Evo ovako. Ja zelim datoteku $r da nalepim na onaj background ($imBack) , ali da bude ova slika ispred kao original.
Code:
<?
//ime datoteke
$r = $_GET['r'];
//load "image.jpg" from disk - you must have it in same dir as script
$imBack = @ImageCreateFromJPEG ('Rback.jpg');
$imBackWidth = ImageSX($imBack);
$imBackHeight = ImageSY($imBack);
$im = @ImageCreateFromJPEG ($r);
$imWidth = ImageSX($im);
$imHeight = ImageSY($im);
//ustvarim nov image prazen
$imNew = @ImageCreate (640, 500);
$imNewWidth = ImageSX($imNew);
$imNewHeight = ImageSY($imNew);
//dolocim barvo
$black = ImageColorAllocate ($imNew, 0, 0, 0);
//prazen image napolnim z crno barvo - ustvarim pravokotnik
ImageFilledRectangle ($imNew, 0, 0, $imNewWidth, $imNewHeight, $black);
//test pozadine
ImageCopy ($imNew, $imBack, 0, 0, 0, 0, $imBackWidth, $imBackHeight);
//copy original image to new image
ImageCopy ($imNew, $im, (($imNewWidth/2)-($imWidth/2)), ((($imNewHeight/2)-($imHeight/2))-20), 0, 0, $imWidth, $imHeight);
//ImageCopy ($imNew, $im, 0, 0, 0, 0, $imWidth, $imHeight);
imagejpeg($imNew,"RbackOutput.jpg", 100);
echo "<h1>Zakaj tak grdo???</h1><br/>";
echo "<h2>Sorce code:</h2><br/>";
show_source($DOCUMENT_ROOT."/".$PHP_SELF);
echo"<br/>";
echo "Predelana:</br>";
echo "<img src=RbackOutput.jpg><br/><br/>";
echo "Original:<br/>";
echo "<img src=".$r."><br/><br/>";
?>
I tako dalje sve do smaka sveta ...
gd quality problem
