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Editorial Team
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Asked: 2026-06-02T06:43:03+00:00

I was trying to send a single file to the server without having to

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I was trying to send a single file to the server without having to change the webpage. I decided to use this plugin: http://cmlenz.github.com/jquery-iframe-transport/
I am very confused on how to proceed for an upload without changing the webpage…
the uploadimg.php returns: 

<br />
<b>Notice</b>:  Undefined index: photo in <b>C:\website\uploadimg.php</b> on line <b>2</b><br />
<br />
<b>Notice</b>:  Undefined index: photo in <b>C:\website\uploadimg.php</b> on line <b>3</b><br />
<img src="userimg/" />

My files:

postAd.php

<html>
<head>
<script src="jquery.js"></script>
<script src="jquery-iframe-transport.js"></script>
<script>
$(document).ready(function(){
    $('#photo1').on('change', function(){
        $.ajax({
            url: "uploadimg.php",
            files: $("#photo1:file"),
            iframe: true,
            processData: false
        }).done(function(data){
            console.log(data);
        });
    });
});
</script>
</head>
<body>
<input type='file' name='photo' id='photo1' /></br>
</body>
</html>

uploadimg.php

<?php 
$file_tmp = $_FILES['photo']['tmp_name'][0];
$file = $_FILES['photo']['name'][0];
move_uploaded_file($file_tmp, "userimg/$file");
echo '<img src="userimg/' .$file .'" />';
?>
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1 Answer

  1. Editorial Team

    You are not submitting a form, so the ‘photo’ field is not going to be sent with the AJAX call. If you have a debugger available, check the $_FILES array to see what got populated. From this, it appears that you might be looking for a ‘userfile’ index, but I’m not sure.

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