I want to upload an image from disk, resize it, and then upload it to Amazon S3.
However, I cant get the proper image output from imagejpeg().
heres my code:
$sourceUrl = $_FILES['path']['tmp_name'];
$thumbWidth = '100';
$thumbid = uniqid();
$img = imagecreatefromjpeg($sourceUrl);
$width = imagesx( $img );
$height = imagesy( $img );
// calculate thumbnail size
$new_width = $thumbWidth;
$new_height = floor( $height * ( $thumbWidth / $width ) );
// create a new temporary image
$tmp_img = imagecreatetruecolor( $new_width, $new_height );
// copy and resize old image into new image
imagecopyresampled($tmp_img, $img, 0, 0, 0, 0, $new_width, $new_height, $width, $height);
// output the image
imagejpeg($tmp_img);
// upload thumbnail to s3
$s3->putObjectFile($tmp_img, "mybucket", $thumbid, S3::ACL_PUBLIC_READ);
Firebug gives me this error :
illegal character
[Break on this error] (�����JFIF���������>CREATOR: g...(using IJG JPEG v62), default quality\n
If I modify imagejpeg this way,
imagejpeg($tmp_img, 'abc.jpg');
then I get the same error. 🙁
Can i get some help here please ?
If you check the documentation of
imagejpegyou can see it outputs the image, it means the way you call it it gets sent to the browser. You can get it to save to a file the second way you call it – by passing a filename in the second parameter.Also,
$tmp_imgis an image resource, not a ready-to-use image file.I don’t know how your upload function works, but: if you need the file contents to upload, do it like this:
if you need a filename to upload: