php getimagesize errors

Question

Im trying to get the image size for an image in my server and the data is in a database or my getimagesize function looks like this

$size = getimagesize(upload/$array['image']);
print_r($size);

I get these errors back...

Warning: Division by zero in /home/content/44/8713044/html/view/home/home.html on line 81

Warning: getimagesize() [function.getimagesize]: Filename cannot be empty in /home/content/44/8713044/html/view/home/home.html on line 81

the image is in the right place...what am I doing wrong?

note that many times, the first error is the error that matters. in this case it is telling you that you attempted to divide a number by zero (see the answers below). It seems obvious that you did not mean to divide. — horatio
– horatio, Commented May 3, 2012 at 15:45

Wiseguy · Accepted Answer · 2012-05-03 15:43:25Z

3

Your path isn't in quotes, so it's not a string.

$size = getimagesize(upload/$array['image']);

It's trying to mathematically divide a constant named upload by $array['image'].

Use quotes and concatenate like this:

$size = getimagesize('upload/' . $array['image']);

answered May 3, 2012 at 15:43

Wiseguy

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ajreal · Accepted Answer · 2012-05-03 15:43:30Z

1

you should quote the string like this :

getimagesize("upload/{$array['image']}");

otherwise, PHP will treat this as a mathematic expression

this url might help :- http://www.php.net/manual/en/language.expressions.php

answered May 3, 2012 at 15:43

ajreal

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