Users with high proportion of Au and Ag badges


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Japanese Language

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select top 2048 userid as [User Link], reputation, bronze, silver, gold
from (
    sum(case when class = 3 then 1 else 0 end) as bronze, 
    sum(case when class = 2 then 1 else 0 end) as silver, 
    sum(case when class = 1 then 1 else 0 end) as gold
  from badges
  group by userid
  ) data JOIN (select id, reputation from users) users ON ( = data.userid)
where (gold > bronze) and (gold + silver + bronze > 2)
order by reputation desc, gold desc, silver desc, bronze desc

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