Q&A for programmers
select top 100 OwnerUserId as [User Link], round(count(case when Score > 0 then 1 else null end) * 100.0 / count(*), 2) as "Helpful %" from Posts where PostTypeId=2 and CreationDate > getdate() - 30 group by OwnerUserId having count(*) >= 10 order by round(count(case when Score > 0 then 1 else null end) * 100.0 / count(*), 2) desc
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