Question

想象一下，我经营一家杂货店，并希望了解我的产品的受欢迎程度。如何从表fruit中获取每个产品的总计，其中1表示订单中存在产品，null - 缺席。

order | apples | oranges | kiwis
================================
    1 |      1 |       1 | 
    2 |        |       1 | 
    3 |        |       1 |     1
    4 |      1 |         |     1
    5 |      1 |       1 |     1

我追求的结果是：

apples:  3
oranges: 4
kiwis:   3

到目前为止我尝试过：

select count(*) from fruit where apples  = '1'
union
select count(*) from fruit where oranges = '1'
union
select count(*) from fruit where kiwis   = '1'

这样可行，但我不知道哪一行对应哪个产品。

Answer 1

您可以使用UNPIVOT将列转换为行，它们会计算每个水果的行数：

SELECT  upvt.FruitName,
        [Count] = COUNT(*)
FROM    fruit AS f
        UNPIVOT
        (   HasFruit
            FOR FruitName IN ([Apples], [Oranges], [Kiwis])
        ) AS upvt
GROUP BY upvt.FruitName;

如果删除聚合函数，您可以看到unpivot正在做什么，以下查询：

SELECT  upvt.*
FROM    (VALUES
            (1,1,1,NULL),
            (2,NULL,1,NULL),
            (3,NULL,1,1),
            (4,1,NULL,1),
            (5,1,1,1)
        ) AS f ([Order], apples, oranges, Kiwis)
        UNPIVOT
        (   HasFruit
            FOR FruitName IN ([Apples], [Oranges], [Kiwis])
        ) AS upvt;

将屈服：

Order   HasFruit    FruitName
------------------------------
1       1           apples
1       1           oranges
2       1           oranges
3       1           oranges
3       1           Kiwis
4       1           apples
4       1           Kiwis
5       1           apples
5       1           oranges
5       1           Kiwis

然后它只是一个简单的组，并计数得到你想要的结果。如果您还可以使用0列来表明它不存在，那么请将您的计数更改为COUNT(NULLIF(upvt.HasFruit, 0))

Answer 2

添加额外的列以识别产品：

select 'apples' as productName, count(*) from fruit where apples  = '1'
union
select 'oranges', count(*) from fruit where oranges = '1'
union
select 'kiwis', count(*) from fruit where kiwis   = '1'

计算产品矩阵中的总计

2 个答案: