imagecopyresampled返回null

Question

I have the following code:

$fullImg = $_POST['img'];
$avatar_name = $log_username.md5(uniqid()).'_avatar.jpg';
$avatar_path = '../user/'.$log_username.'/';
$avatar_new = $avatar_path.$avatar_name;
$target_h = $target_w = 150;
$x = $_POST['x'];
$y = $_POST['y'];
$w = $_POST['w'];
$h = $_POST['h'];
$qual = 90;

$orig_image = imagecreatefromjpeg($fullImg);
$new_image = imagecreatetruecolor($target_w,$target_h);

imagecopyresampled($new_img,$orig_image,0,0,$x,$y,$target_w,$target_h,$w,$h);

imagejpeg($new_image,$avatar_new,$qual);

I have an error in my error_log that says:

Warning: imagecopyresampled() expects parameter 1 to be resource, null given

I'm guessing it has something to do with my $_POST['img'] variable being a full url:

http://domain.com/images/image.jpg

I have set allow_url_fopen to 1 in my php.ini file. i thought this might fix it.

It is only a full url because it is read by js as var img = _('uloadedImg').src; and despite this being set as ../images/image.jpg, var img = _('uloadedImg').src; still returns a full url.

Is there a way to get the src from the image without the full url in js? Or is there something wrong with my script other than that?

Answer 1

There is a type error, you are specifying $new_img instead of $new_image:

$new_image = imagecreatetruecolor($target_w,$target_h);
imagecopyresampled($new_img,$orig_image,0,0,$x,$y,$target_w,$target_h,$w,$h);
                      ^ 

Try below:

imagecopyresampled($new_image,$orig_image,0,0,$x,$y,$target_w,$target_h,$w,$h);