When I pass the variable as below..
var file_path = "/admin/upload/rec1.20130522_003928.f4v";
...
ifrm.src = download.php?file_path=" + file_path;
It shows the link as
_admin_upload_rec1.20130522_003928.f4v
(Capture: http://screencast.com/t/ooQHLdr5cixz)
How can I solve this? Thanks
Presumably, in the script where you send the file, you are also sending a Content-Disposition header that tells the browser that this is a download, not a resource to try and show immediately. I'd presume you're also sending $_GET['file_path'] as the filename in that header.
When you send that header, don't include the directory part. Browsers will do whatever they want with the directory part, from stripping it off to mangling it. Pretty much the only thing they won't do (for security reasons) is actually use it as a directory name.
You can send the base filename using something like
$path_parts = preg_split(':[/\\]+:', $_GET['file_path']);
$filename = addslashes(end($path_parts));
header("Content-Disposition: attachment; filename=\"$filename\"");
Try this: var file_path = "\admin\uploadec1.20130522_003928.f4v"; ... ifrm.src = download.php?file_path=" + file_path; instead.