unknown365 Posted July 7, 2009 Share Posted July 7, 2009 Hi! I have a PHP function which I use to resize the images that I get from the database. These are the functions inside my getImage.php: function resize($image){ $background = imagecreatetruecolor(116,84); $source = imagecreatefromjpeg($image); $width= imagesx($source); $height = imagesy($source); header("Content-type: image/jpeg"); if($width>$height){ imagecopyresampled($background,$source,0,0,0,0,116,84,$width,$height); imagejpeg($background); imagedestroy($background); } else{ imagefill($background,0,0,imagecolorallocate($background, 255, 255, 80)); imagecopyresampled($background,$source,20,0,0,0,75,84,$width,$height); imagejpeg($background); imagedestroy($background); } } function getImage($id){ $query="SELECT i.filepath FROM users u, items i WHERE u.userid=i.userid AND u.userid=$id"; $res = mysql_query($query); while ($obj = mysql_fetch_object($res)) { $file = $obj->filepath; return resize($file); } } Basically, I don't have any problems with resizing the image. My problem is that whenever I try to call this function to output in another page either by using <img src="<?=getImage($id)?>" border="0"> OR <img src="getImage.php" border="0">, the image does not show itself. It's as if I didn't call any image. Please help! By the way, I really need the getImage.php to output the image to another page. Link to comment https://forums.phpfreaks.com/topic/165038-using-libgd-to-output-an-image-in-another-page/ Share on other sites More sharing options...
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