Here is one way:
(probably not the best)
I was able to get almost everything working with your help. Thanks!!! :)
average ages for male/female respondents (database column 'age')
SELECT AVG(age) FROM tablename WHERE sex = 'male';
SELECT AVG(age) FROM tablename WHERE sex = 'female';
I have one small problem remaining. With this code:
$male_average_age = mysql_query("SELECT AVG(age) FROM my_database WHERE sex = 'M'");
..I get a strange "Resource id#10" output. I can't figure out why. What am I missing?
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Hans Vallden
hans@vallden.com