Q&A for professional and enthusiast programmers
distinct u.Id as [User Link], Reputation, DisplayName, WebsiteUrl, AboutMe, Views, COUNT(a.Id) AS [answer count]
from Users u
inner join Posts a on a.OwnerUserId = u.Id
inner join Posts q on a.ParentId = q.Id
u.Location like '%ndia%'
and a.PostTypeId = 2 -- Answers ..
and q.tags LIKE '%<ruby>%' -- to questions with the ruby tag
GROUP BY u.id, Reputation, DisplayName, WebsiteUrl, AboutMe, Views
order by COUNT(a.Id) DESC