I think there has got to be an easy solution to this one:

I have two PHP programs. One (we'll call it progy1) displays the information from a database without conditions. My other program (progy2) has been writen by me to check the database records and check to see if a record should get displayed by progy1 (progy2 goes through all the records using a loop, checks that a field matches the area the user is in, and if there is a match I do a include/require of progy2).

Both programs run fine on their own. However, the call for progy1 from progy2 causes progy1 to die after the first matching record is found, and the following error is returned:

Warning: 1 is not a valid MySQL-Link resource in /home/site/web/poll.php on line 108

Couldn't execute query: 31

poll.php is progy2. Any ideas how I can call/run progy1 in progy2 without running into this problem?

from what little I know.. make sure first that you are calling it at the right location (I suggest first try with them in the same directory, and just putting the filename in the include).. second.. if progy1 is only used when called in progy2, make sure not to have any <?php or ?> commands in progy1...

Put in the full URL so that Prog1 executes before being requested.