Users count by location

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Software Engineering

Q&A for professionals, academics, and students working within the systems development life cycle

select
UPPER(u.Location) [Location],
count(distinct u.Id) [UserCount]
from Users u
inner join Posts p on p.OwnerUserId = u.Id
inner join PostTags pt on pt.PostId = p.Id
inner join Tags t on t.Id = pt.TagId
where UPPER(u.Location) like UPPER('%##Location##%')
group by UPPER(u.Location)
order by count(distinct u.Id) desc

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