Hi all,
I want to display an image with 180 degrees (rotate) an image in my web page, that to getting the image URL from database dynamically, based on select box selection.
But i used the link like this:::
getting $id from select box.
I used like that.
but the same error it is showing on this line.: imagerotate() expects parameter 1 to be resource, string given
$imgFile=imagerotate($image, 180, 0);