Vertica/vSQL“每天选择10行，按顺序排列，并在某个组上

Question

我有Vertica数据库，有很多带有日期(时间戳)和更多属性的记录。例如，'testTable‘看起来像这样

a varchar(255)
b int
timestamp bigint

我需要找到一段时间(比方说1月1日到1月15日)每一天的前10名的sum(b)，这些日期可以由用户指定。

迭代查询将是什么样子的？粗略的方法可能是在中间包含UNION ALL的单个SELECT语句。

select a, sum(b) from testTable where TO_TIMESTAMP( timestamp ) between '2012-01-01 05:10:00' and '2012-01-02 05:10:00' group by a order by sum(b) desc LIMIT 10
UNION ALL
select a, sum(b) from testTable where TO_TIMESTAMP( timestamp ) between '2012-01-02 05:10:00' and '2012-01-03 05:10:00' group by a  order by sum(b) desc LIMIT 10
UNION ALL
select a, sum(b) from testTable where TO_TIMESTAMP( timestamp ) between '2012-01-03 05:10:00' and '2012-01-04 05:10:00' group by a order by sum(b) desc LIMIT 10
..
..
..
UNION ALL
select a, sum(b) from testTable where TO_TIMESTAMP( timestamp ) between '2012-01-14 05:10:00' and '2012-01-15 05:10:00' group by a order by sum(b) desc LIMIT 10 ;

但我希望它更通用，用户可以运行具有两个给定日期的脚本。

Answer 1

语法可能有点不对劲……我没有Vertica来测试。

select day, a, tot
from 
   (
   select 
      *,
      ROW_NUMBER() OVER (PARTITION BY tt4.day) as row_number
   from
      (
      select
         ts as day, 
         tt1.a, 
         sum(tt1.b) as tot
      from 
         testTable tt1, 
         ( select distinct date(TO_TIMESTAMP(tt2.timestamp)) as ts
           from   testTable tt2
           where  date(TO_TIMESTAMP(tt2.timestamp)) between cast('2012/01/01' as date) and cast('2012/01/15' as date) ) as tt3
      where 
         date(TO_TIMESTAMP(tt1.timestamp)) = tt3.ts
      group by 
         date(TO_TIMESTAMP(tt1.timestamp)), 
         tt1.a
      order by 
         date(TO_TIMESTAMP(tt1.timestamp)),
         sum(tt1.b) desc,
         tt1.a
      ) as tt4
   ) as tt5
where 
   tt5.row_number <=10