How about solving both problems at once? :-)
Yes, go with the N:N (the technical term for that car_option table)
relation, *AND* give yourself a weighted search engine to boot!
create table car (car_id auto_increment...);
create table option (option_id auto_increment...);
create table car_option(car_id int4, option_id int4, key(option_id));
insert into car(name) values('Chevy Nova');
insert into car(name) values('Lexus');
insert into option(name) values('a/c');
insert into option(name) values('power window');
insert into option(name) values('power locks');
insert into car_option(car_id, option_id) values(1, 1);
insert into car_option(car_id, option_id) values(2, 1);
insert into car_option(car_id, option_id) values(2, 2);
insert into car_option(car_id, option_id) values(2, 3);
NOTE: The Lexus has all those options (at that price it had better!) and
the Nova, well... It has a/c! :-)
Now, assume the user wants everything, so you have:
$options[1]
$options[2]
$options[3]
all "set" to some value by your checkbox.
#Untested code.
$query = "select count(car_id), car.name from car, car_option ";
$query .= " where car.car_id = car_option.car_id "
$query .= " and (1 = 0";
# 1 = 0 is a "starter yeast" for the following:
while (list($option_id) = each($options)){
$query .= " or option_id = $option_id ";
}
# finish off our options OR list...
$query .= ")";
$query .= " group by car_id ";
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