如何在 j_spring_security_check 中获取密码值

Question

这是我的登录表单

<form class="login-form" action="j_spring_security_check" method="post" >
  <fieldset>
    <legend>Login Here</legend>

    <div class="form-group required">
      <label id="inputEmail" class="control-label col-md-4">Email</label>
      <div class="col-md-8">
        <input type="email" class="form-control" id="username" name="username" placeholder="Enter email" >
      </div>
    </div>
    <div class="form-group">
      <label id="inputpwd" class="control-label col-md-4">Password</label>
      <div class="col-md-8">
        <input type="password" class="form-control" id="pwd" name="password" placeholder="Enter password" >
      </div>
    </div>
    <p><input type="submit" value="Login"/></p>
  </fieldset>
</form>

如何在实现 UserDetailsS​​ervice 的类中检索我们为密码提供的输入

public class LoginUserDetails implements UserDetailsService {

    @Override
    public UserDetails loadUserByUsername(final String username) throws UsernameNotFoundException {

        final User user = DataLayer.queryToGetUserDetails(username);

返回新的 UserDetails() {

        private static final long serialVersionUID = 2059202961588104658L;

        @Override
        public boolean isEnabled() {
            return true;
        }

        @Override
        public boolean isCredentialsNonExpired() {
            return true;
        }

        @Override
        public boolean isAccountNonLocked() {
            return true;
        }

        @Override
        public boolean isAccountNonExpired() {
            return true;
        }

        @Override
        public String getUsername() {
            return user.getUsername();
        }

        @Override
        public String getPassword() {
            return user.getPassword();
        }



        //To DO for authentication.
        @Override
        public Collection<? extends GrantedAuthority> getAuthorities() {
            List<SimpleGrantedAuthority> auths = new java.util.ArrayList<SimpleGrantedAuthority>();
            auths.add(new SimpleGrantedAuthority("patient"));
            return auths;
        }
    };
}

}

我正在返回一个类似上面的值。因此，为了进行自定义身份验证，我需要在该类中获取密码字段 提前致谢

Answer 1

@控制器

公共类 LoginUserDetails 实现 UserDetailsS​​ervice {

@RequestMapping("/j_spring_security_check")
public UserDetails loadUserByUsername(final String username,String password) throws UsernameNotFoundException {

    final User user = DataLayer.queryToGetUserDetails(username,password);

} }

Answer 2

从 loadUserByUsername 返回的 UserDetails 将用于匹配您使用您在 SecurityConfiguration 中设置的 passwordEncoder 传入的密码

@Override
protected void configure(AuthenticationManagerBuilder auth) throws Exception {
    auth.userDetailsService(userDetailsService).passwordEncoder(passwordEncoder);
}

密码编码器接口

public interface PasswordEncoder {
    String encode(CharSequence rawPassword);
    boolean matches(CharSequence rawPassword, String encodedPassword);

rawPassword 参数是你传入的密码

编辑：

你需要有一个如下的 SecurityConfiguration 类：

@Configuration
@EnableWebSecurity
@ComponentScan
public class SecurityConfiguration extends WebSecurityConfigurerAdapter {

    @Inject
    private LoginUserDetails loginUserDetails;

    @Bean
    public BCryptPasswordEncoder passwordEncoder(){
        return new BCryptPasswordEncoder();
    }

    @Override
    protected void configure(HttpSecurity http) throws Exception {
        http
            .authorizeRequests()
                .antMatchers("/login").permitAll()
                .and()
            .authorizeRequests()
                .anyRequest().hasRole("ROLE_USER")
                .and()
            .formLogin()
                .usernameParameter("j_username") //user name form name
                .passwordParameter("j_password") //password form name
                .loginProcessingUrl("/j_spring_security_check") //form submit url
                .loginPage("/login")
                .failureUrl("/")
                .defaultSuccessUrl("")
                .and()
                .logout()
                .logoutSuccessUrl("/");
    }

    @Override
    protected void configure(AuthenticationManagerBuilder auth) throws Exception {
        auth.userDetailsService(loginUserDetails).passwordEncoder(passwordEncoder);
    }