【发布时间】：2016-07-25 18:11:00
【问题描述】：
我有一个基于 SpringSecurity 的非常简单的身份验证系统，与 here 的发现非常相似（有点复杂）。
尽管如此，当我执行登录过程时，SpringSecurity 根据其配置抛出一个 Bad Credentials 错误。
查看代码，我找不到原因，因为在运行时用户名、密码、启用和角色是根据数据库中存储的内容。因此，我认为它可能源于错误的配置或逻辑问题。
根据日志，唯一可能失败的是：WARNING: Encoded password does not look like BCrypt，但我读到每次身份验证都可以正常。
有人可以帮我检查配置吗？谢谢！！
package com.company.config;
import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.beans.factory.annotation.Qualifier;
import org.springframework.context.annotation.Bean;
import org.springframework.context.annotation.Configuration;
import org.springframework.security.config.annotation.authentication.builders.AuthenticationManagerBuilder;
import org.springframework.security.config.annotation.web.builders.HttpSecurity;
import org.springframework.security.config.annotation.web.configuration.EnableWebSecurity;
import org.springframework.security.config.annotation.web.configuration.WebSecurityConfigurerAdapter;
import org.springframework.security.core.userdetails.UserDetailsService;
import org.springframework.security.crypto.bcrypt.BCryptPasswordEncoder;
import org.springframework.security.crypto.password.PasswordEncoder;
@Configuration
@EnableWebSecurity
public class AppSecurityConfig extends WebSecurityConfigurerAdapter {
@Autowired
@Qualifier("userDetailsService")
UserDetailsService userDetailsService;
@Autowired
public void configureGlobal(AuthenticationManagerBuilder auth) throws Exception{
auth.userDetailsService(userDetailsService).passwordEncoder(passwordEncoder());
}
@Bean
public PasswordEncoder passwordEncoder() {
PasswordEncoder encoder = new BCryptPasswordEncoder();
return encoder;
}
@Override
protected void configure(HttpSecurity http) throws Exception {
http.authorizeRequests()
.antMatchers("**/admin/**").access("hasAnyRole('ROLE_ADMIN','ROLE_SUPERADMIN')")
.antMatchers("/superadmin/**").access("hasRole('ROLE_SUPERADMIN')")
.antMatchers("**/user/**").access("hasAnyRole('ROLE_USER','ROLE_ADMIN','ROLE_SUPERADMIN')")
.antMatchers("/resources/**").permitAll()
.antMatchers("/messages/**").permitAll()
.and()
.formLogin()
.loginPage("/login")
.usernameParameter("username")
.passwordParameter("password")
.defaultSuccessUrl("/user/home")
.failureUrl("/403")
.permitAll()
.and()
.exceptionHandling().accessDeniedPage("/403")
.and()
.logout().logoutUrl("/logout")
.and()
.csrf().disable();
}
}
package com.company.service.impl;
import java.util.ArrayList;
import java.util.HashSet;
import java.util.List;
import java.util.Set;
import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.security.core.GrantedAuthority;
import org.springframework.security.core.authority.SimpleGrantedAuthority;
import org.springframework.security.core.userdetails.User;
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.Service;
import org.springframework.transaction.annotation.Transactional;
import com.company.dao.UsuarioDao;
import com.company.model.UserRole;
import com.company.model.Usuario;
@Service("userDetailsService")
public class MyUserDetailsService implements UserDetailsService{
@Autowired
private UsuarioDao usuarioDao;
@Transactional(readOnly=true)
public UserDetails loadUserByUsername(String username)
throws UsernameNotFoundException {
Usuario usuario = usuarioDao.findByChave(username);
List<GrantedAuthority> authorities = buildUserAuthority(usuario.getUserRole());
return buildUserForAuthentication(usuario, authorities);
}
private User buildUserForAuthentication(Usuario user,
List<GrantedAuthority> authorities) {
User usr= new User(user.getUsername(), user.getPassword(),
user.isEnabled(), true, true, true, authorities);
System.out.println(usr.toString());
/*
* Prints: org.springframework.security.core.userdetails.User@ae6e27ef:
* Username: SMITH; Password: [PROTECTED]; Enabled: true;
* AccountNonExpired: true; credentialsNonExpired: true; AccountNonLocked:
* true; Not granted any authorities
*/
return usr;
}
private List<GrantedAuthority> buildUserAuthority(Set<UserRole> userRoles){
Set<GrantedAuthority> setAuths = new HashSet<GrantedAuthority>();
// Build user's authorities
for (UserRole userRole : userRoles) {
setAuths.add(new SimpleGrantedAuthority(userRole.getRole()));
}
List<GrantedAuthority> Result = new ArrayList<GrantedAuthority>(setAuths);
return Result;
}
}
【问题讨论】：
-
如果您收到该消息，则表示您从数据库收到的信息不是预期的，因为密码不是预期的。检查编码，字段长度...检查实际值...显然您没有存储正确的哈希，这就是该消息告诉您的内容。
-
@M.Deinum，发现问题。如果您想查看答案，我认为它很有价值。感谢您的关注。