Question

有人可以告诉我如何获得如下结果。（描述获得前2名现金优惠）

使用dense_rank函数，其中rank＆lt; = 2将给出前2个优惠。

我也希望得到＆total; of_offer＆＃39;这应该是＆quot; offer1＆＃39;的总和。和＆＃39; offer2＆＃39;。当没有offer2（例如：taurus）＆＃39;总报价＆＃39;应该是＆＃39; offer1＆＃39;和＆＃39; null＆＃39;为＆＃39; offer2＆＃39;

输入：

customer    make    zipcode offer notes  
mark        focus   101     250   cash  
mark        focus   101     2500  appreciation cash  
mark        focus   101     1000  cash  
mark        focus   101     1500  cash offer  

henry       520i    21405   500  cash offer  
henry       520i    21405   100  cash  
henry       520i    21405   750  appreciation cash  
henry       520i    21405   100  cash  

mark        taurus  48360   250    appreciation cash  

mark        mustang 730     500  cash  
mark        mustang 730     1000  Cash offer  
mark        mustang 730     1250  appreciation cash

期望的输出：

| CUSTOMER | MAKE    | ZIPCODE | TOP_OFFER1 | NOTES1            | TOP_OFFER2 | NOTES2     | Total_offer   
| henry    | 520i    | 21405   | 750        | appreciation cash | 500        | cash offer | 1250    
| mark     | focus   | 101     | 2500       | appreciation cash | 1500       | cash offer | 4000  
| mark     | mustang | 730     | 1250       | appreciation cash | 1000       | cash offer | 2250    
| mark     | taurus  | 48360   | 250        | appreciation cash | NULL       | 250        |

由于

PS：

下面的链接告诉我，需要执行dense_rank才能获得前2个优惠。（Top 2 offers with sum of all offers）

Answer 1

您将使用条件聚合：

select customer, make, zipcode,
       max(case when seqnum = 1 then offer end) as offer1,
       max(case when seqnum = 1 then notes end) as notes1,
       max(case when seqnum = 2 then offer end) as offer2,
       max(case when seqnum = 2 then notes end) as notes2,
       sum(offer) as total_offer
from (select t.*,
             row_number() over (partition by customer, make, zipcode order by offer desc) as seqnum
      from table t
     ) t
group by customer, make, zipcode;

前2名优惠与所有优惠的总和与描述

1 个答案: