在 SQL 的特定时间查找访问量最大的地点

Question

我有一个用户的 table，其中包含关于 user_id、用户购票地点以及用户购票时间的信息。

用户：

|------------|-------------|----------------------|
|  user_id   |  place      | purchase_time        |
|------------|-------------|----------------------|
|     1      |  New York   | 2021-11-27:17:00:21  |
|     1      |  Chicago    | 2021-11-25:19:00:21  |
|     1      |  Chicago    | 2021-11-23:03:00:21  |
|     1      |  Washington | 2021-11-21:07:00:21  |
|     1      |  Washington | 2021-11-19:12:00:21  |
|     1      |  Washington | 2021-11-17:00:00:21  |
|     1      |  Washington | 2021-11-15:23:00:21  |
|     1      |  Washington | 2021-11-12:21:00:21  |
|     2      |  Chicago    | 2021-09-25:01:00:21  |
|     2      |  Milwaukee  | 2021-09-24:02:00:21  |
|     2      |  Milwaukee  | 2021-09-23:03:00:21  |
|     2      |  New York   | 2021-09-22:19:00:21  |
|     2      |  Chicago    | 2021-09-21:01:00:21  |
|     3      |  Milwaukee  | 2021-10-27:12:31:21  |
|     3      |  Washington | 2021-10-24:07:01:23  |
|     3      |  Chicago    | 2021-10-21:01:78:89  |
|------------|-------------|----------------------|

我想添加一个新列，显示用户购票时最常去的地方。 table 想 (Snowflake):

|------------|-------------|----------------------|---------------------|
|  user_id   |  place      | purchase_time        | most_visited_place  |
|------------|-------------|----------------------|---------------------|
|     1      |  New York   | 2021-11-27:17:00:21  |    Washington       | <--- Washington, because at purchase_time This place was most visited by the user
|     1      |  Chicago    | 2021-11-25:19:00:21  |    Washington       | <--- Washington, because at purchase_time This place was most visited by the user
|     1      |  Chicago    | 2021-11-23:03:00:21  |    Washington       | <--- Washington, because at purchase_time This place was most visited by the user
|     1      |  Washington | 2021-11-21:07:00:21  |    Washington       | <--- Washington, because at purchase_time This place was most visited by the user
|     1      |  Washington | 2021-11-19:12:00:21  |    Washington       | <--- Washington, because at purchase_time This place was most visited by the user
|     1      |  Washington | 2021-11-17:00:00:21  |    Washington       | <--- Washington, because at purchase_time This place was most visited by the user
|     1      |  Washington | 2021-11-15:23:00:21  |    Washington       | <--- Washington, because at purchase_time This place was most visited by the user
|     1      |  Washington | 2021-11-12:21:00:21  |    Washington       | <--- Washington, because at purchase_time This place was most visited by the user
|     2      |  Chicago    | 2021-09-21:01:00:25  |    Chicago          | <-- tie, break. Both Chicago and Milwaukee were most visited then take the recent most visited
|     2      |  Milwaukee  | 2021-09-21:02:00:24  |    Milwaukee        | <--- Milwaukee, because at purchase_time This place was most visited by the user
|     2      |  Milwaukee  | 2021-09-21:03:00:23  |    Milwaukee        | <--- Milwaukee, because at purchase_time This place was most visited by the user
|     2      |  New York   | 2021-09-21:19:00:22  |    New York         | <-- tie, break. Both Chicago and New York were most visited then take the recent most visited
|     2      |  Chicago    | 2021-09-21:01:00:21  |    Chicago          | <--- Chicago, because at purchase_time This place was most visited by the user
|     3      |  Milwaukee  | 2021-10-27:12:31:21  |    Milwaukee        |
|     3      |  Washington | 2021-10-24:07:01:23  |    Washington       |
|     3      |  Chicago    | 2021-10-21:01:78:89  |    Chicago          |
|------------|-------------|----------------------|---------------------|

Answer 1

您想使用 WINDOW version of COUNT to get the "prior rows count" and then join to all the prior counted rows, and filter out the "best" via a QUALIFY

WITH prior_user AS (
    SELECT 
        user_id,
        place,
        purchase_time,
        COUNT(place) OVER (PARTITION BY user_id, place ORDER BY purchase_time ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS place_count
    FROM users
)
SELECT 
    u.user_id,
    u.place,
    u.purchase_time,
    p.place AS most_visited_place
FROM users u
JOIN prior_user p
    ON u.user_id = p.user_id AND u.purchase_time >= p.purchase_time
QUALIFY row_number() OVER (partition by u.user_id, u.purchase_time ORDER BY place_count DESC, p.purchase_time DESC) = 1

*这个 sql 还没有运行。

Answer 2

您可以 lateral 加入 Snowflake。 distinct 的使用有点难看，但我认为您可以用它代替 qualify，甚至可能得到更好的计划。从执行的角度来看，我很好奇这是否等同于其他答案。

select *
from Users u, lateral (
    select distinct first_value(place) over ()
        order by count(*) desc, max(u2.purchase_time) desc) as most_visited_place
    from Users u2
    where u2.user_id = u.user_id and u2.purchase_time <= u.purchase_time
    group by place
    --qualify row_number() over (order by u2.user_id) = 1 
) as mr
order by user_id, purchase_time desc

https://dbfiddle.uk/?rdbms=sqlserver_2019&fiddle=02784df13affab8027f7b052ad942d70

在 SQL 的特定时间查找访问量最大的地点

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