上载到服务器之前裁剪图像

Question

So I've got this PHP script to scale and crop into a square from the center;

<?PHP
//resize and crop image by center
function resize_crop_image($max_width, $max_height, $source_file, $dst_dir, $quality = 80){
    $imgsize = getimagesize($source_file);
    $width = $imgsize[0];
    $height = $imgsize[1];
    $mime = $imgsize['mime'];

    switch($mime){
        case 'image/gif':
            $image_create = "imagecreatefromgif";
            $image = "imagegif";
            break;

        case 'image/png':
            $image_create = "imagecreatefrompng";
            $image = "imagepng";
            $quality = 7;
            break;

        case 'image/jpeg':
            $image_create = "imagecreatefromjpeg";
            $image = "imagejpeg";
            $quality = 80;
            break;

        default:
            return false;
            break;
    }

    $dst_img = imagecreatetruecolor($max_width, $max_height);
    $src_img = $image_create($source_file);

    $width_new = $height * $max_width / $max_height;
    $height_new = $width * $max_height / $max_width;
    //if the new width is greater than the actual width of the image, then the height is too large and the rest cut off, or vice versa
    if($width_new > $width){
        //cut point by height
        $h_point = (($height - $height_new) / 2);
        //copy image
        imagecopyresampled($dst_img, $src_img, 0, 0, 0, $h_point, $max_width, $max_height, $width, $height_new);
    }else{
        //cut point by width
        $w_point = (($width - $width_new) / 2);
        imagecopyresampled($dst_img, $src_img, 0, 0, $w_point, 0, $max_width, $max_height, $width_new, $height);
    }

    $image($dst_img, $dst_dir, $quality);

    if($dst_img)imagedestroy($dst_img);
    if($src_img)imagedestroy($src_img);
}
//usage example
resize_crop_image(100, 100, "test.jpg", "test.jpg");p_image(100, 100, "test.jpg", "test.jpg");
?>

You can simply call the following function:

resize_crop_image(100, 100, "test.jpg", "test.jpg");p_image(100, 100, "test.jpg", "test.jpg");

Added to JSFiddle is my HTML5/JQuery preview file before uploading to the server.

1). Do I need to upload the image to the server before running this script?

2). If needing to prior upload, how can I use my form to upload into a temp location, do the job and move to a specific dir and delete the temp dir?

Answer 1

1) Yes, a copy of the image will need to be on server before you can edit/crop it. 2) Uploaded files are automatically stored in the temp directory (they are usually copied out to use the image, but don't have to be). They can be read as an image from the temp directory by your code and PHP will also automatically clean up the file at the end of the script.

When a file gets uploaded, in the $_FILES array there is a key with the tmp name. You can just pass that into your function as the source file and it should work with imagecopyresampled with no issue. The temp name will be something like $_FILES['nameFromFileFieldOnForm']['tmp_name'] .

Answer 2

if your source file or destination file is

$source_file = www.example.com/storage/packages/image.jpg

change it to

$source_file = ./storage/packages/image.jpg

It works for me in laravel 5