【问题标题】：How to first join tables and then filter rows to keep the greatest per group如何首先连接表然后过滤行以保持每组最大
【发布时间】：2022-11-27 20:22:01
【问题描述】：

我正在尝试写一个 SELECT 声明第一的加入两个表和然后通过保持每组的最大值来过滤行。

例子

以下两个表格描述了动物园中的访客。

visitors 表格包括所有去过动物园的独特人物；每人一排。
activity_log 表格描述了每位游客在参观期间在动物园做了什么；每排一排活动.

生成数据（可重现）

以下 SQL 代码兼容MySQL:

-- visitors
CREATE TABLE visitors(
                      visitor_id       INTEGER  NOT NULL PRIMARY KEY, 
                      country_of_birth VARCHAR(7) NOT NULL
                      );
                      
INSERT INTO visitors(visitor_id, country_of_birth) VALUES
                    (1,          'Bolivia'),
                    (2,          'UK'),
                    (3,          'UK'),
                    (4,          'Bolivia'),
                    (5,          'UK'),
                    (6,          'UK'),
                    (7,          'France'),
                    (8,          'USA'),
                    (9,          'UK'),
                    (10,         'France');

-- activity_log
CREATE TABLE activity_log(
                          visitor_id         INTEGER  NOT NULL,
                          FOREIGN KEY (visitor_id) REFERENCES visitors(visitor_id),
                          activity_time      DATETIME  NOT NULL,
                          activity_name      VARCHAR(14) NOT NULL,
                          what_was_purchased VARCHAR(8)
                          );
                       
INSERT INTO activity_log(visitor_id, activity_time,          activity_name,     what_was_purchased) VALUES
                        (1,          '2020-09-03 11:15:00',  'visit lions',     NULL),
                        (1,          '2020-09-03 10:30:00',  'use restroom',    NULL),
                        (1,          '2020-09-03 10:10:00',  'visit reptiles',  NULL),
                        (1,          '2020-09-03 10:45:00',  'purchase',        'coffee'),
                        (2,          '2021-02-10 15:30:00',  'visit giftshop',  NULL),
                        (2,          '2021-02-10 15:02:00',  'visit zebras',    NULL),
                        (2,          '2021-02-10 15:45:00',  'visit giraffes',  NULL),
                        (3,          '2021-07-07 13:04:00',  'visit reptiles',  NULL),
                        (3,          '2021-07-07 13:50:00',  'visit bears',     NULL),
                        (3,          '2021-07-07 13:40:00',  'purchase',        'icecream'),
                        (3,          '2021-07-07 14:12:00',  'purchase',        'coffee'),
                        (4,          '2021-08-19 11:33:00',  'visit monkeys',   NULL),
                        (4,          '2021-08-19 11:18:00',  'visit lions',     NULL),
                        (4,          '2021-08-19 11:47:00',  'use restroom',    NULL),
                        (5,          '2022-04-12 10:55:00',  'visit zebras',    NULL),
                        (5,          '2022-04-12 11:42:00',  'purchase',        'coffee'),
                        (5,          '2022-04-12 10:45:00',  'purchase',        'hotdog'),
                        (5,          '2022-04-12 11:27:00',  'purchase',        'popcorn'),
                        (6,          '2022-04-12 14:00:00',  'purchase',        'icecream'),
                        (7,          '2022-05-09 12:38:00',  'use restroom',    NULL),
                        (7,          '2022-05-09 12:52:00',  'visit reptiles',  NULL),
                        (7,          '2022-05-09 12:30:00',  'visit zebras',    NULL),
                        (8,          '2022-07-07 15:00:00',  'purchase',        'popcorn'),
                        (8,          '2022-07-07 15:10:00',  'visit birds',     NULL),
                        (9,          '2022-07-11 12:13:00',  'purchase',        'popcorn'),
                        (9,          '2022-07-11 11:23:00',  'purchase',        'coffee'),
                        (9,          '2022-07-11 11:00:00',  'visit lions',     NULL),
                        (9,          '2022-07-11 11:54:00',  'visit monkeys',   NULL),
                        (10,         '2022-08-31 9:30:00',   'use restroom',    NULL);

我要的查询

所有购买过东西的英国游客的表格，以及那是什么。如果一个人购买了不止一件东西，请显示最后购买的物品。因此，一个包含 2 列的表：(1) visitor_id，(2) what_was_purchased。

期望的输出

#> +------------+--------------------+
#> | visitor_id | what_was_purchased |
#> +------------+--------------------+
#> |          3 | coffee             |
#> |          5 | coffee             |
#> |          6 | icecream           |
#> |          9 | popcorn            |
#> +------------+--------------------+#>

我的尝试

我已经走了这么远，甚至这个似乎也不行：

SELECT * 
FROM visitors AS v
LEFT JOIN activity_log AS al ON v.visitor_id = al.visitor_id 
      AND v.country_of_birth = 'UK' 
      AND al.visitor_id IN (
                        SELECT  visitor_id
                        FROM activity_log
                        GROUP BY visitor_id
                        HAVING SUM(CASE WHEN what_was_purchased IS NULL THEN 0 ELSE 1 END)  > 0
                       );

