Accessing info about file uploaded in PHP and modifying with GD

Question

I am uploading an image that is included in the $_FILES object. However, when I attempt to get its size, I get the error that it is not a string or alternatively not a resource. How can I get the size of this image and, in turn, modify it in GD? Do I have to convert it to a string? Or what do these methods want as their input?

$image = $_FILES['uploaded_file'];
var_dump($image);

$oldw = imagesx($image);
$oldh = imagesy($image);

$imagedetails = getimagesize($image);
$width = "width".$imagedetails[0];
$height = "height".$imagedetails[1];
$neww = 512;
$newh = 512;
$temp = imagecreatetruecolor($neww, $newh);
imagecopyresampled($temp, $image, 0, 0, 0, 0, $neww, $newh, $oldw, $oldh);

//Error messages:

imagesx() expects parameter 1 to be resource
imagesy() expects parameter 1 to be resource
getimagesize() expects parameter 1 to be string

//Here is what the var_dump of image 
array(5) {
  ["name"]=>
  string(14) "image.png"
  ["type"]=>
  string(24) "application/octet-stream"
  ["tmp_name"]=>
  string(14) "/tmp/phpLlon22"
  ["error"]=>
  int(0)
  ["size"]=>
  int(2743914)

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– halfer, Commented Jan 12 at 22:37

Musa · Accepted Answer · 2025-01-13 23:59:06Z

2

Both imagesx and imagesy require an opened image, i.e. already loaded into memory.
getimagesize requires a filename but $image is an array. You have to pass $_FILES['uploaded_file']['tmp_name'] to it.
Also, you have to create an image resource from the uploaded image before you can resample it.

$imagefile = $_FILES['uploaded_file']['tmp_name'];
var_dump($image);

$imagedetails = getimagesize($imagefile);
$oldw = $imagedetails[0];
$oldh = $imagedetails[1];

$width = "width".$imagedetails[0];
$height = "height".$imagedetails[1];
$neww = 512;
$newh = 512;
$image = imagecreatefromstring(file_get_contents($imagefile));
$temp = imagecreatetruecolor($neww, $newh);
imagecopyresampled($temp, $image, 0, 0, 0, 0, $neww, $newh, $oldw, $oldh);

edited Jan 13 at 23:59

answered Jan 11 at 22:45

Musa

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2 Comments

user6631314 Jan 12 at 22:34

I will give it a try. Out of curiosity, where is the actual image data?

user6631314 Jan 13 at 17:26

imagedetails in lines 4 and 5 need $ signs but with that correction this worked. I guess, the $_FILES array does not have the actual data, only the location from which you can get it as a string using imagecreatefromstring(file_get_contents($imagefile)

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