http11.AbstractHttp11Processor.process（AbstractHttp11Processor.java:1096）在 org.apache.coyote.AbstractProtocol $ AbstractConnectionHandler.process（AbstractProtocol.java:674）在 org.apache.tomcat.util.net.NioEndpoint $ SocketProcessor.doRun（NioEndpoint.java:1500）在 org.apache.tomcat.util.net.NioEndpoint $ SocketProcessor.run（NioEndpoint.java:1456） at java.util.concurrent.ThreadPoolExecutor.runWorker（Unknown Source） at java.util.concurrent.ThreadPoolExecutor $ Worker.run（Unknown Source）在 org.apache.tomcat.util.threads.TaskThread $ WrappingRunnable.run（TaskThread.java:61）在java.lang.Thread.run（未知来源）引起： java.lang.NullPointerException at com.emusicstore.service.impl.UserServiceImpl.loadUserByUsername（UserServiceImpl.java:33）在 org.springframework.security.authentication.dao.DaoAuthenticationProvider.retrieveUser（DaoAuthenticationProvider.java:102） ......还有35个

请帮助我克服上述错误。

非常感谢。

Answer 1

我相信你的自定义代码中有一个简单的 NullPointerException 。将断点放在33位的UserServiceImpl.java中并检查。

引起：java.lang.NullPointerException at com.emusicstore.service.impl.UserServiceImpl.loadUserByUsername（UserServiceImpl.java:33）

Answer 2

 Users user = userDao.findByUserEmail(useremail);

高于userDao.findByUserEmail(useremail);会返回空值。

Answer 3

对于我来说，this link有所帮助。这是个抢购的问题。

使用SpringSecurity中的UserDetailService通过电子邮件登录身份验证

3 个答案: