1

这是我正在处理的表格的简化版本，即订单

+-------------------+------------------+---------------+
| Order_Base_Number | Order_Lot_Number | Other Cols... |
+-------------------+------------------+---------------+
|                 1 |                3 |               |
|                 1 |                3 |               |
|                 1 |                4 |               |
|                 1 |                4 |               |
|                 1 |                4 |               |
|                 1 |                5 |               |
|                 2 |                3 |               |
|                 2 |                5 |               |
|                 2 |                9 |               |
|                 2 |               10 |               |
+-------------------+------------------+---------------+

我想要做的是根据基数和批号计算唯一条目的数量。我有两组数字，一组是基数，另一组是一组批号。例如，假设这两组是 Base In (1,2,3)，Lot 在 (3,4,20)。

我正在寻找一个 SQL 查询，它可以从两个集合中返回 (Base,Lot) 的所有可能组合，其计数显示在表中找到该组合的次数。我的问题是我想包括所有可能的组合，如果组合不在 Orders 表中，我希望计数显示为零。所以，我正在寻找的输出是这样的。

+------+-----+-----------+
| Base | Lot | Frequency |
+------+-----+-----------+
|    1 |   3 |    2      |
|    1 |   4 |    3      |
|    1 |  20 |    0      |
|    2 |   3 |    1      |
|    2 |   4 |    0      |
|    2 |  20 |    0      |
|    3 |   3 |    0      |
|    3 |   4 |    0      |
|    3 |  20 |    0      |
+------+-----+-----------+

我尝试了很多查询，但从未接近过这个，甚至不确定是否可以完成。现在我正在弄清楚客户端的组合，因此我执行了太多查询来获取频率。

4

3 回答 3

1

也许最清晰的方法是从 CTE 列表开始：

with bases as (
    select 1 as base from dual union all
    select 2 as base from dual union all
    select 3 as base from dual
   ),
     lots as (
    select 3 as lot from dual union all
    select 4 as lot from dual union all
    select 20 as lot from dual
   )
select b.base, l.lot, count(Order_Base_Number) as Frequency
from bases b cross join lots l left outer join
     Orders o
     on o.base = b.base and o.lot = l.lot
group by b.base, l.lot

请注意，这使得cross join明确的，故意不使用,笛卡尔积。

该查询的第一部分也可以写成如下所示（假设每个基数和批次在表中至少有一条记录）：

with bases as (
     select distinct base
     from Orders  -- or some other table, perhaps Orders ?
     where base in (1, 2, 3)
    ),
     select distinct lot
     from Orders  -- or some other table, perhaps Lots ?
     where lot in (3, 4, 20)
    )
. . .

这更简洁，但可能导致查询效率降低。

于 2013-01-10T19:24:11.987 回答
0

您在最里面的子查询中需要的是调用CROSS JOIN，它获取记录的笛卡尔积（所有可能的组合）。这就是你没有JOIN..ON condition也没有的时候得到的WHERE：

SELECT Base.Id as baseid, Lot.Id as lotid FROM Bases, Lots

现在将其放入子查询和LEFT JOIN其他内容中：

SELECT ... FROM 
   (SELECT Base.Id as baseid, Lot.Id as lotid 
           FROM Bases, Lots) baseslots
   LEFT JOIN Orders ON Order_Base_Number = baseid, 
             Order_Lot_Number = lotid ....

有了这个LEFT JOIN，你会得到 NULL 不存在的组合。使用COALESCE（或类似的东西）将它们变成 0。

于 2013-01-10T19:19:31.200 回答
0

我没有 Oracle 来测试它，但这是我要做的：

CREATE TABLE pairs AS
(
SELECT DISTINCT Base.Order_Base_Number, Lot.Order_Lot_Number 
FROM ORDERS Base
CROSS JOIN ORDERS Lot
);

CREATE TABLE counts AS
(
SELECT Order_Base_Number, Order_Lot_Number, Count(*) AS C
FROM ORDERS 
GROUP BY Order_Base_Number, Order_Lot_Number
);

SELECT P.Order_Base_Number, P.Order_Lot_Number, COALESCE(C.C,0) AS [Count]
FROM Pairs P
LEFT JOIN counts C ON P.Order_Base_Number = C.Order_Base_Number 
             AND P.Order_Lot_Number = C.Order_Lot_Number
于 2013-01-10T19:22:22.807 回答