如何使用两个不同的列查询 MAX 子查询的 MIN 值？

Question

我有一张这样的桌子：

+---------------+--------------+------+-----+----------+
| Field         | Type         | Null | Key | Default  |
+---------------+--------------+------+-----+----------+
| id            | smallint(6)  | NO   | PRI | NULL     |
| Book          | tinyint(4)   | NO   |     | NULL     |
| Chapter       | smallint(6)  | NO   |     | NULL     |
| Paragraph     | smallint(6)  | NO   |     | NULL     |
| Text          | text         | YES  |     | NULL     |
| RevisionNum   | mediumint(9) | NO   | PRI | NULL     |
+---------------+--------------+------+-----+----------+

mysql> select id,Book,Chapter,Paragraph,RevisionNum FROM MyTable ORDER BY id LIMIT 11;

+-----+------+---------+-----------+-------------+
| id  | Book | Chapter | Paragraph | RevisionNum |
+-----+------+---------+-----------+-------------+
|   1 |    1 |       1 |         1 |           0 |
|   1 |    1 |       1 |         1 |           1 |
|   1 |    1 |       1 |         1 |           2 |
|   2 |    1 |       2 |         2 |           0 |
|   2 |    1 |       2 |         2 |           1 |
|   2 |    1 |       2 |         2 |           2 |
|   2 |    1 |       2 |         2 |           3 |
|   3 |    1 |       2 |         3 |           0 |
|   4 |    1 |       2 |         4 |           0 |
|   4 |    1 |       2 |         4 |           1 |
|   5 |    1 |       3 |         5 |           0 |
+-----+------+---------+-----------+-------------+

要查找没有未修改段落的书或章节， 我想查询最大值的最小值 该章节或书籍的所有不同ID，或者在 某些时尚确定没有 id 未被编辑（带有 MAX(RevisionNum) 为零)。

迄今为止，我的大部分尝试都以这样的错误告终：

SELECT DISTINCT Book,RecordNum FROM MyTable
    -> WHERE 0 < ALL (SELECT DISTINCT RecordNum,MAX(RevisionNum) 
                      FROM MyTable 
                      WHERE MAX(RevisionNum) > 0);
                      
ERROR 1111 (HY000): Invalid use of group function

...而且我根本没有使用“GROUP BY”功能！

下面的查询会产生结果，但很简单 给出所有 id，实际上并不显示唯一的 根据要求设置账簿记录。怎么会这样？

SELECT DISTINCT Book,id,MAX(RevisionNum) FROM MyTable GROUP BY id LIMIT 5;

+------+----+------------------+
| Book | id | MAX(RevisionNum) |
+------+----+------------------+
|    1 |  1 |               30 |
|    1 |  2 |               16 |
|    1 |  3 |               15 |
|    1 |  4 |               10 |
|    1 |  5 |                9 |
+------+----+------------------+

正确的查询应该是给出更像这样的结果：

+------+-----+-----------------------+
| Book |  id | MIN(MAX(RevisionNum)) |
+------+-----+-----------------------+
|    1 |   5 |                     3 |
|    2 |  17 |                     1 |
|    3 |  33 |                     2 |
|    4 | 147 |                     0 |
|    5 | 225 |                     2 |
+------+-----+-----------------------+

Answer 1

您是否在寻找两个级别的聚合？

select id, book, min(max_revisionnum)
from (select id, book, chapter, paragraph, max(revisionnum) as max_revisionnum
      from mytable
      group by id, book, chapter, paragraph
     ) t
group by id, book;

编辑：

根据您的评论，您可以使用：

select *
from (select id, book, chapter, paragraph, max(revisionnum) as max_revisionnum,
             row_number() over (partition by book order by max(revisionnum) desc) as seqnum
      from mytable
      group by id, book, chapter, paragraph
     ) t
where seqnum = 1;

Here 是一个 dbfiddle。

在旧版本的 MariaDB 中，您可以使用关联子查询：

select t.*
from mytable t
where (id, book, chapter, paragraph, revisionnum) = (select t2.id, t2.book, t2.chapter, t2.paragraph, t2.revisionnum
                                                     from mytable t2
                                                     where t2.book = t.book
                                                     order by t2.revisionnum desc
                                                     limit 1
                                                    );

对于此查询，请尝试在 (book, revisionnum desc) 上添加索引。