我想,但是我不断收到错误显示当前用户的详细信息。我试图从模板访问身份验证的用户,但因为我得到这个错误,没有工作:
方法的getFirstName()不能在org.springframework.security.core.userdetails.User类型中找到
我想从控制器的信息,然后将其保存在一个字符串,passsing字符串到模板但不工作要么。
这里是我SecurityConfig类:
@Configuration
public class SecurityConfig extends WebSecurityConfigurerAdapter {
@Autowired
private UserService userService;
@Override
protected void configure(HttpSecurity http) throws Exception {
http
.authorizeRequests()
.antMatchers(
"/registration",
"/js/**",
"/css/**",
"/img/**",
"/webjars/**").permitAll()
.anyRequest().authenticated()
.and()
.formLogin()
.loginPage("/login")
.permitAll()
.and()
.logout()
.invalidateHttpSession(true)
.clearAuthentication(true)
.logoutRequestMatcher(new AntPathRequestMatcher("/logout"))
.logoutSuccessUrl("/login?logout")
.permitAll();
}
@Bean
public BCryptPasswordEncoder passwordEncoder(){
return new BCryptPasswordEncoder();
}
@Bean
public DaoAuthenticationProvider authenticationProvider(){
DaoAuthenticationProvider auth = new DaoAuthenticationProvider();
auth.setUserDetailsService(userService);
auth.setPasswordEncoder(passwordEncoder());
return auth;
}
@Override
protected void configure(AuthenticationManagerBuilder auth) throws Exception {
auth.authenticationProvider(authenticationProvider());
}
这里是我的UserService类:
public interface UserService extends UserDetailsService {
User findByEmailAddress(String emailAddress);
// User findByFirstName(String firstName);
User save(UserRegistrationDto registration);
}
这里是我UserServiceImpl类:
@Service
public class UserServiceImpl implements UserService {
@Autowired
private UserRepository userRepository;
@Autowired
private BCryptPasswordEncoder passwordEncoder;
@Override
public UserDetails loadUserByUsername(String emailAddress) throws
UsernameNotFoundException {
User user = userRepository.findByEmailAddress(emailAddress);
if (user == null){
throw new UsernameNotFoundException("Invalid username or
password.");
}
return new
org.springframework.security.core.userdetails.User(user.getEmailAddress(),
user.getPassword(),
mapRolesToAuthorities(user.getRoles()));
}
public User findByEmailAddress(String emailAddress){
return userRepository.findByEmailAddress(emailAddress);
}
public User save(UserRegistrationDto registration){
User user = new User();
user.setFirstName(registration.getFirstName());
user.setSurname(registration.getSurname());
user.setEmailAddress(registration.getEmailAddress());
user.setPassword(passwordEncoder.encode(registration.getPassword()));
user.setRoles(Arrays.asList(new Role("ROLE_USER")));
return userRepository.save(user);
}
private Collection<? extends GrantedAuthority>
mapRolesToAuthorities(Collection<Role> roles){
return roles.stream()
.map(role -> new SimpleGrantedAuthority(role.getName()))
.collect(Collectors.toList());
}
}
下面是从模板类,我想获取一些信息的代码:
个:文本= “$ {#authentication.getPrincipal()的getFirstName()}”>
个:文本= “$ {#authentication.getPrincipal()的getUser()的getFirstName()。}”>
这是登录控制器。我注释掉的部分是另一种方式,我试图获取当前用户的详细信息:
@Controller
//@RequestMapping("/login")
public class MainController {
// @GetMapping("/")
// public String root() {
// return "userProfile1";
// }
@GetMapping("/login")
public String login(Model model) {
return "login";
}
// @GetMapping
// public String displayUserAccount(@ModelAttribute("user") @Valid
UserRegistrationDto userDto, BindingResult result, Model model) {
//
//
// model.addAttribute("firstName", ((UserRegistrationDto)
auth).getEmailAddress());
//
// model.addAttribute("emailAddress", userDto.getEmailAddress());
// model.addAttribute("firstName", userDto.getFirstName());
// model.addAttribute("surname", userDto.getSurname());
// model.addAttribute("age", userDto.getAge());
// model.addAttribute("gender", userDto.getGender());
// model.addAttribute("dob", userDto.getDob());
// // return "redirect:/registration?success";
// return "userProfile1";
//
// }
@ResponseBody
public String currentUserName(Authentication auth) {
((UserRegistrationDto) auth).getEmailAddress();
return "userProfile1";
}
}
这是各地对不起的地方!感谢这么多的人谁帮助:d
您可以使用Thymeleaf额外显示身份验证的用户的详细信息。
Thymeleaf Extras Springsecurity4
<div th:text="${#authentication.name} ></div>
问题就在这里:
return new
org.springframework.security.core.userdetails.User(user.getEmailAddress(),
user.getPassword(),
mapRolesToAuthorities(user.getRoles()));
你失去了参考您的User
实体。将其更改为:
return user;
对于这个工作,你需要更新你的User
实体,实现UserDetails接口:
public class User implements UserDetails {
// some new methods to implement
}
然后,你Thymeleaf代码应工作。获得firstName
将是另一种方式:
<span th:text="${#request.userPrincipal.principal.firstName}"></span>
我想出如何解决我的问题。
我创建在控制器此方法:
@Autowired
UserRepository userR;
@GetMapping
public String currentUser(@ModelAttribute("user") @Valid UserRegistrationDto userDto, BindingResult result, Model model) {
Authentication loggedInUser = SecurityContextHolder.getContext().getAuthentication();
String email = loggedInUser.getName();
User user = userR.findByEmailAddress(email);
String firstname = user.getFirstName();
model.addAttribute("firstName", firstname);
model.addAttribute("emailAddress", email);
return "userProfile1"; //this is the name of my template
}
然后我加入这行代码在我的HTML模板:
电子邮箱:到:文本= “$ {电子邮件地址}”
参考(4 Spring Security的话):
https://www.thymeleaf.org/doc/articles/springsecurity.html
添加依赖的pom.xml
<dependency>
<groupId>org.thymeleaf.extras</groupId>
<artifactId>thymeleaf-extras-springsecurity4</artifactId>
</dependency>
和视图(Thymeleaf):
<div sec:authorize="isAuthenticated()">
Authenticated user roles:
Logged user: <span sec:authentication="name"></span> |
Roles: <span sec:authentication="principal.authorities"></span>
</div>
我希望你能为他们服务