Question

I am using an ajax call to pass a value. I have done that and I want to take that value and do a lookup on my tables for a particular image.

When I get to the image to download, it is not working for me. I looked on my php.ini file and my allow_furl_open is off. I think this may be the culprit. Please disregard the mysql bits as I know it should be updated, this is not my code so that will be changed.

Here is the code.

if (isset($_POST['download'])) {
$sql = "SELECT `imageURL` FROM `digital_materials` WHERE `digital_materials`.`id` = '$id'";
$record = mysql_query($sql);
$row = mysql_fetch_assoc($record);

$fileName = $row['imageURL'];
$fileURL = 'images/products/' . $fileName;
header('Content-Type: application/octet-stream');
header("Content-Transfer-Encoding: Binary");
header('Content-Disposition: attachment; filename=' . $row['imageURL']);
readfile($fileURL);
exit();

}

This should be working. Any problems?

Answer 1

As mentioned in the comments by Marc B, you can’t download a file through ajax. Ajax will just return the network request to the sender (i.e. jQuery ajax has a success handler function that contains the result of the request: http://api.jquery.com/jQuery.ajax/). One possible solution is to open a new window with the php url which will trigger the download request. A javascript solution is:

window.open('dynamic-image-url.php');

Another solution is using an anchor tag with the target=”_blank” attribute.

<a href="dynamic-image-url.php" target="_blank">Download Image</a>

The following is a php solution for returning an image taken from php’s readfile documentation:

<?php
$file = 'monkey.gif';

if (file_exists($file)) {
    header('Content-Description: File Transfer');
    header('Content-Type: application/octet-stream');
    header('Content-Disposition: attachment; filename='.basename($file));
    header('Content-Transfer-Encoding: binary');
    header('Expires: 0');
    header('Cache-Control: must-revalidate');
    header('Pragma: public');
    header('Content-Length: ' . filesize($file));
    ob_clean();
    flush();
    readfile($file);
    exit;
}
?>

EDIT ** Added Form example which will post the variable

<form action="dynamic-image.php" method="post" target="_blank">
   <input type="hidden" name="id" value="<?php echo $value; ?>" />
   <input type="submit" value="Download Image" />
</form>

You would then reference the variable as $_POST[‘id’] because that is the value of the name attribute in the form.

Using a $_GET parameter, you would do it as below:

<a href="dynamic-image-url.php?id=<?php echo $id; ?>" target="_blank">Download Image</a>

Use $_GET[‘id’] to reference the value in PHP as id is the name of the query string parameter.