Bir dosyayı silmek için çalışıyor jQuery ve PHP ile ilgili sorunlar ise

0 Cevap php

I have some problems with Jquery and PHP. I want to delete an image from the server. I think I did a little mess with the code... Can someone give me some tips on how to pass variables from PHP Jquery and vice versa?

JQuery

$(document).ready(function() {
$('#load').hide();
});

$(function() {
$(".delete").click(function() {
$('#load').fadeIn();
var container = $(this).parent();
var id = $(this).attr("id");
var string = 'id='+ id ;
var name = $(this).parent().attr("title");

$.ajax({
   type: "GET",
   url: "delete.php",
   data: 'string',
   cache: false,
   success: function(){
 container.slideUp('slow', function() {$(this).remove();});
 $('#load').fadeOut();
  }

 });

 $.ajax({
   method:'GET',
   url: "delete.php",
   data:'name',
   success:function(response){
   if (response === 'deleted') {
      alert('Deleted !!');
   }else{
    document.write(name);
   }
       }
      });


return false;
 });
});

PHP (delete.php)

if (isset($_GET['name']))
 {
   if (unlink($_GET['name']))
   {
   echo 'deleted';
   }else{
   echo "The file ".$_GET['name']." was not deleted"; 
    }
}else {
  echo "The name of the file was not set!";
 }

echo $_GET['id'];

HTML

$dirname = "../images/portraits";
                    $images = scandir($dirname);
                    $ignore = Array(".", "..", ".DS_Store");
     $count=1;

                    foreach($images as $img){
                        if (!in_array($img, $ignore)) {
                            if(!strripos($img, "_t")){
    echo "<div title='$dirname/$img' class='thumb'>$img<a href='#' id='".$count."' class='delete'>x</a>
          <div class='clear'></div><img width='100px' src='$dirname/$img' /></div>";

       }
                        }
                    } 

0 Cevap