|Problem with PHP, 'GetImageSize' function|
GetImageSize("photos/",$name,".jpg") gives 'Wrong parameter count ...'
| 11:03 pm on Jul 13, 2001 (gmt 0)|
I'm using this code:
<?php list($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?
| 7:11 am on Jul 14, 2001 (gmt 0)|
$foo = GetImageSize("photos$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)|
You can do this also
// information is the aray of your data
$information is the width
$information is the height
$information is the Image type GIF=1,JPG=2,PNG=3
$information is height=150 width=200 - for IMG tags
| 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 :)