PhaZZed Posted August 14, 2007 Share Posted August 14, 2007 Greetings, I am writing a small script to convert from .jpg to .gif $filename = $_GET['file']; // always .jpg $type = $_GET['type']; $im = @imagecreatefromjpeg($filename); if ($im === false) { die('Unable to open image'); } if ($type == "gif") { header ('Content-Type: image/gif'); imagegif($im); imagedestroy($im); } This script displays the image on the browser page, whereas I would like to actually have a variable pointing to the relative path, ie. $location = "http://www.xx.com/uploads/image.gif" Any suggestions? I read that imagegif($im,$destination); can be used, but it's giving me errors.. Link to comment https://forums.phpfreaks.com/topic/64812-solved-convert-image-and-save-to-disk/ Share on other sites More sharing options...
PhaZZed Posted August 14, 2007 Author Share Posted August 14, 2007 Solution.. $filename = $_GET['file']; $type = $_GET['type']; $im = @imagecreatefromjpeg($filename); if ($im === false) { die('Unable to open image'); } srand((double)microtime()*1000000); $rand = rand(999999,99999999); $dest = "../temps/" . $rand; if ($type == "gif") { $dest .= ".gif"; imagegif($im,$dest); imagedestroy($im); $filename = $dest; } if ($type == "png") { $dest .= ".png"; imagepng($im,$dest); imagedestroy($im); $filename = $dest; } Link to comment https://forums.phpfreaks.com/topic/64812-solved-convert-image-and-save-to-disk/#findComment-323349 Share on other sites More sharing options...
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