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如何从index.php进行弹出式登录和注册以在mysql中工作和解析数据_Php_Ajax_Forms_Popup_Modal Dialog - Fatal编程技术网

如何从index.php进行弹出式登录和注册以在mysql中工作和解析数据

如何从index.php进行弹出式登录和注册以在mysql中工作和解析数据,php,ajax,forms,popup,modal-dialog,Php,Ajax,Forms,Popup,Modal Dialog,我有: Index.php、login.php、register.php 在index.php中,我有一个模式弹出窗口,我希望用户可以从弹出窗口登录和注册 我尝试使用ajax,但不起作用。 这是我的密码: ajax.php: function validLogin(){ var email=$('.email').val(); var password=$('.password').val(); $.ajax({ type: "POST", url: "si

我有: Index.php、login.php、register.php

在index.php中,我有一个模式弹出窗口,我希望用户可以从弹出窗口登录和注册

我尝试使用ajax,但不起作用。 这是我的密码: ajax.php:

 function validLogin(){
  var email=$('.email').val();
  var password=$('.password').val();
  $.ajax({
      type: "POST",
      url: "sign-in.inc",
     data: "name="+username+"&pwd="+password,
      success: function(html){
           if(html=='true'){
            alert ("Correct");
          }
          else{
              alert ("Wrong");
          }
}
  });
}

这是弹出式窗体:

<div id="login">
    <form class="form login-form" name="login-popup" >
        <a class="btn-facebook" href="#"><span><i class="fa fa-facebook"></i>
</span><span>Login with Facebook</span></a>
        <a class="btn-google" href="#"><span><i class="fa fa-google-plus"></i>
</span>
        <span>Login with Google</span></a>
        <p class="fieldset">
            <label class="image-replace email icon-log" for="signin-email">E-mail</label>
            <input class="full-width has-padding has-border email" id="signin-email" name="email-login" type="email" placeholder="E-mail" name="email" required>
            <span class="error_box"></span>
          </p>
          <p class="fieldset">
            <label class="image-replace password" for="signin-password">Password</label>
            <input class="full-width has-padding has-border password" id="signin-password"  type="password"  placeholder="Password"
            name="password" required>
            <a href="#0" class="hide-password">Show</a>
            <span class="error-message"></span>
          </p>
          <p class="fieldset">
            <input type="checkbox" id="remember-me" checked>
            <label for="remember-me">Remember me</label>
          </p>
          <p class="fieldset">
            <input class="full-width orange" id="login-submit" type="submit" name="submit" onclick="validLogin()" value="Login">
          </p>
        </form>
        <p class="form-bottom-message"><a href="#0">Forgot your password?</a></p>
        <!-- <a href="#0" class="close-form">Close</a> -->
      </div>

电子邮件

密码

记得我吗


首先,我建议将JQuery改为对选择器使用ID,而不是类。第二,你的
url
看起来不对劲。除非服务器上正在处理或映射
.inc
,否则将无法处理该请求。看一看:

function validLogin(){
    var email = $('#signin-email').val();
    var password = $('#signin-password').val();
    $.ajax({
        type: "POST",
        url: "sign-in.php",
        data: "name="+email+"&pwd="+password,
        success: function(html){
            if(html=='true'){
                alert ("Correct");
            } else {
                alert ("Wrong");
            }
        }
    });
}

你能为我们定义“不工作”吗?我看到的第一个问题是,你有
email
password
类,用于超过1个输入…对^使用id。其次,你不能发布到“sign-in.inc”,除非你的Web服务器中有*.inc映射到PHP。我怀疑你会。您将希望发布到一个*.php或一些可以处理该发布的文件。“sign-in.inc”??这是什么?