Spring Security - 405请求方法' POST'不支持

时间:2017-02-13 14:55:12

标签: java spring spring-mvc

我已经为我的项目实施了Spring Security,但是当我尝试登录时,我获得状态405.我已经在csrf中添加了form令牌。

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

Spring版本:4.0.2.RELEASED

<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;
    }

几乎关于此问题的每个主题都说我们需要添加csrf令牌,但我已经添加了。我错过了什么吗?

6 个答案:

答案 0 :(得分:9)

从Spring 4.0开始,首先csrfenabled by default,因此无需自己明确启用它。

其次,您没有端点可以验证您的登录信息。您正在做的是向/login发送请求,该请求只需GET次请求。您可以创建另一个控制器方法来接收POST请求并进行身份验证,或者您可以使用UserDetailsService

SecurityConfiguration

protected void configure(HttpSecurity http) throws Exception {
        http
                .authorizeRequests()
                    .antMatchers("/login-form")
                        .anonymous()
                    .and()
                .formLogin()
                    .loginPage("/user-login") 
                    .defaultSuccessUrl("/admin", true) // the second parameter is for enforcing this url always
                    .loginProcessingUrl("/login")
                    .failureUrl("/user-login")
                    .permitAll();
}

@Autowired 
private UserDetailsService userDetailsService;  

@Autowired
public void configureGlobal(AuthenticationManagerBuilder auth) throws Exception {
    BCryptPasswordEncoder pe = new  BCryptPasswordEncoder();
    auth.userDetailsService(userDetailsService).passwordEncoder(pe);
}

此处我们的视图页面为/user-login,处理网址为/login,这意味着您需要在控制器中移除/login的映射并添加以下内容:

控制器

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

改变你的观点。

查看(login-form.jsp)

<c:url value="/login" var="loginUrl"/>
<form action="${loginUrl}" method="post" modelAttribute="user">
    Username: <input type="text" id="username" name="username" placeholder=""><br>
    Password: <input type="password" id="password" name="password" placeholder=""><br>

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

答案 1 :(得分:2)

您可以为一个网址设置两个端点。但是您无法根据需要设置任何请求参数。当我看到您的登录请求地图时,您可以设置您的请求方法:

CGRect

答案 2 :(得分:1)

当我添加successForwardUrl时,我开始得到相同的东西,并发现成功登录的响应是对该端点的POST或未设置的“/”。一旦我在定义的端点上启用了POST以及GET,一切都很好。

答案 3 :(得分:0)

您正在呼叫POST但仅定义了GET方法。将您的端点更改为RequestMethod.POST

答案 4 :(得分:0)

检查您可能忘记保留“ securityFilterChain”的web.xml文件

在web.xml文件中使用此代码

<!-- Security configuration goes here -->
<filter>
    <filter-name>springSecurityFilterChain</filter-name>
    <filter-class>org.springframework.web.filter.DelegatingFilterProxy</filter-class>
</filter>
<filter-mapping>
    <filter-name>springSecurityFilterChain</filter-name>
    <url-pattern>/*</url-pattern>
</filter-mapping>

答案 5 :(得分:0)

如果您使用的是JSP / JSTL

更改

<form action="${loginUrl}" method="post"></form> 

<form:form action="${loginUrl}" method="post" </form:form>

带有标签声明

<%@ taglib prefix="form" uri="http://www.springframework.org/tags/form"%>

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