Re: MySQL Query Help

SELECT * FROM products p LEFT JOIN criteria_values cv ON p.key=cv.key LEFT JOIN criteria c ON cv.key=c.key WHERE c.value IS NOT NULL Hard to answer without more detail, but I am guessing the answer will be something like the above. Your question makes it hard to understand whether c or cv is joined to p. So swap em around if I misunderstood. iPhone 4. It rocks! On Nov 21, 2010, at 1:37 AM, Simcha Younger <simcha@syounger.com> wrote: > On Sat, 20 Nov 2010 13:54:29 -0700 > "Ben Miller" <biprellim@gmail.com> wrote: > >> Hi, >> >> I'm building a website for a client in which I need to compare their >> products, side-by-side, but only include criteria for which all selected >> products have a value for that criteria. >> >> In my database (MySQL), I have a tables named "products","criteria" and >> "criteria_values" >> >> If I have something like >> >> $selected_product = array("1"=>"Product >> 1","2"=>"Product 2"...) // All >> products selected for comparison by the user >> >> I need to get only rows from "criteria" where there is a row in >> "criteria_values" matching "criteria.criteria_id" for each >> $selected_product >> - in other words, if any of the $selected_product does not have a row in >> "criteria_values" that matches "criteria.criteria_id", that criteria >> would >> not be returned. I hope that makes sense. > > It would be a lot easier to think about this if you could provide the table structure or create > table statements. > > If I understood correctly, you have products which reference a criteria ID which has no > matching value. If this is the problem you have a to first take care of the integrity of your data, > as this should never happen. > > > -- > Simcha Younger <simcha@syounger.com> > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php >