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.
$imgFile=imagerotate('images/'.$id.'.gif', 180, 0);
then it will shows an error like this:
Warning: imagerotate() expects parameter 1 to be resource, string given.
how to use this function? can any one help regarding this.....
in same page i want to display another select box and another image as well without any disturbance in the other images.