Question

我在我的网络应用程序中使用Spring security 3.2.0。

我有两个安全角色：具有存储在数据库中的凭据的用户和具有存储在配置文件中的凭据的管理员。

在我打开记住我身份验证之前，它工作正常。这是我的security.xml配置：

<?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:security="http://www.springframework.org/schema/security"
       xsi:schemaLocation="http://www.springframework.org/schema/beans
      http://www.springframework.org/schema/beans/spring-beans-4.0.xsd
   http://www.springframework.org/schema/security
   http://www.springframework.org/schema/security/spring-security-3.2.xsd">

    <security:global-method-security secured-annotations="enabled"/>

    <security:http auto-config="true">
        <security:remember-me key="myAppKey" user-service-ref="adminSrv"/>
        <security:remember-me key="myAppKey" user-service-ref="jdbcUserSrv"/>
        <security:form-login login-page="/login" default-target-url="/"
            authentication-failure-url="/loginfailed" />
        <security:logout logout-success-url="/logout"/>
    </security:http>

    <security:authentication-manager>
        <security:authentication-provider>
            <security:jdbc-user-service data-source-ref="dataSource" id="jdbcUserSrv"
                                        users-by-username-query="/* SQL query */"
                                        authorities-by-username-query="/* Another SQL query */"
                    />
        </security:authentication-provider>
        <security:authentication-provider>
            <security:user-service id="adminSrv">
                <security:user name="admin" authorities="ROLE_ADMIN" password="passwd"/>

            </security:user-service>
        </security:authentication-provider>
    </security:authentication-manager>
</beans>

当我以管理员身份登录时，它就可以了。

当我尝试以用户身份登录时，它会在j_spring_security_check上抛出异常：

HTTP ERROR 500

Problem accessing /mercury/j_spring_security_check. Reason:

    user1
Caused by:

org.springframework.security.core.userdetails.UsernameNotFoundException: user1
    at org.springframework.security.provisioning.InMemoryUserDetailsManager.loadUserByUsername(InMemoryUserDetailsManager.java:113)
    at org.springframework.security.web.authentication.rememberme.TokenBasedRememberMeServices.onLoginSuccess(TokenBasedRememberMeServices.java:178)
    at org.springframework.security.web.authentication.rememberme.AbstractRememberMeServices.loginSuccess(AbstractRememberMeServices.java:262)
    at org.springframework.security.web.authentication.AbstractAuthenticationProcessingFilter.successfulAuthentication(AbstractAuthenticationProcessingFilter.java:324)
    at org.springframework.security.web.authentication.AbstractAuthenticationProcessingFilter.successfulAuthentication(AbstractAuthenticationProcessingFilter.java:298)
    at org.springframework.security.web.authentication.AbstractAuthenticationProcessingFilter.doFilter(AbstractAuthenticationProcessingFilter.java:235)
    at org.springframework.security.web.FilterChainProxy$VirtualFilterChain.doFilter(FilterChainProxy.java:342)
    at org.springframework.security.web.authentication.logout.LogoutFilter.doFilter(LogoutFilter.java:110)
    at org.springframework.security.web.FilterChainProxy$VirtualFilterChain.doFilter(FilterChainProxy.java:342)
    at org.springframework.security.web.context.SecurityContextPersistenceFilter.doFilter(SecurityContextPersistenceFilter.java:87)
    at org.springframework.security.web.FilterChainProxy$VirtualFilterChain.doFilter(FilterChainProxy.java:342)
    at org.springframework.security.web.FilterChainProxy.doFilterInternal(FilterChainProxy.java:192)
    at org.springframework.security.web.FilterChainProxy.doFilter(FilterChainProxy.java:160)
    at org.springframework.web.filter.DelegatingFilterProxy.invokeDelegate(DelegatingFilterProxy.java:344)
    at org.springframework.web.filter.DelegatingFilterProxy.doFilter(DelegatingFilterProxy.java:261)
    at org.mortbay.jetty.servlet.ServletHandler$CachedChain.doFilter(ServletHandler.java:1157)
    at org.mortbay.jetty.servlet.ServletHandler.handle(ServletHandler.java:388)
    at org.mortbay.jetty.security.SecurityHandler.handle(SecurityHandler.java:216)
    at org.mortbay.jetty.servlet.SessionHandler.handle(SessionHandler.java:182)
    at org.mortbay.jetty.handler.ContextHandler.handle(ContextHandler.java:765)
    at org.mortbay.jetty.webapp.WebAppContext.handle(WebAppContext.java:418)
    at org.mortbay.jetty.handler.ContextHandlerCollection.handle(ContextHandlerCollection.java:230)
    at org.mortbay.jetty.handler.HandlerCollection.handle(HandlerCollection.java:114)
    at org.mortbay.jetty.handler.HandlerWrapper.handle(HandlerWrapper.java:152)
    at org.mortbay.jetty.Server.handle(Server.java:326)
    at org.mortbay.jetty.HttpConnection.handleRequest(HttpConnection.java:542)
    at org.mortbay.jetty.HttpConnection$RequestHandler.content(HttpConnection.java:943)
    at org.mortbay.jetty.HttpParser.parseNext(HttpParser.java:756)
    at org.mortbay.jetty.HttpParser.parseAvailable(HttpParser.java:218)
    at org.mortbay.jetty.HttpConnection.handle(HttpConnection.java:404)
    at org.mortbay.io.nio.SelectChannelEndPoint.run(SelectChannelEndPoint.java:410)
    at org.mortbay.thread.QueuedThreadPool$PoolThread.run(QueuedThreadPool.java:582)

