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Problem with PHP, 'GetImageSize' function

GetImageSize("photos/",$name,".jpg") gives 'Wrong parameter count ...'



11:03 pm on Jul 13, 2001 (gmt 0)

10+ Year Member

I'm using this code:

<?php list($width, $height) = GetImageSize("photos/",$name,".jpg"); ?> <----[line 3]

<img src="photos/<?php echo $name; ?>.jpg" width="<?php echo $width; ?>" height="<?php echo $height; ?>" id="photo">

And getting this error:

Warning: Wrong parameter count for getimagesize() in c:\-= web-sites =-\in progress\photography\site\photo.php on line 3

If I change the ("photos/",$name,".jpg") part to ("photos/THEFILENAME.jpg"), it works fine. Is there something else I should be using instead of commers to seperate the parts? Or does it need to be done another way?



7:11 am on Jul 14, 2001 (gmt 0)

WebmasterWorld Senior Member 10+ Year Member

Try this:

$foo = GetImageSize("photos$name.jpg");

There's no need to split up the path, filename and extension with commas - PHP can interprate a variable within a string just fine.

Does that work?


2:06 pm on Jul 14, 2001 (gmt 0)

10+ Year Member

You can do this also
// information is the aray of your data

$information[0] is the width
$information[1] is the height
$information[2] is the Image type GIF=1,JPG=2,PNG=3
$information[3] is height=150 width=200 - for IMG tags


11:04 pm on Jul 14, 2001 (gmt 0)

10+ Year Member

Thanks, guys. I managed to find a solution at php.net (spose that's where I should have looked in the first place). And replaced the commers with fullstops, so it counts as one string. There seems to be about 20 ways to string a thing :)

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