按 2 个表之间的 ID 和日期匹配，或最后已知的匹配 ID答案

【问题标题】：Match by id and date between 2 tables, OR last known match id按 2 个表之间的 ID 和日期匹配，或最后已知的匹配 ID
【发布时间】：2023-01-12 05:29:38
【问题描述】：

尝试进行以下工作：

T1：取id per dt，其中name = A是most recent by load_id
- 在 23 年 1 月 5 日通知 2 条记录，其中 load_id 为 2 和 3 => 采用 load_id = 3
T2：并显示对应的id per dt for each param行，最近的load_id
- 仅注意 load_id = 13 保留在 05-Jan-23
T2：如果日期现在在T1 可用，请保留T2 行匹配最后已知的id

小提琴：https://dbfiddle.uk/-JO16GSj

我的 SQL 似乎有点乱。可以简化吗？

SELECT t2.dt, t2.param, t2.load_id, t2.id FROM
(SELECT 
  dt, 
  param,
  load_id,
  MAX(load_id) OVER (PARTITION BY dt, param) AS max_load_id,
  id
FROM table2) t2

LEFT JOIN
(SELECT * FROM
(SELECT
  dt, 
  id,
  load_id,
  MAX(load_id) OVER (PARTITION BY dt) AS max_load_id
FROM table1
WHERE name = 'A') t1_prep
WHERE t1_prep.load_id = t1_prep.max_load_id) t1

ON t1.dt = t2.dt and t1.id = t2.id
  
WHERE t2.load_id = t2.max_load_id

ORDER BY 1, 2

【问题讨论】：

标签： sql oracle

【解决方案1】：

您的查询可以重写为：

SELECT t2.*
FROM   ( SELECT *
         FROM   table2
         ORDER BY RANK() OVER (PARTITION BY dt, param ORDER BY load_id)
         FETCH FIRST ROW WITH TIES
       ) t2
       LEFT OUTER JOIN
       ( SELECT *
         FROM   table1
         WHERE  name = 'A'
         ORDER BY RANK() OVER (PARTITION BY dt ORDER BY load_id DESC)
         FETCH FIRST ROW WITH TIES
       ) t1
       ON t1.dt = t2.dt and t1.id = t2.id
ORDER BY t2.dt, t2.param

但是，由于来自 t1 的列永远不会输出并与 LEFT OUTER JOIN 连接（并且每个 dt 只会输出单行），因此是否找到与 t1 和该表的匹配无关紧要可以从查询中删除，将其简化为：

SELECT *
FROM   (
  SELECT *
  FROM   table2
  ORDER BY RANK() OVER (PARTITION BY dt, param ORDER BY load_id)
  FETCH FIRST ROW WITH TIES
)
ORDER BY dt, param;

或使用您的查询：

SELECT dt, param, load_id, id
FROM   (
  SELECT dt, param, load_id, id,
         MAX(load_id) OVER (PARTITION BY dt, param) AS max_load_id
  FROM   table2
)
WHERE  load_id = max_load_id
ORDER BY dt, param

fiddle

【讨论】：