但是，如果它已成功验证，并且我打开了我的应用程序索引页面，则会显示为“user1”。

如果jdbcUserSrv的remember-me配置在adminSrv的配置之前，那么“admin”用户会出现错误。

当前配置在没有remember-me选项的情况下正常工作，但我需要启用它们。

请告诉我如何避免这些错误。

Answer 1

有两个记住我的元素只是相互重写。拥有多个UserDetailsService实例也会让您遇到麻烦。例如，我假设您可能有与您的用户相关的数据。如果内存和jdbc UserDetailsService实例中都存在名为admin的用户，那么您如何知道该数据属于哪个管理员用户？因此，最好使用单个UserDetailsService。

或者，您可以创建一个委托给多个UserDetailsService实例的UserDetailsService。例如：

public class DelegatingUserDetailsService implements UserDetailsService {
    private final List<UserDetailsService> userDetailsServices;

    public DelegatingUserDetailsService(
            List<UserDetailsService> userDetailsServices) {
        this.userDetailsServices = userDetailsServices;
    }

    public UserDetails loadUserByUsername(String username)
            throws UsernameNotFoundException {
        RuntimeException last = null;
        for(UserDetailsService uds : userDetailsServices) {
            try {
                return uds.loadUserByUsername(username);
            } catch(RuntimeException error) {
                last = error;
            }
        }
        throw last;
    }
}

然后你可以记住 - 我改为使用它：

<bean id="delegateUds" class="DelegatingUserDetailsService">
     <constructor-arg>
         <list>
             <ref bean="jdbcUserSrv"/>
             <ref bean="adminSrv"/>
         </list>
     </constructor-arg>
</bean>
<security:http auto-config="true">
    <security:remember-me key="myAppKey" user-service-ref="delegateUds"/>
    <security:form-login login-page="/login" default-target-url="/"
        authentication-failure-url="/loginfailed" />
    <security:logout logout-success-url="/logout"/>
</security:http>

Answer 2

如果您使用Spring Security Java Configuration（Spring 5.0.6），请记住我

from sklearn.linear_model import LogisticRegression
from sklearn.datasets import load_iris
import os
import json
import codecs
import numpy as np

iris = load_iris()
X, y = iris.data, iris.target

lr = LogisticRegression()
lr.fit(X, y)

attr = lr.__dict__
New_attr = attr 
keys = New_attr.keys()


# --------------------
# Converting array to list  
# --------------------
New_attr['coef_'] = attr['coef_'].tolist()
New_attr['classes_'] = attr['classes_'].tolist()
New_attr['n_iter_'] = attr['n_iter_'].tolist()
New_attr['intercept_'] = attr['intercept_'].tolist()

# --------------------
# Writing the JSON file... 
# --------------------
json_file = "file.json" 
json.dump(New_attr, codecs.open(json_file, 'w', encoding='utf-8'), 
    sort_keys=True, indent=4)


# --------------------
# Reading the JSON file... 
# --------------------
obj_text = codecs.open(json_file, 'r', encoding='utf-8').read()
b_new = json.loads(obj_text)

lr = LogisticRegression()
print(b_new)
lr.__dict__ = dict(b_new)
lr.predict(X)

和多个这样的身份验证提供程序：

@Override
protected void configure(HttpSecurity http) throws Exception {
    http
            //... nothingImportantToThisCase
            .rememberMe();
}

将引发相同的异常，但仍将登录。但令人惊讶的是，它将起作用：

public void configureGlobal(AuthenticationManagerBuilder auth,
                            UserDetailsService userDetailsService) Exception {
        auth
            .inMemoryAuthentication()
                .withUser("ADMIN")
                    .password("{noopassword")
                    .roles("ADMIN", "USER")
                    .and()
                .withUser("USER")
                    .password("{noop}password)
                    .roles("USER");
        auth
            .userDetailsService(userDetailsService)
            .passwordEncoder(new BCryptPasswordEncoder());
}

重要的是，即使它们具有不同的方法参数，如果只需要一个参数，则两个参数都必须具有@Autowired

Spring安全 - 使用多个身份验证提供程序进行身份验证

2 个答案: