Welcome to WebmasterWorld Guest from 54.234.8.146

Forum Moderators: coopster & jatar k

Message Too Old, No Replies

Problem with PHP, 'GetImageSize' function

GetImageSize("photos/",$name,".jpg") gives 'Wrong parameter count ...' Soupisgoodfood 11:03 pm on Jul 13, 2001 (gmt 0) I'm using this code: -- <?php list($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?

Thanks,
Justin.

sugarkane

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

Try this:

$foo = GetImageSize("photos$name.jpg");
$width=$foo[0];
$height=$foo[1];

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?

David

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

You can do this also
$photo="photo/filename.jpg"; getimagesize($photo,information);
// 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

Soupisgoodfood

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

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 :)