Running PHP Program From Another PHP Program

[bytech]

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?

[insitedev]

Can you post up some code for us to look at? It could be something along the lines of your query not being valid, or your overwriting your result variable. That, and it is saying "1 is not a valid...". Are you trying to set it to a numeric variable?

It could also be that your connection resource variable isn't defined in the function that you are using (assuming that you are using functions).

for instance, if you are using $dbh for your mysql_connect variable inside of a function, then calling something that uses $dbh from another function, it might not have access to it. If that is the case, add global $dbh; to the top of that second function.

If we could see the code that you are using, it would really help out.

[LazyJim]

If you're trying to loop through all the records and only display certain ones, then there's probably better ways.

Specifically, take the part of progy1 that makes the decision and make it into a function (coded into progy2) that is called from each iteration of the loop.

(that is if it cannot be done in SQL)