select rank() over(partition by null order by [Number of ...

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select
rank() over(partition by null order by [Number of Posts] desc) [Contribution Rank],
*
from (
  SELECT
  COUNT(P.Id) [Number of Posts],
  U.Location,
  U.DisplayName

  from Posts P
  join Users U on U.Id = P.OwnerUserId

  where
  UPPER(U.Location) like '%SOUTH AFRICA%'

  group by
  U.Location,
  U.DisplayName
) T

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