Re: Re: query problem..

For those that are curious, here's the working query: SELECT options. * , count( option_items.optionID ) AS n FROM options LEFT JOIN option_links USING ( optionID ) LEFT JOIN option_items ON options.optionID = option_items.optionID WHERE option_links.productID = '$productID' GROUP BY options.optionID HAVING n > 0 -Micah On Wednesday 14 July 2004 12:07, Micah Stevens wrote: > It is a column.. ;) > > And the docs say you can refer to an aliased aggregate column in the where > clause: > > http://dev.mysql.com/doc/mysql/en/SELECT.html > > I'll take this to the mysql list.. Thanks for your help.. > > -Micah > > On Wednesday 14 July 2004 12:01, Torsten Roehr wrote: > > "Micah Stevens" <micah@raincross-tech.com> wrote in message > > news:200407141153.40622.micah@raincross-tech.com... > > > > > Thank you Torsten , > > > > > > They're backticks, and I get the same error after removing them. > > > > > > Any other ideas? > > > > Then I guess you can't use num in your WHERE clause - MySQL thinks it's a > > column. > > > > Torsten > > > > > -Micah > > > > > > On Wednesday 14 July 2004 11:48, Torsten Roehr wrote: > > > > "Micah Stevens" <micah@raincross-tech.com> wrote in message > > > > > > > > news:200407141141.15947.micah@raincross-tech.com... > > > > > > > > > Hi, > > > > > > > > > > I'm getting an unknown colum num in > > > > > where clause error with this > > > > query: > > > > > $options = mysql_query("select options.*, > > > > > count(option_items.optionID) as num > > > > > from options > > > > > left join option_links using (optionID) > > > > > left join option_items on options.optionID = option_items.optionID > > > > > where option_links.productID = '$productID' > > > > > and num > 0"); > > > > > > > > > > But I'm defining it with the agregate function in the second line.. > > > > Any > > > > > > idea > > > > > > > > > why it's popping the error? > > > > > > > > > > -Micah > > > > > > > > Remove the quotes around num and try again. > > > > > > > > Regards, Torsten Roehr