[英]Get MAX value of each record in group by query
I have a query in SQL looks like that: 我在SQL中的查询看起来像这样:
select fldCustomer, fldTerminal, COUNT(fldbill)
from tblDataBills
group by fldCustomer, fldTerminal
order by fldCustomer
results looks like: 结果看起来像:
fldCustomer fldTerminal (number of bills)
0 1 19086
0 2 10
0 5 236
1 1 472
1 5 3
1 500 19
2 1 292
2 500 22
how can i get the MAX count of each customer so i get results like 我如何获得每个客户的最大数量,所以我得到如下结果
0 1 19086
1 1 472
2 1 292
Thanks in advance! 提前致谢!
Use a subquery with row_number()
: 使用带有row_number()
的子查询:
select fldCustomer, fldTerminal, cnt
from (select fldCustomer, fldTerminal, COUNT(*) as cnt,
row_number() over (partition by fldCustomer order by count(*) desc) as seqnum
from tblDataBills
group by fldCustomer, fldTerminal
) db
where seqnum = 1
order by fldCustomer ;
Note that in the event of ties, this will arbitrarily return one of the rows. 请注意,在发生平局的情况下,这将任意返回行之一。 If you want all of them, then use rank()
or dense_rank()
. 如果要全部使用,请使用rank()
或dense_rank()
。
This might require a little trickery with the RANK()
function 使用RANK()
函数可能需要一些技巧
SELECT fldCustomer, fldTerminal, [(number of bills)] FROM ( SELECT fldCustomer, fldTerminal, COUNT(fldbill) [(number of bills)], RANK() OVER (PARTITION BY fldCustomer ORDER BY COUNT(fldbill) DESC) Ranking FROM tblDataBills GROUP BY fldCustomer, fldTerminal ) a WHERE Ranking = 1
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