做一个计数MySQL查询？

Question

我有下表:

UID | ID  | Type
1   | 1   | product
1   | 2   | product
1   | 3   | service
1   | 4   | product
1   | 5   | product
2   | 6   | service
1   | 7   | order
2   | 8   | invoice
2   | 9   | product

我想最终得到:

UID | product | service | invoice | order
1   |  4      |  1      |  0      |  1
2   |  1      |  1      |  1      |  0

SQL查询会是什么样的？或者至少,最充足的一个？

Answer 1

如果您真的只需要这四种类型,那么您可以按如下方式对值进行硬编码:

select UID,
    count(case when type='product' then 1 else null end) as product,
    count(case when type='service' then 1 else null end) as service,
    count(case when type='invoice' then 1 else null end) as invoice,
    count(case when type='order' then 1 else null end) as order
from MyTable
group by UID
order by UID    

Answer 2

您想要做的是一个数据透视操作,SQL语法不直接支持它.但是,它并不太复杂,概念上涉及两个步骤:

1. 将数据"爆炸"成许多列,原始数据集中每行一行.这通常使用CASE WHEN ... ELSE ... END或偶尔使用函数(如oracle中的decode())来完成.我将在下面的示例中使用CASE WHEN,因为它对大多数RDBMS都同样有效
2. 使用GROUP BY和聚合函数(SUM,MIN,MAX等)将许多行折叠到所需的输出行集中.

我正在使用此数据集作为示例:

mysql> select * from foo;
+------+------+---------+
| uid  | id   | type    |
+------+------+---------+
|    1 |    1 | product | 
|    1 |    2 | product | 
|    1 |    3 | service | 
|    1 |    4 | product | 
|    1 |    5 | product | 
|    2 |    6 | service | 
|    1 |    7 | order   | 
|    2 |    8 | invoice | 
|    2 |    9 | product | 
+------+------+---------+

第1步是"炸毁"数据集:

select uid
     , case when type = 'product' then 1 else 0 end as is_product
     , case when type = 'service' then 1 else 0 end as is_service
     , case when type = 'invoice' then 1 else 0 end as is_invoice
     , case when type = 'order' then 1 else 0 end as is_order
  from foo;

这使:

+------+------------+------------+------------+----------+
| uid  | is_product | is_service | is_invoice | is_order |
+------+------------+------------+------------+----------+
|    1 |          1 |          0 |          0 |        0 | 
|    1 |          1 |          0 |          0 |        0 | 
|    1 |          0 |          1 |          0 |        0 | 
|    1 |          1 |          0 |          0 |        0 | 
|    1 |          1 |          0 |          0 |        0 | 
|    2 |          0 |          1 |          0 |        0 | 
|    1 |          0 |          0 |          0 |        1 | 
|    2 |          0 |          0 |          1 |        0 | 
|    2 |          1 |          0 |          0 |        0 | 
+------+------------+------------+------------+----------+

接下来,我们在每个日期的输出中折叠为一行,并将每个is_*列相加,使用或初始查询作为内联视图(也称为"子查询"):

select uid
     , sum(is_product) as count_product
     , sum(is_service) as count_service
     , sum(is_invoice) as count_invoice
     , sum(is_order)   as count_order
  from (
         select uid
              , case when type = 'product' then 1 else 0 end as is_product
              , case when type = 'service' then 1 else 0 end as is_service
              , case when type = 'invoice' then 1 else 0 end as is_invoice
              , case when type = 'order' then 1 else 0 end as is_order
           from foo
       ) x
 group by uid;

(另请注意,您可以将这两个查询合并为一个,但为了清楚起见,我在这里单独显示它们;至少在MySQL中,这似乎导致更简单的执行计划,这通常意味着更快的执行 - 一如既往,测试你在现实数据集上的SQL性能,不要相信我的话!)

这给了我们:

+------+---------------+---------------+---------------+-------------+
| uid  | count_product | count_service | count_invoice | count_order |
+------+---------------+---------------+---------------+-------------+
|    1 |             4 |             1 |             0 |           1 | 
|    2 |             1 |             1 |             1 |           0 | 
+------+---------------+---------------+---------------+-------------+

这是期望的结果.