Question

我如何发布ID，我不了解其逻辑。为什么不更新？没有错误信息，请帮帮我！我对ajax不太了解。 :(
。。。。 index.php

<?php
require "header.php";

$con= mysqli_connect("localhost","root","","propanel_001");
mysqli_set_charset($con, 'utf8');

if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
mysqli_close($con);
}
$gelen_id=@$_GET['id'];
$sql="SELECT * FROM tb_haber where id='$gelen_id'";
$result=mysqli_query($con,$sql);
$row=mysqli_fetch_assoc($result);
?>

<form id="my_form">
<label>Haber Başlığı</label>
<input value="<?php echo $row['haber_title'] ?>" type="text" name="haber_title">
<label>Haber İçeriği</label>
<textarea name="haber_desc" class="ckeditor text-justify"><?php echo $row['haber_desc'];?></textarea>
<button type="submit" name="btn_haber_update">UPDATE</button>

</form>

<script>
    $(document).ready(function(){
      $("form[id=my_form]").submit( function(up_haber) {
          for (instance in CKEDITOR.instances) {
            CKEDITOR.instances[instance].updateElement();
          }
          var form_data = new FormData(this);
          SwalUp_Haber(form_data);
          up_haber.preventDefault();
      });
  }); 

  function SwalUp_Haber(form_data){
     $.ajax({
      url: "update-haber.php",
      type: "POST",
      data: form_data,
      dataType: "json",
      contentType: false,
      cache: false,
      processData: false,        
  })

   .always(function(response_up_haber) {
      swal({
          title: response_up_haber.titles, 
          text: response_up_haber.message,
          type: response_up_haber.status,
          showConfirmButton: false,
          timer:3000,
      });
     });
  }
</script>

update-haber.php

<?php
header('Content-type: application/json; charset=UTF-8');
$response_up_haber = array();

$id_haber_up    = $_GET['id'];
$haber_title    = mysqli_real_escape_string($con, $_POST['haber_title']);
$haber_desc     = mysqli_real_escape_string($con, $_POST['haber_desc']);


$query = "UPDATE tb_haber SET haber_title='$haber_title', haber_desc='$haber_desc' where id='$id_haber_up'";
$update=mysqli_query($con, $query);

if ($update>0) {
$response_up_haber['status']  = 'success';
$response_up_haber['titles']  = 'BAŞARILI';
$response_up_haber['message'] = 'Haber başarılı bir şekilde güncellendi..!';
}else{
$response_up_haber['status']  = 'error';
$response_up_haber['titles']  = 'HATA';
$response_up_haber['message'] = 'Bir sorun oluştu ...
<br>Lütfen bu durumu ProPANEL yetkililerine bildirin..!';
}
echo json_encode($response_up_haber);
?>

PHP与Ajax表单输入数据更新

0 个答案: