[如果验证失败，如何使用ajax和php以模态形式显示错误，如果成功则如何重定向到其他页面

Question

我正在登录页面（这是模式表单，当用户单击登录按钮时，将出现模式表单）在这里

<!-- Signin Window Code -->
            <div class="modal fade" id="signup" tabindex="-1" role="dialog" aria-labelledby="myModalLabel2" aria-hidden="true">
                <div class="modal-dialog">
                    <div class="modal-content">
                        <div class="modal-body">
                     <form action ="login.php" method="POST" id="frmLogin">
                            <div class="new-logwrap">   
                                <div class="form-group">
                                <label>Email</label>
                                <div class="input-with-icon">
                                <input type="email" class="form-control" name="login_email" id = "email_login" placeholder="Enter Your Email" required>
                                <i class="theme-cl ti-email"></i>
                                    </div>
                                </div>
                                <div class="form-group">
                                    <label>Password</label>
                                    <div class="input-with-icon">
                                        <input type="password" class="form-control" name="login_pass" id = "pass_login" placeholder="Enter Your Password" required>
                                        <i class="theme-cl ti-lock"></i>
                                    </div>
                                </div>
                                <div class="form-groups">
                                    <button type="submit" name="login" id="logBtn" class="btn btn-primary theme-bg full-width .login">Login</button>
                                </div>                  
                        <!-- error message will show here -->
                                <div id="ack"></div>

并且我正在使用ajax来验证用户是否通过验证

        $('button#logBtn').click(function(){

            if($("#email_login").val() == "" || $("#pass_login").val() == "")
             $("div#ack").html("please enter username or password");
            else
            $.post($("#frmLogin").attr("action"),
            $("#frmLogin :input").serializeArray(),
            function(data) {    
                $("div#ack").html(data);
            });

            $("#frmLogin").submit(function (){
              return false;
            });
   });

当用户验证失败时，它会在模式表单上显示错误，但问题是，如果验证成功，则要重定向的页面会在模式表单上被覆盖，即，它不会重定向，而是在模式本身上显示下一页这是我的PHP代码

<?php
// for testing purpose
require 'dbserver.inc.php';
$email     = mysqli_real_escape_string($conn,$_POST['login_email']);
$password  = mysqli_real_escape_string($conn,$_POST['login_pass']); 
$sql = "SELECT count(*) from registergac1 WHERE (email='$email ' AND password = '$password')";
$res  = mysqli_query($conn,$sql);
$row = mysqli_fetch_array($res);
if($row[0] > 0)
{
    header("location: register.php"); //some page to redirect
}
else
{
    echo '<div class="alert alert-danger text-center" role="alert">
    Enter correct email or password!
  </div>';
}

?>

Answer 1

您无法使用ajax直接从php重定向，响应将发送回ajax的success-function。现在，要解决此问题，请执行以下操作：