Select 具有每个位置最近日期的行，并使用 MariaDB 按位置将每行的最近日期递增 1

Question

我有一个 table 的位置 'Date column'。我必须按每组 locationID 查找最近的日期，例如locationID 1 的最近日期为“2022 年 5 月 31 日”。从 locationID 组中找到最近的日期后，我必须在最近的日期中添加 14 天并将其存储在 NewDate 列中。并在该组 locationID 的其他行的新日期中添加 + 1。

我的table是：

id  locationID    Date        NewDate   
1       1       31 May 2022
2       1       16 May 2022
3       1       28 Apr 2021
4       2       29 Mar 2022
5       2       22 Feb 2022
6       3       14 Jun 2022
7       3       27 Oct 2021
8       4       01 Feb 2022
9       4       04 May 2022
10      4       14 Jun 2021
11      5       01 Jun 2022
12      5       29 May 2022
13      5       20 Sep 2022
14      5       11 Aug 2022
15      5       03 Aug 2022

答案应该如下：

例如对于 locationID = 1

   id     locationID       Date             NewDate   
    1       1         31 May 2022       14 Jun 2022    // Recent Date + 14 Days - 31 May + 14 Days
    2       1         16 May 2022       15 Jun 2022    // Recent Date + 15 Days - 31 May + 15 Days
    3       1         28 Apr 2021       16 Jun 2022    // Recent Date + 16 Days - 31 May + 16 Days

我遇到过几个类似的 post 并且发现最近的日期是这样的：

SELECT L.*
FROM Locations L 
INNER JOIN 
    (SELECT locationID, MAX(Date) AS MAXdate
     FROM Locations 
     GROUP BY locationID) groupedL
     ON L.locationID = groupedL.locationID
     AND L.Date = groupedL.MAXdate

使用上面的代码我可以找到每个位置的最近日期，但是如何添加和增加所需的天数并将其存储到 NewDate 列？我是MariaDB新手，请推荐类似postlink，任何参考文档或博客。我是否应该创建一些函数来执行此逻辑并调用该函数以将所需日期存储在 NewDate 列中？我不确定请建议。谢谢。

结果应如下所示：

id     locationID       Date             NewDate   
    1       1         31 May 2022       14 Jun 2022    // Recent Date for locationid 1 + 14 Days - 31 May + 14 Days
    2       1         16 May 2022       15 Jun 2022    // Recent Date for locationid 1 + 15 Days - 31 May + 15 Days
    3       1         28 Apr 2021       16 Jun 2022    // Recent Date for locationid 1 + 16 Days - 31 May + 16 Days
    4       2         29 Mar 2022       12 APR 2022    // Recent Date for locationid 2 + 14 Days
    5       2         22 Feb 2022       13 APR 2022    // Recent Date for locationid 2 + 15 Days
    6       3         14 Jun 2022       28 JUN 2022    // Recent Date for locationid 3 + 14 Days
    7       3         27 Oct 2021       29 JUN 2022    // Recent Date for locationid 3 + 15 Days
    8       4         01 Feb 2022       18 MAY 2022    // Recent Date for locationid 4 + 14 Days
    9       4         04 May 2022       19 MAY 2022    // Recent Date for locationid 4 + 15 Days
    10      4         14 Jun 2021       20 MAY 2022    // Recent Date for locationid 4 + 16 Days
    11      5         01 Jun 2022       04 OCT 2022    // Recent Date for locationid 5 + 14 Days
    12      5         29 May 2022       05 OCT 2022    // Recent Date for locationid 5 + 15 Days
    13      5         20 Sep 2022       06 OCT 2022    // Recent Date for locationid 5 + 16 Days
    14      5         11 Aug 2022       07 OCT 2022    // Recent Date for locationid 5 + 17 Days
    15      5         03 Aug 2022       08 OCT 2022    // Recent Date for locationid 5 + 18 Days

Answer 1

您可以使用 cte:

with cte as (
   select l1.*, l2.m, (select sum(l4.id < l1.id and l4.locationid = l1.locationid) from locations l4) inc from locations l1 
   join (select l3.locationid, max(l3.dt) m from locations l3 group by l3.locationid) l2 on l1.locationid = l2.locationid
)
select c.id, c.locationid, c.dt, c.m + interval 14 + c.inc day from cte c

Answer 2

您可以使用解析 window 函数并通过加入 sub-query（适用于 MariaDB）来更新原始 table：

update t
join (
  select Id, 
    Date_Add(First_Value(date) over(partition by locationId order by date desc), 
      interval (13 + row_number() over(partition by locationId order by date desc)) day 
  ) NewDate
  from t
)nd on t.id = nd.id
set t.Newdate = nd.NewDate;

参见DB<>Fiddle示例

Select 具有每个位置最近日期的行，并使用 MariaDB 按位置将每行的最近日期递增 1

Select row with most recent date per location and increment recent date by 1 for each row by location using MariaDB

sql

mariadb-10.2