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Question

我们正在使用SAP HANA 1.0 SPS12。

我们有如下的日表-

从表_1中选择trans_date，article，measure1，measure2

表格数量〜500万行

我们需要查看以下数据-

select 'day-1',sum(measure1),sum(meaure2) from table1 where trans_date=add_days(current_date,-1) group by 'day-1'
union all
select 'day-2',sum(measure1),sum(meaure2) from table1 where trans_date>=add_days(current_date,-2) group by 'day-2' 
union all
select 'WTD',sum(measure1),sum(meaure2) from table1 where trans_date>=add_days(current_date,-7) group by 'WTD'
union all
select 'WTD-1',sum(measure1),sum(meaure2) from table1 where trans_date>=add_days(current_date,-15) and trans_Date <= add_days(current_date,-7) group by 'WTD-1'

MTD，MTD-1，MTD-2，YTD等等。

在性能方面，最好与WITH CLAUSE一起使用并将数据保存一年，然后根据时间范围进行拆分？ 或最好对每个时间范围使用单独的汇总，如上所示。

据我了解，在Oracle等RDBMS中，WITH CLAUSE实现结果并从内存中使用它。 SAP HANA是内存数据库本身。 在SAP HANA中使用WITH CLAUSE是否会带来独特的性能优势？

使用WITH CLAUSE查询-

WITH t1 as
(
select trans_date,sum(measure1),sum(meaure2) from table1 where trans_date>=add_days(current_date,-365)
)
select 'day-1',sum(measure1),sum(meaure2) from t1 where trans_date=add_days(current_date,-1) group by 'day-1'
union all
select 'day-2',sum(measure1),sum(meaure2) from t1 where trans_date>=add_days(current_date,-2) group by 'day-2' 
union all
select 'WTD',sum(measure1),sum(meaure2) from t1 where trans_date>=add_days(current_date,-7) group by 'WTD'
union all
select 'WTD-1',sum(measure1),sum(meaure2) from t1 where trans_date>=add_days(current_date,-15) 
                                                  and trans_Date <= add_days(current_date,-7) 
                                                  group by 'WTD-1'

Answer 1

如果您关心性能，那么将数据放在一行中应该会更好：

select sum(case when trans_date = add_days(current_date, -1) then measure1 end) as measure1_day1,
       sum(case when trans_date = add_days(current_date, -1) then measure2 end) as measure2_day1,
       sum(case when trans_date = add_days(current_date, -2) then measure1 end) as measure1_day2,
       sum(case when trans_date = add_days(current_date, -2) then measure2 end) as measure2_day2,
       . . .       
from table1
where trans_date >= add_days(current_date, -15);

如果确实需要在单独的行中输入值，则可以在以后取消显示结果。

或者，您可以执行以下操作：

select days, sum(measure1), sum(measure2)
from (select 1 as days from dummy union all
      select 2 from dummy union all
      select 7 from dummy union all
      select 15 from dummy
     ) d left join
     table1 t
     on t.trans_date = add_days(current_date, - d.days)
group by days
order by days;