與左右連接組合的MySQL子查詢 - 優化

Question

MySQL似乎無法使用GROUP BY子查詢來優化選擇，並且以較長的執行時間結束。對於這種常見的情況必須有一個已知的優化。

假設我們試圖從數據庫返回所有訂單，並帶有一個標誌，指示它是否是客戶的第一筆訂單。

CREATE TABLE orders (order int, customer int, date date); 

檢索客戶的第一個訂單是超快。

SELECT customer, min(order) as first_order FROM orders GROUP BY customer; 

然而，一旦我們使用子查詢

SELECT order, first_order FROM orders LEFT JOIN ( 
    SELECT customer, min(order) as first_order FROM orders GROUP BY customer 
) AS first_orders ON orders.order=first_orders.first_order; 

我希望有我們缺少一個簡單的一招加入這個與全單組就變得很慢，因爲否則的話將約1000倍快做

CREATE TEMPORARY TABLE tmp_first_order AS 
    SELECT customer, min(order) as first_order FROM orders GROUP BY customer; 
CREATE INDEX tmp_boost ON tmp_first_order (first_order) 

SELECT order, first_order FROM orders LEFT JOIN tmp_first_order 
    ON orders.order=tmp_first_order.first_order; 

編輯：
通過@ruakh啓發提出d選項3，使用INNER JOIN和UNION確實有一個不太難看的解決方法，它具有可接受的性能，但不需要臨時表。但是，這是有點特定於我們的情況，我想知道是否存在更通用的優化。

SELECT order, "YES" as first FROM orders INNER JOIN ( 
    SELECT min(order) as first_order FROM orders GROUP BY customer 
) AS first_orders_1 ON orders.order=first_orders_1.first_order 
UNION 
SELECT order, "NO" as first FROM orders INNER JOIN ( 
    SELECT customer, min(order) as first_order FROM orders GROUP BY customer 
) AS first_orders_2 ON first_orders_2.customer = orders.customer 
    AND orders.order > first_orders_2.first_order; 

Answer 1

這裏有一些事情你可以試試：

1. 去除子查詢的字段列表customer，因爲它沒有做任何事情反正：

   SELECT order, 
        first_order 
       FROM orders 
       LEFT 
       JOIN (SELECT MIN(order) AS first_order 
         FROM orders 
         GROUP 
         BY customer 
        ) AS first_orders 
       ON orders.order = first_orders.first_order 
   ; 
   

2. 相反，添加customer到ON條款，所以它實際上爲您做了一些事情：

   SELECT order, 
        first_order 
       FROM orders 
       LEFT 
       JOIN (SELECT customer, 
          MIN(order) AS first_order 
         FROM orders 
         GROUP 
         BY customer 
        ) AS first_orders 
       ON orders.customer = first_orders.customer 
       AND orders.order = first_orders.first_order 
   ; 
   

3. 與以前相同，但使用INNER JOIN代替LEFT JOIN的，和你原來的ON條款轉換爲CASE表達：

   SELECT order, 
        CASE WHEN first_order = order THEN first_order END AS first_order 
       FROM orders 
   INNER 
       JOIN (SELECT customer, 
          MIN(order) AS first_order 
         FROM orders 
         GROUP 
         BY customer 
        ) AS first_orders 
       ON orders.customer = first_orders.customer 
   ; 
   

4. 與不相關IN -subquery更換整個JOIN方法在CASE表達式中：

   SELECT order, 
        CASE WHEN order IN 
           (SELECT MIN(order) 
            FROM orders 
           GROUP 
            BY customer 
          ) 
         THEN order 
        END AS first_order 
       FROM orders 
   ; 
   

5. 與相關EXISTS -subquery更換整個JOIN方法在CASE表達：

   SELECT order, 
        CASE WHEN NOT EXISTS 
           (SELECT 1 
            FROM orders AS o2 
           WHERE o2.customer = o1.customer 
            AND o2.order < o1.order 
          ) 
         THEN order 
        END AS first_order 
       FROM orders AS o1 
   ; 
   

（這很可能是上面的一些將實際執行糟糕，但我覺得他們都值得嘗試。）

Answer 2

我希望使用一個變量，而不是離開時，這是更快的連接：

SELECT 
    `order`, 
    If(@previous_customer<>(@previous_customer:=`customer`), 
    `order`, 
    NULL 
) AS first_order 
FROM orders 
JOIN (SELECT @previous_customer := -1) x 
ORDER BY customer, `order`; 

這是我對SQL Fiddle回報什麼例子：

CUSTOMER ORDER FIRST_ORDER 
1   1  1 
1   2  (null) 
1   3  (null) 
2   4  4 
2   5  (null) 
3   6  6 
4   7  7