saving imagejpeg output the file

Question

Is there something I am missing here? If I change imagejpeg($thumb, $newImage); to imagejpeg($thumb); it echos a load of unreadable characters. The thumb image directory exists.

My ultimate aim is to copy an image at a lower quality.

$filename = $imageDirectory . '/1.jpg';
$percent = 0.5;
$newImage = $imageDirectory . '/thumbs/1.jpeg';    
echo "image: <br>";
echo $filename;
list($width, $height) = getimagesize($filename);
$newwidth = $width * $percent;
$newheight = $height * $percent;    
// Load
$thumb = imagecreatetruecolor($newwidth, $newheight);
$source = imagecreatefromjpeg($filename);    
// Resize
imagecopyresized($thumb, $source, 0, 0, 0, 0, $newwidth, $newheight, $width, $height);    
// Output
imagejpeg($thumb, $newImage);

UPDATE: I now realize that the second parameter must be an image location with the new name. So... I have redefined $newImage. The path is fine... if I upload an image named 1.jpg to that location manually it exists at that path.

Answer 1

You'll have to add the correct Content-Type header for the response if you want to output the image to the browser:

header('Content-Type: image/jpeg');
imagejpeg($yourimage);

Please check the first example in the documentation for more information. If you want to decrease the quality before outputting to your browser check the third example.