javascript - Ajax 成功消息不起作用

Question

我想提交表单而不刷新页面并在表单下显示成功/错误消息，以及在成功重置表单的情况下。但我想，我在 ajax 代码中做错了什么，php 工作正常。 Ajax :

$("#usignupform").validate({
    submitHandler:function() {
        var $form = $(form),
        $messageSuccess = $('#mesSuccess'),
        $messageError = $('#mesError');
        $.ajax({
            url: "investor-invite.php",
            type: "POST",
            data: $(this).serialize(),
            complete: function(data) {
                if (data.status == 'success') {
                    $messageSuccess.removeClass('hidden');
                    $messageError.addClass('hidden');

                    // Reset Form
                    $form.reset();

                    return;

                } else {
                    $messageError.removeClass('hidden');
                    $messageSuccess.addClass('hidden');
                }
            }
        });
        return false;
    }
});

PHP:

function NewUser(){ 
    $data = mysql_query($query)or die(mysql_error()); 

    if($data){
        $response_array['status'] = 'success';  
    } else {
        $response_array['status'] = 'error';  
    }
}
if(isset($_POST['submit'])){ 
    NewUser();
    }

HTML:

<form>
    --all fields--

    <button type="submit" name="submit" id="submitButton">
    Request Invite</button>
    <div class="alert alert-success hidden" id="mesSuccess">
        <strong>Success!</strong> Your message has been sent to us.
    </div>
    <div class="alert alert-danger hidden" id="mesError">
        <strong>Error!</strong> There was an error sending your message.
    </div>      
</form>

Answer 1

您需要回显 php 页面上的内容，然后这将返回成功消息。如果 php 文件或任何目标文件中有 html 或某些打印数据，Ajax 调用将获得响应。

尝试这样

    function NewUser(){ 
         $data = mysqli_query($query)or die(mysql_error()); 

         if($data){
            $response_array['status'] = 'success';  
               }else {
            $response_array['status'] = 'error';  
          }
return $response_array;
        }
    $data=NewUser();
    echo json_encode($data);

并更新ajax函数

$("#usignupform").validate({
                  submitHandler:function() {
                  var $form = $(form),
                   $messageSuccess = $('#mesSuccess'),
                   $messageError = $('#mesError');
             $.ajax({
                   url: "investor-invite.php",
                   type: "POST",
                   data: $(this).serialize(),
                    dataType: 'json',                  //add this to get the json response
                   complete: function(data) {
                        if (data.status == 'success') {

                        $messageSuccess.removeClass('hidden');
                        $messageError.addClass('hidden');

                        // Reset Form
                        $form.reset();


                    return;

                } else {

                    $messageError.removeClass('hidden');
                    $messageSuccess.addClass('hidden');
                }
            }
  });
  return false;
}
});

更新请从您的代码中删除 Depriated mysql_* 并使用 mysqli_*或pdo查询