使用PHP生成多个图像

Question

Code

$files = scandir("images");
$exclude = array(".", "..");

$images = array_diff($files, $exclude);

foreach($images as $image) {

    $original_image  = imagecreatefromjpeg("images/{$image}");
    $original_width  = imagesx($original_image);
    $original_height = imagesy($original_image);

    $new_width  = 180;
    $new_height = floor($original_height * ($new_width/$original_width));
    $new_image  = imagecreatetruecolor($new_width, $new_height);

    imagecopyresampled($new_image, $original_image, 0, 0, 0, 0, $new_width, $new_height, $original_width, $original_height);

    header("Content-Type: image/jpeg");

    imagejpeg($new_image);
}

Problem

The resizing part of the code works just fine, but it only outputs the first resized image. How can I make it so that it dumps all of the resized images?

Answer 1

You can't do such the way you are doing it. imagejpeg "prints" the bytes of the image (which is why, combined with the proper header, you see that picture in your browser). However, printing multiple images (plus trying to change the header) won't work, as you are "modifying" the bytes of the first image, and not appending new images to the output. If you want to use a solution like that, you could:

1. Save every image in your server and then access those images in another way (either by printing <img> tags pointing to the pictures or throwing some kind of structure listing the images).

2. If you don't want to save them, you can always encode the bytes using base64 and then using that as an <img src>:

   ob_start ();
   imagejpeg ($new_image);
   $image_data = ob_get_contents ();
   ob_end_clean ();
   $image_data_base64 = base64_encode ($image_data);
   echo '<img src="data:image/jpg;base64,'.$image_data_base64.'" />';
   

You should do so every time.

And very important: always use imagedestroy($new_image) after outputting (or finishing your work with) an image object in php: it will release resources, so you won't receive a "Memory exhausted" error that easy.