Re: Displaying Results

Sorry, I just noticed that the count() function will do at least the first part of my question, i.e. SELECT year, COUNT(*) FROM cars GROUP BY year But the second part still has me a bit stumped. I know that you can pass a variable using something like script.php?year=1991, but doesn't that assume that in your script you have something like: $query = ("SELECT * FROM mytable WHERE year=$year") So the $year is the only variable that gets replaced by the script.php? year=1991 you called it with. How could I make it so that the entire 'year=$year' part of the query gets replaced by what comes after the ? in script.php?year=1991 ? Thanks in advance Jason ----- Original Message ----- From: Jason Soza <solja@gci.net> Date: Tuesday, April 16, 2002 2:49 pm Subject: [PHP] Displaying Results > Suppose I want a script that goes into a table, looks up all > records in > a given field, and groups and displays the results like so: > match1 (1) > match2 (5) > match3 (6) > > So basically it looked at the field 'matches' and found a total of > 12 > records, 1 record for match1, 5 records for match2, and 6 records > for > match3. It would also be handy for the output to be a link so when > you > clicked on it, it would pass a query on to the database querying > for > just that particular field, like: > SELECT * FROM table WHERE match="match1" > > Can someone point me in the right direction for this? Or show me > an > example script that would accomplish this? Any help would be > great. > Thanks a lot! > > Jason > > > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php > >