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Home/ Questions/Q 8547335
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Editorial Team
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Editorial Team
Asked: June 11, 20262026-06-11T13:14:40+00:00 2026-06-11T13:14:40+00:00

I have the following code which is supposed to return a random string back

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I have the following code which is supposed to return a random string back to the form once a file is uploaded. Previously to adding the onUploadSuccess code, it was working OK, but if I tried to upload multiple files, each file would have the same ID. I fixed this using the PHP to generate the random ID, but now I’m having issues getting the form to update.

The ID I’m trying to update is “data”, but when I upload the files, I get a Uncaught TypeError: Object [object Object] has no method 'call' error.

If someone could point me in the right direction, it would be appreciated. If you need more info, let me know.

The Javascript:

<script>
        var sessid = ''; 
        $(document).ready(function() {

            $('#myModal').modal({show: false});
            $('#mm').modal({show: false});

                $('#file_upload').uploadify({
                    'fileObjName': 'file',
                    'fileSizeLimit': '8MB',
                    'buttonText': 'BROWSE FILE(S)...',
                    'fileTypeExts': '*.JPEG; *.GIF; *.PNG; *.APNG; *.TIFF; *.BMP; *.PDF; *.XCF',
                    'cancelImg': 'uploadify-cancel.png',
                    'swf': 'uploadify.swf',
                    'uploader': 'uploadify.php',
                    'auto': false,
                    'onUploadSuccess': $("data").livequery(function(file, data, response){
                            document.getElementById("data").innerHTML=data;
                    })
                });
          });
    </script>

The PHP:

<?php
if ( is_uploaded_file( $_FILES['file']['tmp_name'])) {

$tempFile = $_FILES['file']['tmp_name'];
$fileParts = pathinfo($_FILES['file']['name']);
$randName = substr(sha1_file($_FILES['file']['tmp_name']), rand(0,30), 7);
$targetFile = 'uploads/'.$randName.'.' . $fileParts['extension'];

move_uploaded_file($tempFile,$targetFile);
echo $randName;
} else {
echo 'Malformed data';
}
?>
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-11T13:14:41+00:00Added an answer on June 11, 2026 at 1:14 pm

    I think that your problem is that you need to supply the function definition to the “onUploadSuccess” option; get rid of the $(“data”).livequery( ).

    Use:

    'onUploadSuccess': function(file, data, response){ 
                                document.getElementById("data").innerHTML=data; 
                        }
    
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