MySQL兩列動態交叉表

Question

女士們，先生們，

我在MySQL 5.1中遇到以下問題。這是我的桌子。

Sale | Store | Product | Discount 
--------------------------------- 
1 | 1 | B  | Yes 
2 | 1 | B  | Yes 
3 | 1 | B  | No 
4 | 1 | D  | Yes 
5 | 1 | A  | No 
6 | 2 | A  | No 
7 | 3 | B  | No 
8 | 3 | B  | No 
9 | 1 | D  | Yes 
10 | 2 | A  | No 

現在，這當然是一個非常不方便的方式來組織數據，但這是該數據如何恰巧進來（並會繼續進來）。

我需要能夠拋出一個完全任意的產品類型列表（如果您願意，用逗號分隔）;比方說A,B,C。最後，我需要一個商店賣什麼東西的清單。這是一箇中間結果，以便於理解。

Store | Product | Sales | Discounts 
----------------------------------- 
    1 | A  | 1 | 0 
    1 | B  | 3 | 2 
    1 | C  | 0 | 0 
    2 | A  | 2 | 0 
    2 | B  | 0 | 0 
    2 | C  | 0 | 0 
    3 | A  | 0 | 0 
    3 | B  | 2 | 0 
    3 | C  | 0 | 0 

我需要什麼，到底，是這樣的：

Store | Product types sold | Discounts given on product types: 
------------------------------------------------------------- 
    1 |   2   |  1 
    2 |   1   |  0 
    3 |   1   |  0 

第三列說，有多少賣（和查詢）產品類型分別給予至少一個折扣。我試過了各種各樣的動態交叉表生成查詢（是的，我已經看過全部），但是我沒有足夠的精神狀態RAM來一次性將所有東​​西都包裹在頭上。你如何以最有效的方式來處理這個問題？臨時表格/存儲過程等都可以，粗略的想法概要非常受歡迎。謝謝！

Answer 1

您可以將數據分組的存儲和產品生產的中間結果：

mysql> SELECT Store, Product, 
COUNT(1) AS Sales, 
SUM(IF(Discount='Yes', 1, 0)) AS Discounts 
FROM input 
WHERE Product IN ('A', 'B', 'C', 'D') 
GROUP BY Store, Product 
+-------+---------+-------+-----------+ 
| Store | Product | Sales | Discounts | 
+-------+---------+-------+-----------+ 
|  1 | A  |  1 |   0 | 
|  1 | B  |  3 |   2 | 
|  1 | D  |  2 |   2 | 
|  2 | A  |  2 |   0 | 
|  3 | B  |  2 |   0 | 
+-------+---------+-------+-----------+ 
5 rows in set (0.00 sec) 

然後您可以將這個結果只是由商店將它轉化成你的最終結果：

mysql> SELECT Store, 
COUNT(1) AS `Product types sold`, 
SUM(Discounts) AS `Discounts given on product types` 
FROM (
SELECT Store, Product, 
COUNT(1) AS Sales, 
SUM(IF(Discount='Yes', 1, 0)) AS Discounts 
FROM input 
WHERE Product IN ('A', 'B', 'C', 'D') 
GROUP BY Store, Product 
) AS intermediate 
GROUP BY Store 
+-------+--------------------+----------------------------------+ 
| Store | Product types sold | Discounts given on product types | 
+-------+--------------------+----------------------------------+ 
|  1 |     3 |        4 | 
|  2 |     1 |        0 | 
|  3 |     1 |        0 | 
+-------+--------------------+----------------------------------+ 
3 rows in set (0.00 sec) 

請注意，第一個查詢是第二個查詢的子查詢。

Answer 2

我覺得是剛剛分離的查詢是最簡單的：

select Store, 
    (Select count(distinct Product) from products as p2 where p2.Store = p1.Store), 
    (Select count(distinct Product) from products as p3 where p3.Store = p1.Store and p3.Discount = "Yes") 
from products p1 
group by Store 

（如果它是關於大表，一定要測試你的表現）