一起使用 GROUP BY 和 PARTITION BY

Question

这是我的第一个查询，returns 下图中的结果

我从 3 个 table 中获取“餐厅名称”、“客户名称”、“订单计数”和“日期”，即“订单”、“客户”和“餐厅”。然后我按餐厅名称、顾客姓名和日期对其进行分组。

SELECT R.name AS name_of_r, C.name AS name_of_c, COUNT(O.id) AS Orders, date,
FROM orders O
INNER JOIN restaurants R ON R.id = O.restaurant_id AND R.country = O.country 
INNER JOIN customers C ON C.id = O.customer_id AND C.country = O.country
GROUP BY R.name, C.name, date

name_of_r       |   name_of_c    |       Orders       |       date
Howdy           | John Almagro   |          1         |     2021-01-07
Howdy           | John Almagro   |          1         |     2021-07-01
Howdy           | Luke Pablo     |          1         |     2021-01-01
Howdy           | Steven Smith   |          1         |     2021-09-01
JFC             | Monty Tron     |          1         |     2021-01-01
JFC             | Steven Smith   |          1         |     2021-05-04
JFC             | Steven Smith   |          1         |     2021-06-01
KFC             | Luke Pablo     |          1         |     2021-01-09
KFC             | Michael Wren   |          1         |     2021-01-01
KFC             | Monty Tron     |          1         |     2021-04-01
KFC             | Steven Smith   |          1         |     2021-01-01
KFC             | Steven Smith   |          1         |     2021-03-01
KFC             | Steven Smith   |          1         |     2021-08-01
Kitchen Cuisine | Luke Pablo     |          1         |     2021-01-05
Kitchen Cuisine | Luke Pablo     |          1         |     2021-04-01
Kitchen Cuisine | Steven Smith   |          1         |     2021-03-01
Kitchen Cuisine | Steven Smith   |          1         |     2021-06-01
McDonald's      | Arthur Chen    |          1         |     2021-01-01
McDonald's      | Arthur Chen    |          1         |     2021-03-02
McDonald's      | Arthur Chen    |          1         |     2021-05-03
McDonald's      | Arthur Chen    |          1         |     2021-07-01
McDonald's      | Arthur Chen    |          1         |     2021-08-01
McDonald's      | Arthur Chen    |          1         |     2021-09-01

现在，我想做的是按“name_of_r”（餐厅名称）按“日期”排序，并给每个行号一个行号，这样我就可以得到每个 window.

实际上，这是我的子查询，我希望从中获取在每家餐厅下第 3 个订单的顾客的姓名。

我尝试添加 row_number() 和分区依据，但它没有用，给我一个语法错误

SELECT R.name AS name_of_r, C.name AS name_of_c, COUNT(O.id), date,
ROW_NUMBER() OVER(PARTITION BY R.name ORDER BY date) AS row_num
FROM orders O
INNER JOIN restaurants R ON R.id = O.restaurant_id AND R.country = O.country
INNER JOIN customers C ON C.id = O.customer_id AND C.country = O.country
GROUP BY R.name, C.name, date

我想要的最终输出是在每家餐厅下第 3 份订单的顾客的姓名。喜欢下面的table：

name_of_restaurant       |  name_of_customer_who_placed_the_3rd_order
Howdy                    |      Luke Pablo
JFC                      |      Steven Smith
KFC                      |      Monty Tron
Kitchen Cuisine          |      Steven Smith
McDonald's               |      Arthur Chen

我知道这个，因为我第一次 table 订购日期

Answer 1

我想你需要一个HAVING COUNT

SELECT
    R.name AS name_of_r,
    C.name AS name_of_c,
    COUNT(O.id),
    date
  FROM 
    orders O
  INNER JOIN restaurants R ON R.id = O.restaurant_id AND R.country = O.country
  INNER JOIN customers C   ON C.id = O.customer_id   AND C.country = O.country
  GROUP BY R.name, C.name, date
  HAVING COUNT(O.id) >= 3
;

PS: 但对当天下第3单的人有效。否则，日期必须从分组中排除。

更新： 添加了餐厅中每三位顾客的选择请求。

SELECT name_of_r, name_of_c, date
  FROM (
    SELECT
        R.name AS name_of_r,
        C.name AS name_of_c,
        date,
        ROW_NUMBER() OVER (PARTITION BY R.name ORDER BY date) AS nc
      FROM 
        orders O
      INNER JOIN restaurants R ON R.id = O.restaurant_id AND R.country = O.country
      INNER JOIN customers C   ON C.id = O.customer_id   AND C.country = O.country
  ) t
  WHERE t.nc = 3
;

见ROW_NUMBER Function

一起使用 GROUP BY 和 PARTITION BY

Using GROUP BY and PARTITION BY together

mysql

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