Norman_4

how to split the ajax output response in different dropdown???

I have problem updating the mySQL table i have simple loop to display the rows in the table in HTML FORM but when i try to update the row notting happens, no row is updated! here is the code <form action="" method="post"> <?php $sql = "SELECT ID, votemodelName, votemodelImage, modelLink, …

Hello i have a problem. It seems that the value of this this input `<input type="hidden" id="uploadUid" value="<?php echo $sessionUid;?>" name="update_uid">` in here `formData.append('uploadUid', uploadUid.value);` When i put an alert `alert(uploadUid.value);` on Firefox it returns just fine but on Chrome returns undefined Any guess why?