манипуляции с изображениями
с мануала взял код он уменьшает размер изображения и выводит на экран как jpg картинку
а как сделать чтобы он сохранял эту картинку
Код:
<?php
// File and new size
$filename = 'test.jpg';
$percent = 0.5;
// Content type
header('Content-type: image/jpeg');
// Get new sizes
list($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);
?>
// File and new size
$filename = 'test.jpg';
$percent = 0.5;
// Content type
header('Content-type: image/jpeg');
// Get new sizes
list($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);
?>
а как сделать чтобы он сохранял эту картинку
RTFM твою налево!!!!
Description
bool imagejpeg ( resource image [, string filename [, int quality]] )
imagejpeg() creates the JPEG file in filename from the image image. The image argument is the return from the imagecreatetruecolor() function.
The filename argument is optional, and if left off, the raw image stream will be output directly. To skip the filename argument in order to provide a quality argument just use a NULL value. By sending an image/jpeg content-type using header(), you can create a PHP script that outputs JPEG images directly.
Цитата:
Description
bool imagejpeg ( resource image [, string filename [, int quality]] )
imagejpeg() creates the JPEG file in filename from the image image. The image argument is the return from the imagecreatetruecolor() function.
The filename argument is optional, and if left off, the raw image stream will be output directly. To skip the filename argument in order to provide a quality argument just use a NULL value. By sending an image/jpeg content-type using header(), you can create a PHP script that outputs JPEG images directly.