Question

大师，我偶然发现了蜂巢排名过程，我喜欢在每天排名交易（对于相同的trx值没有重复的排名值）

date      hour trx  rnk
18/03/2018  0   1   24
18/03/2018  1   2   23
18/03/2018  2   3   22
18/03/2018  3   4   21
18/03/2018  4   5   20
18/03/2018  5   6   19
18/03/2018  6   7   18
18/03/2018  7   8   17
18/03/2018  8   9   16
18/03/2018  9   10  15
18/03/2018  10  11  14
18/03/2018  11  12  13
18/03/2018  12  13  12
18/03/2018  13  14  11
18/03/2018  14  15  10
18/03/2018  15  16  9
18/03/2018  16  17  8
18/03/2018  17  18  7
18/03/2018  18  19  6
18/03/2018  19  20  5
18/03/2018  20  21  4
18/03/2018  21  22  3
18/03/2018  22  23  2
18/03/2018  23  24  1
17/03/2018  0   1   24
17/03/2018  1   2   23
17/03/2018  2   3   22
17/03/2018  3   4   21
17/03/2018  4   5   20
17/03/2018  5   6   19
17/03/2018  6   7   18
17/03/2018  7   8   17
17/03/2018  8   9   16
17/03/2018  9   10  15
17/03/2018  10  11  14
17/03/2018  11  12  13
17/03/2018  12  13  12
17/03/2018  13  14  11
17/03/2018  14  15  10
17/03/2018  15  16  9
17/03/2018  16  17  8
17/03/2018  17  18  7
17/03/2018  18  19  6
17/03/2018  19  20  5
17/03/2018  20  21  4
17/03/2018  21  22  3
17/03/2018  22  23  2
17/03/2018  23  24  1

这是我的代码

select a.date, a.hour, trx, rank() over (order by a.trx) as rnk from(
select date,hour, count(*) as trx from  smy_tb
group by date, hour
)a
limit 100;

问题是： 1.使用相同的trx值重复排名值 2.等级值继续到下一个日期（应按日期和小时分组，因此每个日期只返回24等级值）

需要建议，谢谢

Answer 1

您应该partition by日期列并使用特定的排序。

rank() over (partition by a.date order by a.hour desc)

Answer 2

由@BKS解释

这是已解决的代码

select a.date, a.hour, trx, row_number() over (partition by a.date order by a.trx desc) as rnk from(
select date,hour, count(*) as trx from  smy_tb
group by date, hour
)a
limit 100;

蜂巢排名超过分组值

2 个答案: