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：

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

SQL查询会是什么样的？或者至少是最充足的一个？

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：

将数据“爆破”到多个列中，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都同样适用
使用GROUP BY和聚合函数（SUM，MIN，MAX等）将多行折叠到您想要的输出行集中。

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;

给出：

+------+------------+------------+------------+----------+
| 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 - 如总是，在真实的数据集上测试你的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。

Answer 3

您必须在MySQL中使用CASE statements将行数据转换为列（反之亦然）：

  SELECT t.uid,
         SUM(CASE WHEN t.type = 'product' THEN COUNT(*) END) as PRODUCT,
         SUM(CASE WHEN t.type = 'service' THEN COUNT(*) END) as SERVICE,
         SUM(CASE WHEN t.type = 'invoice' THEN COUNT(*) END) as INVOICE,
         SUM(CASE WHEN t.type = 'order' THEN COUNT(*) END) as ORDER
    FROM TABLE t
GROUP BY t.uid, t.type

Answer 4

你正在寻找一种称为“数据透视表”的东西，这不是MySQL本身可以做的事情。

在这种情况下，您可能最好转换应用程序中的数据，并使用GROUP BY和/或DISTINCT的某种组合来收集您正在寻找的数据。这个查询可能有效，我还没有测试过：

SELECT Type, COUNT(ID), UID
  FROM tablename
 GROUP BY UID, Type

Answer 5

从mytable中选择计数（产品，服务，发票，订单）

做一个计数MySQL查询？

5 个答案: