MySQL 中没有会话变量的排序分组数据?

Rank order grouped data in MySQL without session variables?

我的数据是这样的:

Table Name = sales_orders

Customer_id| Order_id| Item_Id
-------------------------------
1          | 1      | 10
1          | 1      | 24
1          | 1      | 37
1          | 2      | 11
1          | 2      | 15
1          | 3      | 28
2          | 4      | 37
4          | 6      | 10
2          | 7      | 10

但是,我需要它看起来像这样:

Customer_id| Order_id| Item_Id |Order_rank
------------------------------------------
1          | 1      | 10       |    1   
1          | 1      | 24       |    1
1          | 1      | 37       |    1
1          | 2      | 11       |    2
1          | 2      | 15       |    2
1          | 3      | 28       |    3
2          | 4      | 37       |    1
4          | 6      | 10       |    1
2          | 7      | 10       |    2

Customer_Id 是一个独一无二的人

Order_id是唯一的顺序

item_id是产品代码

为了进一步说明,前三行来自客户 #1 的第一笔订单 (order_id = 1),此人订购了 3 件不同的商品(10、24 和 37)。然后他们购买了另一个订单 (order_id =2) 和另外两种产品。 customer_id =2 的人有 2 个唯一订单(4 和 6),而 ID 为“4”的客户有一个唯一订单(order_id =6)

本质上,我需要做的是按 customer_id 和订单 ID 对这些订单进行排名,这样我就可以说 "Order_id = 7 is the second order for customer_id = 2, because Order_rank = 2"

这里的挑战是我不能在 MySQL 查询

中使用会话变量(例如 @grp := customer_id )

例如,不允许这样的查询:

SELECT 
customer_id,
order_id,
@ss := CASE WHEN @grp = customer_id THEN @ss + 1 ELSE 1 END AS 
order_rank,
@grp := customer_id 
FROM 
(
SELECT 
customer_id,
order_id
FROM sales_orders
GROUP BY customer_id, order_id
ORDER BY customer_id, order_id ASC
) AS t_1
CROSS JOIN (SELECT @ss := 0, @grp = NULL)ss

ORDER BY customer_id asc

感谢您的帮助!

您可以使用相关子查询:

select so.*,
       (select count(*)
        from sales_orders so2
        where so2.Customer_id = so.Customer_id and
              so2.order_id <= so.order_id
       ) as rank_order
from sales_orders so;

或 MySQL 8+:

select so.*,
       dense_rank() over (partition by Customer_Id order by Order_Id) as rank_order
from sales_orders so;

Correlated Subquery 中,我们可以 Count(..) 特定行的 customer_id 唯一值和先前的 order_id 值,并且order_id确定排名。

我们需要计算 唯一 值,因为每个订单有多行(由于有多个项目)。


查询

SELECT 
  t1.Customer_id, 
  t1.Order_id, 
  t1.Item_Id, 
  (SELECT COUNT(DISTINCT t2.Order_id) 
   FROM sales_orders t2 
   WHERE t2.Customer_id = t1.Customer_id AND 
         t2.Order_id <= t1.Order_id
  ) AS Order_rank 
FROM sales_orders AS t1;

结果

| Customer_id | Order_id | Item_Id | Order_rank |
| ----------- | -------- | ------- | ---------- |
| 1           | 1        | 10      | 1          |
| 1           | 1        | 24      | 1          |
| 1           | 1        | 37      | 1          |
| 1           | 2        | 11      | 2          |
| 1           | 2        | 15      | 2          |
| 1           | 3        | 28      | 3          |
| 2           | 4        | 37      | 1          |
| 4           | 6        | 10      | 1          |
| 2           | 7        | 10      | 2          |

View on DB Fiddle