我正在构建一个Angular 2 Web客户端,尝试使用SpringBoot Security对服务器进行POST。我该如何编写Spring安全配置?
My Angular对身份验证的要求:
public login(username, password) {
let body = JSON.stringify({username: username, password: password});
let headers = new Headers({'Content-Type': 'application/json'});
let options = new RequestOptions({headers: headers});
this.http.post("http://localhost:8080/login", body, options)
.subscribe(
res => this.loggedIn = true,
err => console.error("failed authentication: " + err),
() => console.log("tried authentication")
);
}
身份验证失败并显示错误:
{"时间戳":1487007177889,"状态":401,"错误":"未授权""消息&# 34;:"身份验证失败:空用户名","路径":" / login"}
我的春季安全配置:
@Configuration
public class WebSecurityConfigurer extends WebSecurityConfigurerAdapter {
@Autowired
private RestAuthenticationEntryPoint restAuthenticationEntryPoint;
@Autowired
private RestAuthenticationSuccessHandler restAuthenticationSuccessHandler;
@Autowired
private RestAuthenticationFailureHandler restAuthenticationFailureHandler;
@Autowired
private RestLogoutSuccessHandler restLogoutSuccessHandler;
@Override
protected void configure(HttpSecurity http) throws Exception {
http.csrf().disable()
.exceptionHandling()
.authenticationEntryPoint(restAuthenticationEntryPoint)
.and().formLogin()
.loginProcessingUrl("/login")
.usernameParameter("username")
.passwordParameter("password")
.successHandler(restAuthenticationSuccessHandler)
.failureHandler(restAuthenticationFailureHandler)
.permitAll()
.and().logout()
.logoutUrl("/logout")
.logoutSuccessHandler(restLogoutSuccessHandler)
.permitAll()
.and().authorizeRequests().anyRequest().authenticated()
;
}
@Override
public void configure(AuthenticationManagerBuilder builder) throws Exception {
// This configuration has been tested, it works.
// It has been removed for better readability
}
@Bean
public LdapContextSource contextSource() {
// This configuration has been tested, it works.
// It has been removed for better readability
}
}
答案 0 :(得分:1)
您应该使用application/x-www-form-urlencoded
参数进行表单登录,而不是JSON。这就是错误导致用户名丢失的原因,因为Spring Security试图从HttpServletRequest#getParameters中获取用户名。要在Angular中发送表单参数,您可以执行
import { URLSearchParams } from '@angular/http';
let params = new URLSearchParams();
params.set('username', username);
params.set('password', password);
如果你把它设置为Http请求的body参数,它应该(从我记忆中)自动被序列化为正确的格式,即
username=xxx&password=xxx
我认为您不需要将标头Content-Type
设置为applicatio/x-www-form-urlencoded
。我认为,当Angular将URLSearchParams
视为正文时,也应该为您设置。