spring-security 使用 AJAX 的Spring安全性-记录的用户详细信息

dzhpxtsq  于 2022-11-11  发布在  Spring
关注(0)|答案(1)|浏览(151)

我已经通过 AJAX 使用AuthenticationEntryPoint和SimpleUrlAuthenticationSuccessHandler实现了用户身份验证。
现在,我需要将登录用户名输入到脚本变量中。
有人能帮我吗?

我的验证成功处理程序

public class MyAuthenticationSuccessHandler extends SimpleUrlAuthenticationSuccessHandler {

    private Log log = LogFactory.getLog(MyAuthenticationSuccessHandler.class);

    @Override
    public void onAuthenticationSuccess(HttpServletRequest request, HttpServletResponse response,
            Authentication authentication) throws IOException, ServletException {

        log.info("point-2-->"+authentication.getName()); //this prints what I need.

        // This is actually not an error, but an OK message. It is sent to avoid redirects.
        response.sendError(HttpServletResponse.SC_OK);

    }
}

我的JavaScript函数

$("#login").on('click', function(e) {
        e.preventDefault();
        $.ajax({url: getHost() + "/j_spring_security_check",
            type: "POST",
            beforeSend: function(xhr) {
                xhr.withCredentials = true;
            },
            data: $("#loginForm").serialize(),
            success: function(response, options) {
                    // We get a success from the server
                    //I need to get user name here
            },
            error: function(result, options) {
                    // We get a failure from the server...
                $(".error").remove();
                $('#j_username').before('<div class="error">Login failed, please try again.</div>');
        }

        });
    });

我已经为不同的问题附上了所有相关的文件。请访问下面的链接检查他们。
Spring Security; custom-filter and user-service-ref not working together

ifsvaxew

ifsvaxew1#

我找到了解决办法。
我只是进行了另一个 AJAX 调用,并捕获了登录用户。

我的JavaScript函数

success: function(response, options) {
                    // We get a success from the server
                    //I need to get user name here

    $.post('/api/loggeduser', function(data) {
           $('#loggeduser').html(data);
        }); 
},

登录用户控制器

@RequestMapping(value = "/api/loggeduser", method = RequestMethod.POST)
       public String printWelcome(ModelMap model, Principal principal ) {

        try {

            String name = null;

            if (principal!=null) {

           name = principal.getName();

          }

                model.addAttribute("username", name);

    } catch (Exception e) {
        e.printStackTrace();
    }
  }

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