How to retrieve a file name - php

Question

How can I get it so that when the user clicks on the "Cancel" button, then in the cancel button function where it contains the variable var image_file_name, it retrieves the name of the file which was entered within the file input? Then hopefully the $GET method in the cancelimage.php page can be used to retrieve the file name from the function?
Below is my attempt but it is not retrieving a file name. It is just retrieving a blank.
Below is the form code and javascript function which controls the "Cancel" button:
var $fileImage = $("<form action='imageupload.php'
method='post'enctype='multipart/form-data' target='upload_target' onsubmit='return
startImageUpload(this);' class='imageuploadform' ><label>" +
"Image File: <input name='fileImage' type='file' class='fileImage' /></label><br/>
<br/><label class='imagelbl'>" +
"<input type='submit' name='submitImageBtn' class='sbtnimage' value='Upload'/>
</label>" +
"<input type='button' name='imageCancel' class='imageCancel' value='Cancel'/>
</label>" +
"<iframe class='upload_target' name='upload_target' src='#' style='width:0;height:0;border:0px;solid;#fff;'></iframe></form>")
...
var _cancelimagecounter = cancelimagecounter;
$(".imageCancel").click(function() {
var image_file_name = <?php echo json_encode($image_file_name); ?>;
jQuery.ajax("cancelimage.php?fileImage=" + image_file_name)
.done(function(data) {
$(".imagemsg" + _cancelimagecounter).html(data);
});
return stopImageUpload();
});
Below is the cancelimage.php script where it uses the $GET method to get the file name from the var image_file_name:
<?php
...
$image_file_name = $_GET["fileImage"];
echo "File Upload was Canceled";
$imagecancelsql = "DELETE FROM Image
WHERE ImageFile = 'ImageFiles/". mysql_real_escape_string($image_file_name)."'";
mysql_query($imagecancelsql);
mysql_close();
?>

Answer 1

Filename is far form the issue here.... But your fileName is not displaying because you are trying to use $_GET["fileImage"];
The Form was submitted via enctype='multipart/form-data and <input name='fileImage' type='file'
You should use $_FILES instead ...
Your file name would be
$image_file_name = $_FILES['fileImage']['name'] ;