【发布时间】:2018-12-23 00:49:07
【问题描述】:
我正在尝试在我的项目中实现 JWTAuthentication。我已经这样设置了我的实体:
@Entity
public class ApplicationUser {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
Long id;
String username,password;
public Long getId() {
return id;
}
public void setId(Long id) {
this.id = id;
}
public String getUsername() {
return username;
}
public void setUsername(String username) {
this.username = username;
}
public String getPassword() {
return password;
}
public void setPassword(String password) {
this.password = password;
}
}
然后我像这样设置 AuthenticationFilter:
public class JWTAuthenticationFilter extends UsernamePasswordAuthenticationFilter {
private AuthenticationManager authenticationManager;
public JWTAuthenticationFilter(AuthenticationManager authenticationManager) {
this.authenticationManager = authenticationManager;
}
@Override
public Authentication attemptAuthentication(HttpServletRequest request, HttpServletResponse response) throws AuthenticationException {
try {
System.out.println("request "+request.getPathInfo());
ApplicationUser user = new ObjectMapper().readValue(request.getInputStream(),ApplicationUser.class);
//this is where the exception is
return authenticationManager.authenticate(new UsernamePasswordAuthenticationToken(user.getUsername(),user.getPassword(),new ArrayList<>()));
} catch (IOException e) {
throw new RuntimeException(e);
}
}
@Override
protected void successfulAuthentication(HttpServletRequest request, HttpServletResponse response, FilterChain chain, Authentication authResult) throws IOException, ServletException {
String token = Jwts.builder()
.setSubject(((ApplicationUser) authResult.getPrincipal()).getUsername())
.setExpiration(new Date(System.currentTimeMillis() + EXPIRATION_TIME))
.signWith(SignatureAlgorithm.HS512, SECRET.getBytes())
.compact();
response.addHeader(HEADER_STRING, TOKEN_PREFIX + token);
}
}
这是我的安全配置:
@EnableWebSecurity
public class SecurityConfig extends WebSecurityConfigurerAdapter {
private BCryptPasswordEncoder bCryptPasswordEncoder;
private UserDetailsService userDetailsService;
public SecurityConfig(AppUserDetailService userDetailsService, BCryptPasswordEncoder bCryptPasswordEncoder){
this.bCryptPasswordEncoder=bCryptPasswordEncoder;
this.userDetailsService=userDetailsService;
}
@Override
protected void configure(HttpSecurity http) throws Exception {
http.cors().and().csrf().disable().authorizeRequests()
.antMatchers(HttpMethod.POST, SIGN_UP_URL).permitAll()
// .antMatchers(HttpMethod.POST,LOGIN_URL).permitAll()
.anyRequest().authenticated()
.and()
.addFilter(new JWTAuthenticationFilter(authenticationManager()))
.addFilter(new JWTAuthorizationFilter(authenticationManager()))
// this disables session creation on Spring Security
.sessionManagement().sessionCreationPolicy(SessionCreationPolicy.STATELESS);
}
@Override
public void configure(AuthenticationManagerBuilder auth) throws Exception {
auth.userDetailsService(userDetailsService).passwordEncoder(bCryptPasswordEncoder);
}
@Bean
CorsConfigurationSource corsConfigurationSource() {
final UrlBasedCorsConfigurationSource source = new UrlBasedCorsConfigurationSource();
source.registerCorsConfiguration("/**", new CorsConfiguration().applyPermitDefaultValues());
return source;
}
}
注册工作正常,但是当我尝试使用相同的凭据登录时,出现以下异常:
原因:com.fasterxml.jackson.databind.exc.MismatchedInputException:由于输入结束,没有要映射的内容 在 [来源:(org.apache.catalina.connector.CoyoteInputStream);行:1,列:0] 在 com.fasterxml.jackson.databind.exc.MismatchedInputException.from(MismatchedInputException.java:59) ~[jackson-databind-2.9.6.jar:2.9.6] 在 com.fasterxml.jackson.databind.ObjectMapper._initForReading(ObjectMapper.java:4145) ~[jackson-databind-2.9.6.jar:2.9.6] 在 com.fasterxml.jackson.databind.ObjectMapper._readMapAndClose(ObjectMapper.java:4000) ~[jackson-databind-2.9.6.jar:2.9.6] 在 com.fasterxml.jackson.databind.ObjectMapper.readValue(ObjectMapper.java:3070) ~[jackson-databind-2.9.6.jar:2.9.6] 在 com.project.pq.security.JWTAuthenticationFilter.attemptAuthentication(JWTAuthenticationFilter.java:35) ~[classes/:na] ...省略了53个常用框架
我正在使用 Postman 访问 API,在那里我将 application/json 设置为内容类型,并将用户名和密码作为 post 参数发送。我正在关注this 教程。我究竟做错了什么?任何帮助,将不胜感激。
谢谢
【问题讨论】:
-
不要使用过滤器中的 POST 内容,尤其是适用于所有 URL 的过滤器。如果登录需要一个带有 JSON 内容的指定 URL 的 POST,那么它应该由一个 Servlet 处理,而不是一个过滤器。身份验证过滤器用于处理标头值(例如
Authentication),因此可以在处理任意请求期间即时进行身份验证。
标签: java spring-boot spring-security jwt jackson-databind