mysql - 我知道 PIVOT 命令可以转换数据集，这是正确的方法吗？

Question

我有一个看起来像这样的数据集:

+----+-------------+
| ID | StoreVisit  |
+----+-------------+  
| 1  |  Home Depot |
| 2  |  Lowes      | 
| 3  |  Home Depot |
| 2  |  ACE        |
| 2  |  Lowes      |
| 1  |  Home Depot |
| 4  |  ACE        |
| 5  |  ACE        |
| 4  |  Lowes      |
+----+-------------+

我是 SQL 的新手，我知道我可以全选，然后使用 Excel(数据透视表/函数/特殊粘贴)或 R (tidyr) 进行转置。但是，如果我有很多数据，这样效率不高。下面的查询是否正确？如果是这样，如果有数千种类型的商店而无需在查询中逐一键入，我如何定义 StoreVisit 的所有值？

select * from Stores
pivot (COUNT(StoreVisit) for StoreVisit in ([ACE],[Lowes],[Home Depot])) as StoreCounts

+----+-------+-----------+-----+
| ID | Lowes | HomeDepot | ACE |
+----+-------+-----------+-----+
| 1  |   0   |     2     |  0  |
| 2  |   2   |     0     |  0  |
| 3  |   0   |     1     |  0  |
| 4  |   1   |     0     |  1  |
| 5  |   0   |     0     |  1  |
+----+-------+-----------+-----+

请原谅这篇文章的格式!非常抱歉。

Answer 1

使用条件聚合:

select id,
       sum(storevisit = 'Lowes') as lowes,
       sum(storevisit = 'HomeDepot') as HomeDepot,
       sum(storevisit = 'Ace') as ace
from t
group by id;