[PHP] Problem setting variables.

[Jason Martyn]

I'm creating a script that filters the table of results from a database based on the 
configuration set by the user (config stored in database as well).

For example a table has 30 columns. However the user need only see 5. Every user is 
different so a user configuration is stored in the database.

The database has 2 columns: field_name and show_field. 
field_name is the name of the field (obviously) that is shown in the table.
show_field can either be a 1 or a 0. 1 = show field. 0 = don't show field.

I'm getting to a certain point. But I'm not sure how I can go about setting variables 
for each field_name that has a show_field value of 1.

Heres the code.
<?php
 function Interactive()
 {
 //Database Variables
 include ( "include/db_config.inc" );
 
 //Database connection.
 $connect = mysql_connect($host, $login, $passwd)
  OR die("Could not connect to MySQL Database: ".mysql_error()."");
 mysql_select_db("admin", $connect)
  OR die("Could not select Database: ".mysql_error()."");
 
 //Find User Configuration
 $sql = mysql_query("SELECT field_name FROM config WHERE show_field=1")
  OR die("Could not query database: ".mysql_error()."");
 $result = mysql_fetch_array($sql, MYSQL_ASSOC); 

 //Make sure there are fields to be shown.
 if(count($result) > 0)
 {


    //*****************************************************//
    // Here is where I would like to set variables.
    // I first thought of using a for statement, but
    // realized that variables cannot begin with a
    // number. Is there any possible way to set
    // variables for the array results of my query?
    //*****************************************************//

 }

else
 {
  die("Please select at least 1 field to be shown.");
 } 

 }
 
?>

Thanks,
Jason