Question

我有一个表，代表一组集合，这些集合由该组中称为序列的列排序：

| GroupId | Collection | Sequence |
|---------|------------|----------|
| 002     | A.2018     | 1        |
| 002     | A.2017     | 2        |
| 003     | P.2018     | 1        |
| 003     | L.2018     | 2        |
| 003     | R.2018     | 3        |
| 003     | M.2018     | 4        |

在另一个表格中，我按收集和细分列出了客户销售额：

| Collection | Segment | Customer | Sales  |
|------------|---------|----------|--------|
| A.2018     | 002     | C001030  | 304.30 |
| A.2017     | 002     | C001030  | 493.10 |
| L.2018     | 002     | C001030  | 232.33 |
| L.2018     | 010     | C001030  | 343.12 |
| R.2018     | 002     | C001030  | 434.23 |
| M.2018     | 002     | C001030  | 121.12 |

我想通过GroupID获取客户在该细分市场中销售的第一个集合（按集合顺序排序）。

| GroupID | Collection | Segment | Customer | Sales  |
|---------|------------|---------|----------|--------|
| 002     | A.2018     | 002     | C001030  | 304.30 |
| 003     | L.2018     | 002     | C001030  | 232.33 |
| 003     | L.2018     | 010     | C001030  | 343.12 |

Answer 1

SELECT * FROM (
with t1 as (
SELECT '002' as groupid,      'A.2018' as collection, 1 as sequence FROM DUAL UNION ALL
SELECT '002','A.2017',2 FROM DUAL UNION ALL
SELECT '003','P.2018',1 FROM DUAL UNION ALL
SELECT '003','L.2018',2 FROM DUAL UNION ALL
SELECT '003','R.2018',3 FROM DUAL UNION ALL
SELECT '003','M.2018',4 FROM DUAL
),
    t2 as (
SELECT 'A.2018' as collection,'002' as segment ,'C001030' as customer,304.30 as sales  FROM DUAL UNION ALL
SELECT 'A.2017','002','C001030',493.10  FROM DUAL UNION ALL
SELECT 'L.2018','002','C001030',232.33  FROM DUAL UNION ALL
SELECT 'L.2018','010','C001030',343.12  FROM DUAL UNION ALL
SELECT 'R.2018','002','C001030',434.23  FROM DUAL UNION ALL
SELECT 'M.2018','002','C001030',121.12  FROM DUAL
) 
SELECT groupid,
       t1.collection,
       segment,
       customer,
       sales,
       row_number() over (partition by groupid,segment,customer order by t1.collection ASC) rn
FROM   t2,
       t1
WHERE  t1.collection = t2.collection
)
WHERE rn = 1

输出：

GROUPID COLLECTION  SEGMENT CUSTOMER    SALES   RN
002 A.2017  002 C001030 493.1   1
003 L.2018  002 C001030 232.33  1
003 L.2018  010 C001030 343.12  1

但是：

您期望的002组的输出是

002     | A.2018     | 002     | C001030  | 304.30 |

您写道：

“第一个集合（按集合顺序排序）”

所以我认为这里的第一个集合是A.2017。

Answer 2

种类

select GroupId, Collection, Segment, Customer, Sales
from(
  select t1.GroupId, t1.Collection, t2.Segment, t2.Customer, t2.Sales,
     row_number() over(partition by t2.Segment, t2.Customer order by t1.Sequence) rn
   from t1
   join t2 on t1.Collection = t2.Collection
   ) t
where rn=1

Answer 3

如果我理解正确，则可以使用窗口功能：

select cs.*
from (select c.*, s.segment, s.customer, s.sales,
           row_number() over (partition by c.collection, s.segment, s.customer order by c.sequence) as seqnum
    from collections c join
         sales s
         on c.collection = s.collection
     ) cs
where seqnum = 1;

有序查询的第一组值

3 个答案: