I am using following fucntion to get proportionate width for user specific height

However I noticed that in case image name is having space it gives an error
Example Function works ok for image name say "pict1.jpg"

However gives error for
pict 1.jpg (space in name beforer 1)

list($width, $height, $type, $attr) = getimagesize(addslashes($imagename));
if ($width ==0)
{$imgstr="No Image"; }
else
{
$h=$height*($reqwidth/$width) ;
if ($width > $reqwidth)
{ $imgstr="<img src='".addslashes($imagename)."' width=".$reqwidth." height=".$h." border=0 >";}
else
{ $imgstr="<img src='".addslashes($imagename)."' width=".$width." height=".$height." border=0 >";}
}
return $imgstr;
}

The space in the name is most likely making the file inaccessible. Is the problem with actually getting the file or when php returns the $imgstr ?

Either way, try adding a rawurlencode() in there http://uk2.php.net/manual/en/function.rawurlencode.php

Did u try getimagesize($imagename) without addslashes()?