如何显示使用Spring Boot Thymeleaf的当前登录用户? [英] How can I display the current logged in User with Spring Boot Thymeleaf?
问题描述
我正在尝试显示当前用户的详细信息，但是我一直遇到错误.我尝试从模板访问经过身份验证的用户，但是由于出现此错误，该操作不起作用:
I am trying to display the details of the current user however I keep getting errors. I tried accessing the authenticated user from the template but that did not work as I was getting this error:
在org.springframework.security.core.userdetails.User类型上找不到方法getFirstName()
Method getFirstName() cannot be found on org.springframework.security.core.userdetails.User type
我试图从控制器获取信息，然后将其保存在字符串中并将该字符串传递给模板，但这都不起作用.
I was trying to get the information from a controller and then saving it in a string and passsing the string to a template but that wasn't working either.
这是我的SecurityConfig类:
Here is my SecurityConfig class:
@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());
}
这是我的UserService类:
Here is my UserService Class:
public interface UserService extends UserDetailsService {
User findByEmailAddress(String emailAddress);
// User findByFirstName(String firstName);
User save(UserRegistrationDto registration);
}
这是我的UserServiceImpl类:
Here is my UserServiceImpl class:
@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());
}
}
这是我试图获取信息的模板类中的一些代码:
Here is some code from the template class where I'm trying to get the information:
th:text ="$ {#authentication.getPrincipal().getFirstName()}">
th:text ="${#authentication.getPrincipal().getFirstName()}">
th:text ="$ {#authentication.getPrincipal().getUser().getFirstName()}">
th:text ="${#authentication.getPrincipal().getUser().getFirstName()}">
这是登录控制器.我注释掉的部分是我尝试获取当前用户详细信息的另一种方式:
This is the login controller. The parts I have commented out was another way I was trying to get the current users details:
@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";
}
}
对不起，这到处都是！非常感谢任何帮助:D
This is all over the place I'm sorry! Thanks so much for anyone who helps :D
推荐答案
我想出了解决问题的方法.
I figured out how to fix my problem.
我在控制器中创建了此方法:
I created this method in a controller:
@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
}
，然后在我的html模板中添加以下代码行:
and then I added this line of code in my html template:
电子邮件:th:text ="$ {emailAddress}"
Email: th:text="${emailAddress}"
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