I Have A Screwball PHP Problem (Resolved...Sort Of)

I know enough PHP to be dangerous. I am no expert, but I can do some
basic stuff.

Anyway, here's the problem.

I have a script that I created on a server that I've been using for several
years now. It works fine.

I wanted to take the script and move it over to another server on another
host, create an identical table and run it for another domain.

The code on both is identical. Yet, on the new server, I get the following
error:

Warning: extract() [function.extract]: First argument should be an array in /home/xxxxx/public_html/track.php on line 27

Now, I know what the problem is.

It's here:

$row=mysql_fetch_array($result);
extract($row);

$result is coming back empty. It shouldn't. I've gone over the whole code
and even put in echo test prints throughout the whole thing.

Bottom line: Things are not getting retrieved from the database.

Given that the code is the same, I then looked at my PHP version on
each. They say they're the same BUT... I am getting the following
differences in the server versions.

The one that works:

Server version: 5.0.67-community

The one that doesn't work

Server version: 5.0.51a-community

Everything else, including the MySQL version (4.1.22) is the same.

I have gone over everything. It's the only variable between the two
setups as the code is identical. It just won't run on the other server.

So...I have to figure out what changes I have to make in order to get
this code to work.

I don't even know where to begin.

I learned PHP 4 (I think 4.14 but not sure, been so long), it's what I am
using, and I don't know what I have to change the syntax of anything
(or everything) on my new server.

If somebody has a QUICK thing that I can look at, just throw some
suggestions in this thread.

Thanks.