Re: Displaying Results

On Wednesday 17 April 2002 08:21, Jason Soza wrote: > 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 ? You can do something like: script.php?qry=year%3D1991 %3D is the url encoding for '='. Then in your script: $query = ("SELECT * FROM mytable WHERE $qry") -- Jason Wong -> Gremlins Associates -> www.gremlins.com.hk Open Source Software Systems Integrators * Web Design & Hosting * Internet & Intranet Applications Development * /* The chief danger in life is that you may take too many precautions. -- Alfred Adler */