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在 SSMS 2016 中，我有一个带有各种连接的 select 语句，它给了我以下数据：

| box_barcode | order_number | order_shipment_id | item     | qty |
|-------------|--------------|-------------------|----------|-----|
|  3330000001 | 0000105      | FP001             | tshirt-S | 1   |
|  3330000001 | 0000105      | FP001             | tshirt-M | 2   |
|  3330000001 | 0000105      | FP001             | tshirt-L | 2   |
|  3330000005 | 0000108      | FP002             | shorts-S | 2   |
|  3330000005 | 0000108      | FP002             | shorts-M | 1   |
|  3330000005 | 0000120      | FP002             | shorts-S | 1   |
|  3330000010 | 0000120      | FP003             | shirts-M | 2   |
|  3330000010 | 0000120      | FP003             | shirts-L | 2   |
|  3330000010 | 0000121      | FP003             | shirts-S | 3   |
|  3330000010 | 0000121      | FP003             | shirts-M | 3   |
|  3330000010 | 0000122      | FP003             | shirts-S | 2   |

我想添加一列来计算每个 box_barcode 的不同 order_numbers 的数量，以获得所需的结果：

| box_barcode | order_number | order_shipment_id | item     | qty | count |
|-------------|--------------|-------------------|----------|-----|-------|
|  3330000001 | 0000105      | FP001             | tshirt-S | 1   | 1
|  3330000001 | 0000105      | FP001             | tshirt-M | 2   | 1
|  3330000001 | 0000105      | FP001             | tshirt-L | 2   | 1
|  3330000005 | 0000108      | FP002             | shorts-S | 2   | 2
|  3330000005 | 0000108      | FP002             | shorts-M | 1   | 2
|  3330000005 | 0000120      | FP002             | shorts-S | 1   | 2
|  3330000010 | 0000120      | FP003             | shirts-M | 2   | 3
|  3330000010 | 0000120      | FP003             | shirts-L | 2   | 3
|  3330000010 | 0000121      | FP003             | shirts-S | 3   | 3
|  3330000010 | 0000121      | FP003             | shirts-M | 3   | 3
|  3330000010 | 0000122      | FP003             | shirts-S | 2   | 3

我正在努力找出如何最好地实现这一目标。我知道 count(distinct..)，但是我是否必须将当前查询放入子查询中，以便计数首先与该查询的结果相抵触？

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2 回答 2

1

唉，SQL Server 不支持count(distinct)作为窗口函数。但是很容易模仿：

select t.*,
       sum(case when seqnum = 1 then 1 else 0 end) over (partition by box_barcode) as distinct_count
from (select t.*,
             row_number() over (partition by box_barcode, order_numbers order by box_barcode) as seqnum
      from t
     ) t;
于 2017-06-07T21:40:20.420 回答
1

dense_rank和的另一种选择max。

select t.*,
       max(rnk) over(partition by box_barcode) as distinct_count
from (select t.*,
      dense_rank() over(partition by box_barcode order by order_numbers) as rnk
      from t
     ) t

排名最高的行（使用dense_rank）将是每个box_barcode 的不同数量的订单号。

于 2017-06-07T21:53:24.397 回答