getimagesize returns empty array

[Alex_B]

Hello,
I am having problem with getimagesize() function.
In the script I was trying to get image size( height and width)

My code looked this way
$size = getimagesize($image_name);
$width = $size[0];
$height = $size[1];

both $width and $height returns empty string.

I have looked at

http://us3.php.net/manual/en/function.getimagesize.php

Also I tried to use function that they offer
sgetimagesize(filename) and that did not work either.

Images are not large
Any other suggestions?

[paulsbooker]

does your form tag have an enctype which allow the uploading of images ?

paul

[Weedpacket]

Is $image_name the name of a file? And is that file an image? Is the image a recognised format?

[Alex_B]

Thank you for the reply,
I am not using forms there at al. I am using images as links to somewhere else. DOes that function works only with forms?

Also
$image_name is the name of the file. The file is image and the format of it is jpeg format.

[macewan]

list($width, $height)=getimagesize("$image_name");

then echo it out

<img src=\"$image_name\" width=\"$width\" height=\"$height\" alt=\" \" border=\"0\" />

[Weedpacket]

Try
if(!file_exists($image_name)) echo "File not found.";

(I.e., is PHP looking where you want it to?)

Then

$size=getimagesize($image_name);
var_dump($size);

I have seen earlier versions of PHP (I forget which; it was a while ago) that weren't able to recognise certain jpegs (from Casio cameras in this case). That's the only other reason I can think of right now for a failure.

[Alex_B]

It does not matter if I use list(height, weight). The problem is that function getimagesize() returns null.

[Alex_B]

FOund the problem. As always - syntax error. Thank you for all the help
Kindly
Alex