Javascript AJAX响应不会在第二次单击按钮时显示

Javascript AJAX响应不会在第二次单击按钮时显示,javascript,php,jquery,ajax,Javascript,Php,Jquery,Ajax,我有一个简单的登录表单,其中通过AJAX调用传递了值。问题是当我第一次输入错误的电子邮件或密码时,它会向我显示错误消息。第二次,如果输入错误,则不会显示错误。我哪里做错了请提供建议/帮助 表格 <?php if (isset($_SESSION['login_email']) && !empty($_SESSION['login_email'])) { //header('Location:profile.php'); ?> <script> lo

我有一个简单的登录表单,其中通过AJAX调用传递了值。问题是当我第一次输入错误的电子邮件或密码时,它会向我显示错误消息。第二次,如果输入错误,则不会显示错误。我哪里做错了请提供建议/帮助

表格

<?php
if (isset($_SESSION['login_email']) && !empty($_SESSION['login_email'])) {
    //header('Location:profile.php');
?>
<script> location.replace("profile.php"); </script>
<?php
} else {
?>
  <div class="login_form">
    <h1 class="login_heading">Login</h1>
    <div class="alert-error"></div>
    <div class="alert-success"></div>
    <div class="login">
      <form method="post" action="">
        <label >Email</label>
        <input class="inputs_login" type="email" name="email" id="email" placeholder="email" >
        <label>Password</label>
        <input class="inputs_login" type="password" name="password" id="password" placeholder="password"><br>
        <input type="button" name="login_submit" id="login_submit" value="login">
      </form>
    </div>
  </div>
<?php
}
?>

location.replace(“profile.php”);
登录
电子邮件
密码

Ajax

<script>
  $(document).ready(function() {
    $('#login_submit').click(function(e){
      //e.preventDefault();
      var  email = $("#email").val(),
          password = $("#password").val();
      var proceed = true;
      if(proceed){
        post_data= { 'Email': email, 'Password': password};
        $.post('login_index.php', post_data, function(response){
          //load json data from server and output message
          if(response.type == 'error')
          {
            output=$('.alert-error').html(response.text);
          }else{
            location.href="profile.php";
          }
          $(".alert-error").delay(3200).fadeOut(300);
        }, 'json');
      }
    });
  });
</script>

$(文档).ready(函数(){
$(“#登录_提交”)。单击(函数(e){
//e、 预防默认值();
var email=$(“#email”).val(),
密码=$(“#密码”).val();
var=true;
如果(继续){
post_数据={'Email':Email,'Password':Password};
$.post('login\u index.php',post\u数据,函数(响应){
//从服务器加载json数据并输出消息
如果(response.type==“error”)
{
输出=$('.alert error').html(response.text);
}否则{
location.href=“profile.php”;
}
$(“.alert error”)。延迟(3200)。淡出(300);
}“json”);
}
});
});
php

<?php
include "db/db.php";
session_start();
if ($_POST) {
    if (!isset($_SERVER['HTTP_X_REQUESTED_WITH']) AND strtolower($_SERVER['HTTP_X_REQUESTED_WITH']) != 'xmlhttprequest') {

        //exit script outputting json data
        $output = json_encode(array(
            'type' => 'error',
            'text' => 'Request must come from Ajax'
        ));

        die($output);
    }
    if (isset($_POST['Email']) && isset($_POST['Password'])) {
        $email = filter_var($_POST["Email"], FILTER_SANITIZE_STRING);
        $pwd   = filter_var($_POST["Password"], FILTER_SANITIZE_STRING);
        $query = mysqli_query($con, "select * from customers where email='$email' and password='$pwd'");
        $count = mysqli_num_rows($query);
        $row   = mysqli_fetch_array($query, MYSQLI_ASSOC);
        if ($row) {

            $_SESSION['login_email'] = $row['email'];
            $output                  = json_encode(array(
                'type' => 'message',
                'text' => 'Hi ' . $email . ' You are successfully login'
            ));
            die($output);

        } else {
            $output = json_encode(array(
                'type' => 'error',
                'text' => 'Could not Login! Please check your email/password OR <a href="index.php?page=register_account">REGISTER FREE ACCOUNT</a> .'
            ));
            die($output);
        }
    }

}
?>

尝试给出$(“.alert error”).html(“”)$(“.alert error”).show()在//e.preventDefault()之后@Anopll yes it Worked您了解这个问题吗?请尝试给出$(“.alert error”).html(“”)$(“.alert error”).show()在//e.preventDefault()之后@是的,成功了你明白这个问题吗?