select count(t.y), count(t.m) from (select *, case when t...

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Portuguese Language

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select count(t.y), count(t.m) 
from (select *,
case when t.pr>0 then '>0' end as diff
from (select t.y as y, t.m as m, 
((lag(t.c) over (order by t.y, t.m))-t.c)*100/(lag(t.c) over (order by t.y, t.m)) 
as pr
from (select year(creationdate) as y, month(creationdate) as m, count(*) as c 
from posts group by year(creationdate), month(creationdate)) as t) as t) as t
where t.diff='>0'

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