Q:

PHP get image size

I'm using the following code but it gives me an error about an unexpeted " in the statement at this line, but I'm not sure how the syntax should go:
list($width,$height,$type,$attr) = getimagesize("' . $SESSION_["html_folder"] . '/uploadedfiles/' . $row['logo'] . '");

echo "<p>This logo is ".$width; x $.height; echo "pixels in size.</p>";

PHP says the error is on that first line.

A:

Try this
list($width,$height,$type,$attr) = getimagesize($SESSION_['html_folder'] . '/uploadedfiles/' . $row['logo']);

The error in your code was that you opened the string "twice", one with " and one with '. If you use a variable as parameter where a string is expected, you do not need to set the variable in quotes.
You should make sure that in $SESSION_['html_folder'] there's no malicious code, e.g. with
if(!is_dir($SESSION_['html_folder']))
   die("ERROR");

