如何根据两个字段的最大值选择记录作为第三个字段的子集？

Question

给定以下简单表格：

+-----+-------------+-----------+---------+----+
| id_ | match_op_id | bookie_id | version | p1 |
+-----+-------------+-----------+---------+----+
|   1 |           1 |         1 |       1 |  1 |
|   2 |           1 |         2 |       1 |  5 |
|   3 |           1 |         1 |       2 |  3 |
|   4 |           1 |         2 |       2 |  4 |
|   5 |           2 |         1 |       1 |  5 |
|   6 |           2 |         2 |       1 |  3 |
|   7 |           2 |         2 |       2 |  4 |
+-----+-------------+-----------+---------+----+

我想构建一个查询，从每个 @ 的最大 version 中为每个 match_op_id 选择单个记录的 match_op_id、bookie_id 和 p1 字段（无关紧要） 987654327@，然后是最大值p1。所以从上面我会得到输出：

+-------------+-----------+----+
| match_op_id | bookie_id | p1 |
+-------------+-----------+----+
|           1 |         2 |  4 |
|           2 |         1 |  5 |
+-------------+-----------+----+

从我得到的另一个问题的答案来看：

SELECT o1.match_op_id, o1.bookie_id, MAX(o1.p1) p1
FROM odds_op o1
WHERE o1.version = (SELECT MAX(o2.version) FROM odds_op o2 WHERE o2.match_op_id = o1.match_op_id)
GROUP BY o1.match_op_id

但是，我不知道如何通过bookie_id 实现最大version。任何帮助将不胜感激。

---

编辑：

为了解决 cmets 中提出的问题...对于每个 match_op_id，我正在寻找查询以查找最高 version by bookie_id，然后返回最高 p1。所以在上面的例子中 match_op_id "2" 那么bookie_id "1" 的最大 version 是 "1"，而 p1 是 "5"。 version 的最大值为 bookie_id“2”为“2”，p1 为“4”。因此查询应该返回bookie_id“1”和“5”的p1。

Answer 1

SQL

WITH max_versions AS
(SELECT match_op_id, bookie_id, MAX(version) AS version
 FROM tbl
 GROUP BY match_op_id, bookie_id),
mv_full AS
(SELECT t.match_op_id, t.bookie_id, t.p1
 FROM max_versions mv
 JOIN tbl t
   ON mv.match_op_id = t.match_op_id
  AND mv.bookie_id = t.bookie_id
  AND mv.version = t.version),
max_p1s AS
(SELECT match_op_id, MAX(p1) AS p1
 FROM mv_full
 GROUP BY match_op_id)
SELECT mf.*
FROM mv_full mf
JOIN max_p1s mp
ON mp.match_op_id = mf.match_op_id
AND mp.p1 = mf.p1

演示

DB Fiddle Demo

Answer 2

SELECT DISTINCT match_op_id, bookie_id, max_version, max_p1
FROM (
SELECT distinct st1.match_op_id match_op_id, st2.bookie_id, st2.max_version, st3.max_p1
        FROM SIMPLE_table st1
        JOIN ( -- version
        SELECT match_op_id, bookie_id, MAX(`version`) max_version FROM simple_table GROUP BY match_op_id, bookie_id) st2
        ON st1.match_op_id = st2.match_op_id
        JOIN ( -- p1
        SELECT match_op_id, bookie_id, `version`, MAX(p1) max_p1 FROM simple_table GROUP BY match_op_id, bookie_id, `version`) st3
        ON st2.max_version=st3.`version` AND st2.bookie_id=st3.bookie_id AND st1.match_op_id=st3.match_op_id) all_
GROUP BY max_version HAVING MAX(max_p1)

给出：

"match_op_id"   "bookie_id" "max_version"   "max_p1"
"2"             "1"         "1"             "5"
"1"             "2"         "2"             "4"

你觉得这对吗？

Answer 3

您可以使用两次ROW_NUMBER()窗口函数。
首先使用它为match_op_id 和bookie_id 的每个组合获取最大version 的行，然后对于返回的行，使用它为每个match_op_id 获取最大p1 的行：

SELECT match_op_id, bookie_id, p1
FROM (
  SELECT *, ROW_NUMBER() OVER (PARTITION BY match_op_id ORDER BY p1 DESC) rn2
  FROM (
    SELECT *, ROW_NUMBER() OVER (PARTITION BY match_op_id, bookie_id ORDER BY version DESC) rn1
    FROM odds_op
  ) t  
  WHERE rn1 = 1
) t
WHERE rn2 = 1

或者用ROW_NUMBER()和FIRST_VALUE：

SELECT DISTINCT match_op_id, 
       FIRST_VALUE(bookie_id) OVER (PARTITION BY match_op_id ORDER BY p1 DESC) bookie_id, 
       MAX(p1) OVER (PARTITION BY match_op_id) p1
FROM (
  SELECT *, ROW_NUMBER() OVER (PARTITION BY match_op_id, bookie_id ORDER BY version DESC) rn
  FROM odds_op
) t  
WHERE rn = 1

请参阅demo。
结果：

match_op_id	bookie_id	p1
1	2	4
2	1	5