-- +------------+------------------+------------+---------------------+----------------+--------------------+
-- | visitor_id | country_of_birth | visitor_id | activity_time       | activity_name  | what_was_purchased |
-- +------------+------------------+------------+---------------------+----------------+--------------------+
-- |          1 | Bolivia          |       NULL | NULL                | NULL           | NULL               |
-- |          2 | UK               |       NULL | NULL                | NULL           | NULL               |
-- |          3 | UK               |          3 | 2021-07-07 13:04:00 | visit reptiles | NULL               |
-- |          3 | UK               |          3 | 2021-07-07 13:50:00 | visit bears    | NULL               |
-- |          3 | UK               |          3 | 2021-07-07 13:40:00 | purchase       | icecream           |
-- |          3 | UK               |          3 | 2021-07-07 14:12:00 | purchase       | coffee             |
-- |          4 | Bolivia          |       NULL | NULL                | NULL           | NULL               |
-- |          5 | UK               |          5 | 2022-04-12 10:55:00 | visit zebras   | NULL               |
-- |          5 | UK               |          5 | 2022-04-12 11:42:00 | purchase       | coffee             |
-- |          5 | UK               |          5 | 2022-04-12 10:45:00 | purchase       | hotdog             |
-- |          5 | UK               |          5 | 2022-04-12 11:27:00 | purchase       | popcorn            |
-- |          6 | UK               |          6 | 2022-04-12 14:00:00 | purchase       | icecream           |
-- |          7 | France           |       NULL | NULL                | NULL           | NULL               |
-- |          8 | USA              |       NULL | NULL                | NULL           | NULL               |
-- |          9 | UK               |          9 | 2022-07-11 12:13:00 | purchase       | popcorn            |
-- |          9 | UK               |          9 | 2022-07-11 11:23:00 | purchase       | coffee             |
-- |          9 | UK               |          9 | 2022-07-11 11:00:00 | visit lions    | NULL               |
-- |          9 | UK               |          9 | 2022-07-11 11:54:00 | visit monkeys  | NULL               |
-- |         10 | France           |       NULL | NULL                | NULL           | NULL               |
-- +------------+------------------+------------+---------------------+----------------+--------------------+
-- 19 rows in set (0.00 sec)

解释我的语法

我做了
```
LEFT JOIN activity_log AS al ON v.visitor_id = al.visitor_id 
      AND v.country_of_birth = 'UK'
```
基于 this answer，在连接之前已经只有 UK 行。如您所见，效果并不理想，因为我还有其他国家/地区的NULL。但我想我可以用 WHERE 子句过滤那些。（但是，我不知道为什么它没有像参考答案中那样被删除）。

我做了

AND al.visitor_id IN (
                        SELECT  visitor_id
                        FROM activity_log
                        GROUP BY visitor_id
                        HAVING SUM(CASE WHEN what_was_purchased IS NULL THEN 0 ELSE 1 END)  > 0
                       );

过滤在加入之前至少购买过一次的人。在这里，ID 为2 的访客也是空的，应该被删除。

怎么办？

为了这个问题，让我们忽略 NULL 行并假装通过“仅UK”和“至少一次购买”过滤成功：

-- pseudo result I manually edited
-- +------------+------------------+------------+---------------------+----------------+--------------------+
-- | visitor_id | country_of_birth | visitor_id | activity_time       | activity_name  | what_was_purchased |
-- +------------+------------------+------------+---------------------+----------------+--------------------+
-- |          3 | UK               |          3 | 2021-07-07 13:04:00 | visit reptiles | NULL               |
-- |          3 | UK               |          3 | 2021-07-07 13:50:00 | visit bears    | NULL               |
-- |          3 | UK               |          3 | 2021-07-07 13:40:00 | purchase       | icecream           |
-- |          3 | UK               |          3 | 2021-07-07 14:12:00 | purchase       | coffee             |           |
-- |          5 | UK               |          5 | 2022-04-12 10:55:00 | visit zebras   | NULL               |
-- |          5 | UK               |          5 | 2022-04-12 11:42:00 | purchase       | coffee             |
-- |          5 | UK               |          5 | 2022-04-12 10:45:00 | purchase       | hotdog             |
-- |          5 | UK               |          5 | 2022-04-12 11:27:00 | purchase       | popcorn            |
-- |          6 | UK               |          6 | 2022-04-12 14:00:00 | purchase       | icecream           |
-- |          9 | UK               |          9 | 2022-07-11 12:13:00 | purchase       | popcorn            |
-- |          9 | UK               |          9 | 2022-07-11 11:23:00 | purchase       | coffee             |
-- |          9 | UK               |          9 | 2022-07-11 11:00:00 | visit lions    | NULL               |
-- |          9 | UK               |          9 | 2022-07-11 11:54:00 | visit monkeys  | NULL               |
-- +------------+------------------+------------+---------------------+----------------+--------------------+

我怎样才能让每个人只获得与上次购买对应的行（如activity_time列所示）？请注意，人体内的时间是混乱的。我已经看到 this answer，它似乎就在现场，但我不知道如何将它合并到现有查询中。

【问题讨论】：

标签： mysql sql group-by greatest-n-per-group

【解决方案1】：

假设您使用的是最新版本的 MySql，典型的方法是使用行号窗口函数方法：

with l as (
    select l.visitor_id, l.what_was_purchased, 
      Row_Number() over(partition by l.visitor_id order by l.activity_time desc) rn
  from activity_log l
  join visitors v on v.visitor_id = l.visitor_id and v.country_of_birth = 'UK'
  where l.activity_name = 'purchase'
)
select visitor_id, what_was_purchased
from l
where rn = 1;

见Demo Fiddle

【讨论】：