我需要拿起最高的数量

Question

这是我在SQL中使用的查询

select distinct  d.npa, n.time_zone, count(n.time_zone) as tz_count 

from  #duplicates d

inner join psi.npanxx n
on d.npa = n.npa

where n.time_zone is not null and n.time_zone <> '0'

group by --d.npanxx_row_id, d.npa, n.time_zone

order by npa

我得到的结果是这样的

npa        time_zone                tz_count
208           5                      10600
208           4                      31300
219           6                       7882
219           9                       7446

我需要获得最高计数，所以看起来像这样

npa        time_zone                tz_count
208           5                      10600
219           6                       7882

如果有任何我可以使用的功能，请告诉我。 提前感谢您的帮助。

Answer 1

而不是COUNT，正确的SQL关键字是MAX

select distinct  d.npa, n.time_zone, max(n.time_zone) as tz_count 

from  #duplicates d

inner join psi.npanxx n
on d.npa = n.npa

where n.time_zone is not null and n.time_zone <> '0'

group by --d.npanxx_row_id, d.npa, n.time_zone

order by npa

这是一个小例子。它使用不同的数据集，但使用相同的MAX-Group BY公式。

http://sqlfiddle.com/#!9/2d871e/1/0

Answer 2

如果我理解正确，您可以使用row_number()：

select d.npa, d.time_zone, d.tz_count
from (select d.npa, n.time_zone, count(*) as tz_count,
             row_number() over (partition by npa order by count(*) desc) as seqnum
      from #duplicates d inner join
           psi.npanxx n
           on d.npa = n.npa
      where n.time_zone is not null and n.time_zone <> '0'
      group by d.npa, n.time_zone
     ) d
where seqnum = 1;