第二次单击按钮时不显示AJAX响应

Question

我有一个简单的登录表单，我通过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，它会显示错误消息。第二次，如果我输入错误，它不会显示错误。我在哪里做错了任何建议/帮助。

表格

template <typename T, typename U> struct MyClass : Base<T, U> {
    using Base::Base;
};

template <typename U> struct MyClass<char32_t, U> : Base<char32_t, U> {
    using Base::Base;

    MyClass(char32_t init) { /* your constructor here */ }
};

的Ajax

<?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
}
?>

PHP

<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>