Jpeg image Creating Problems with PHP GD library

Question

If i am trying to create new image from existing .jpeg image will gives black(blank) image.

My Code:

$new_path = 'post/'.time().'-myimage.jpeg';
$src="post/myimage.jpeg";
$jpeg_quality = 90;
$img_r = imagecreatefromjpeg($src);
$dst_r = ImageCreateTrueColor($my_post['w'], $my_post['h']);    
   imagecopyresampled($dst_r,$img_r,0,0,$my_post['x'],$my_post['y'],$my_post['w'],$my_post['h'],$my_post['w'],$my_post['h']);
    header('Content-Type: image/jpeg');
    imagejpeg($dst_r, $new_path, $jpeg_quality);

score 0 · Answer 1 · answered Sep 08 '14 at 12:34

0

If you use a filename in imagejpeg the jpeg isn't outputed to stdout - it's saved to the given file name. You have to use null as file name.

imagejpeg($dst_r, null, $jpeg_quality);

if you want also to save the picture, you have to call the function twice or redirect the file contents to the server with some function like readfile.

answered Sep 08 '14 at 12:34

Philipp

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when i trying to save jpg it works but not with jpeg and i also want to save .jpeg to new path but not works. – Rahul K Sep 08 '14 at 12:56
jpg und jpeg is the same – Philipp Sep 08 '14 at 13:16

Jpeg image Creating Problems with PHP GD library

1 Answers1