合并 MySQL SELECT 语句

Question

我有以下tabletbl_crps，注意crp列出现了不止一次, 但 id 和 starts_on 在具有相同 crp 的记录中总是不同的.

id |  crp | starts_on  |
------------------
1  |  20  | 2020-02-09 |
2  |  31  | 2019-06-05 |
3  |  20  | 2018-01-10 |
4  |  31  | 2021-07-03 |
5  |  58  | 2022-01-01 |
6  |  58  | 2025-02-02 |

我需要提取符合以下条件的记录：

如果对于crp一个或多个过去starts_on的记录存在，那么与具有 crp 的那些记录中最大的 id 必须是 returned
如果 crp 不存在具有过去日期的记录，则在具有该 crp[ 的所有记录中具有最大 ID 的记录=50=]

我可以成功 select 记录符合以下条件的第一个条件。

SELECT * FROM tbl_crps AND id IN (SELECT MAX(b.id) FROM tbl_crps b WHERE b.starts_on <= "2020-10-02")

将 return 记录 ID 为 3 和 2 的记录。

我需要实现的是将 starts_on 不是过去的记录和 crp[=50= 的记录添加到结果中] 不在第一个结果集中。

可能吗？也许是这样？

SELECT * FROM tbl_crps a WHERE id IN ( SELECT MAX(a.id) FROM tbl_crps a WHERE a.starts_on <= "2020-10-02" GROUP BY a.crp ) UNION ( SELECT * FROM tbl_crps b WHERE id IN ( SELECT MAX(c.id) FROM tbl_crps c WHERE c.starts_on > "2020-10-02" AND c.crp NOT IN ( SELECT d.crp FROM tbl_crps d WHERE id IN ( SELECT MAX(e.id) FROM tbl_crps e WHERE e.starts_on <= "2020-10-02" GROUP BY e.crp) ) ) )

我正在使用 MySQL 5.7

Answer 1

我不确定我是否完全理解您的要求，但我认为此查询可以满足您的要求，或者非常接近您的要求：

WITH cteData AS (SELECT b.CRP, MAX(b.ID) AS ID
                   FROM tbl_crps b
                   WHERE b.starts_on <= "2020-10-02"
                   GROUP BY b.CRP)
SELECT *
  FROM tbl_crps a
  WHERE a.ID IN (SELECT ID
                   FROM (SELECT b.CRP, b.ID
                           FROM cteData b
                         UNION
                         SELECT d.CRP, d.ID
                           FROM tbl_crps d
                           WHERE d.starts_on > "2020-10-02" AND
                                 d.CRP NOT IN (SELECT crp
                                                 FROM cteData)) c)

这个returns

id  crp starts_on
2   31  2019-06-05
3   20  2018-01-10
5   58  2022-01-01
6   58  2025-02-02

根据我对要求的理解，这似乎是正确的。简而言之

1 - NOT included because its date is in the past ('current' date = 2020-10-02) 
    and its ID is NOT the greatest among records in the past with that CRP.
2 - included because its date is in the past and its ID is the greatest
    among records with that CRP which have dates in the past.
3 - date is in the past, ID is greatest for that CRP with date in the past
4 - NOT included because date is in the future, but CRP is included in recs
    from first subquery.
5 - date is in the future, and CRP does not occur in first subquery
6 - date is in the future, and CRP does not occur in first subquery

db<>fiddle here

Answer 2

至于 Akina 对我的新问题的回答（已关闭），这就可以了。

SELECT id, crp, starts_on
FROM tbl_crps
NATURAL JOIN (SELECT crp, MAX(id) id
              FROM tbl_crps
              WHERE starts_on < CURRENT_DATE
              GROUP BY crp) x
UNION ALL
SELECT id, crp, starts_on
FROM tbl_crps
NATURAL JOIN (SELECT crp, MAX(id) id
              FROM tbl_crps
              GROUP BY crp
              HAVING MIN(starts_on) >= CURRENT_DATE) x;

fiddle

我的解决方案（性能不佳）

SELECT * FROM tbl_crps a
WHERE id IN (
     SELECT MAX(a.id) FROM tbl_crps a WHERE a.starts_on <= "2020-10-02"
 )
UNION
(
    SELECT * FROM tbl_crps b
        WHERE id IN (
             SELECT MAX(c.id) FROM tbl_crps c WHERE c.starts_on > "2020-10-02"
             AND c.crp NOT IN (
                  SELECT d.crp FROM tbl_crps d
                  WHERE id IN (
                       SELECT MAX(e.id) FROM tbl_crps e WHERE e.starts_on <= "2020-10-02")
                   )
        )
)

合并 MySQL SELECT 语句

Merge MySQL SELECT statements

mysql

mysql-5.7