Spring Security-405请求方法'POST'不支持


问题内容

我已经为项目实现了Spring Security,但是尝试登录时状态为405。我已经在中添加了csrf令牌form

这是我发送用户名和密码时遇到的错误: HTTP Status 405 - Request method 'POST' not supported

spring版本:4.0.2。发布

<div class="login-form">
    <c:url var="loginUrl" value="/login" />
    <form action="${loginUrl}" method="post" class="form-horizontal">
        <c:if test="${param.error != null}">
            <div class="alert alert-danger">
                <p>Invalid username and password.</p>
            </div>
        </c:if>
        <c:if test="${param.logout != null}">
            <div class="alert alert-success">
                <p>You have been logged out successfully.</p>
            </div>
        </c:if>
        <div class="input-group input-sm">
            <label class="input-group-addon" for="username">
                <i class="fa fa-user"></i>
            </label>
            <input type="text" class="form-control" id="username"
                name="clientusername" placeholder="Enter Username" required>
        </div>
        <div class="input-group input-sm">
            <label class="input-group-addon" for="password">
                <i class="fa fa-lock"></i>
            </label>
            <input type="password" class="form-control" id="password"
                name="clientpassword" placeholder="Enter Password" required>
        </div>

        <input type="hidden" name="${_csrf.parameterName}"
            value="${_csrf.token}" />

        <div class="form-actions">
            <input type="submit" class="btn btn-block btn-primary btn-default"
                value="Log in">
        </div>
    </form>
</div>

安全配置:

@Configuration
@EnableWebSecurity
public class SecurityConfiguration extends WebSecurityConfigurerAdapter {

    @Autowired
    @Qualifier("G2BUserDetailsService")
    UserDetailsService userDetailsService;

    @Autowired
    public void configureGlobalSecurity(AuthenticationManagerBuilder auth) throws Exception {
        auth.userDetailsService(userDetailsService);
    }

    @Override
    protected void configure(HttpSecurity http) throws Exception {
      http.authorizeRequests()
        .antMatchers("/", "/home").permitAll()
        .antMatchers("/admin/**").access("hasRole('ADMIN')")
        .and().formLogin().loginPage("/login")
        .usernameParameter("clientusername").passwordParameter("clientpassword")
        .and().csrf()
        .and().exceptionHandling().accessDeniedPage("/Access_Denied");
//        .and().csrf().disable();
    }

控制器:

@RequestMapping(value = "/login", method = RequestMethod.GET)
public ModelAndView loginPage() {
    return new ModelAndView("login");
}

@RequestMapping(value="/logout", method = RequestMethod.GET)
public String logoutPage (HttpServletRequest request, HttpServletResponse response) {
    Authentication auth = SecurityContextHolder.getContext().getAuthentication();
    if (auth != null){    
        new SecurityContextLogoutHandler().logout(request, response, auth);
    }
    return "redirect:/login?logout";
}

 @RequestMapping(value = "/Access_Denied", method = RequestMethod.GET)
    public ModelAndView accessDeniedPage(ModelMap model) {
        model.addAttribute("user", getPrincipal());
        return new ModelAndView("accessDenied");
    }

 @RequestMapping(value = "/admin", method = RequestMethod.GET)
    public ModelAndView adminPage(ModelMap model) {
        model.addAttribute("user", getPrincipal());
        return new ModelAndView("admin");
    }

 private String getPrincipal(){
        String userName = null;
        Object principal = SecurityContextHolder.getContext().getAuthentication().getPrincipal();

        if (principal instanceof UserDetails) {
            userName = ((UserDetails)principal).getUsername();
        } else {
            userName = principal.toString();
        }
        return userName;
    }

关于此问题的几乎每个主题都说我们需要添加csrf令牌,但是我已经添加了。我想念什么吗?


问题答案:

您可以为一个网址设置两个端点。但是您不能根据需要设置任何请求参数。当我看到您的登录请求映射时,可以这样设置请求方法:

@RequestMapping(value = "/login", method = { RequestMethod.GET, RequestMethod.POST })
public ModelAndView loginPage() {
    return new ModelAndView("login");
}