mysql - 查询是否可以优化 : Get a records max date then join the max date's values-6ren

mysql - 查询是否可以优化 : Get a records max date then join the max date's values

转载作者：IT老高更新时间：2023-10-29 00:12:13

我已经创建了一个返回我想要的结果的查询，但我觉得必须有更好的方法来执行此操作。任何指导将不胜感激。

我正在尝试获取特定 session 的所有项目并加入他们的最大 session 日期 < X 并加入最大日期的委员会首字母缩写词。 X 是当前 session 日期。

我已经尝试了几个不同的查询，但除了下面的查询之外，没有一个始终返回预期的结果。

您可以转到 rextester 查看此查询的运行情况。 .

DROP TABLE IF EXISTS `committees`;
CREATE TABLE committees
    (`id` int, `acronym` varchar(4))
;

INSERT INTO committees
    (`id`, `acronym`)
VALUES
    (1, 'Com1'),
    (2, 'Com2'),
    (3, 'Com3')
;

DROP TABLE IF EXISTS `meetings`;
CREATE TABLE meetings
    (`id` int, `date` datetime, `committee_id` int)
;

INSERT INTO meetings
    (`id`, `date`, `committee_id`)
VALUES
    (1, '2017-01-01 00:00:00', 1),
    (2, '2017-02-02 00:00:00', 2),
    (3, '2017-03-03 00:00:00', 2)
;

DROP TABLE IF EXISTS `agenda_items`;
CREATE TABLE agenda_items
    (`id` int, `name` varchar(6))
;

INSERT INTO agenda_items
    (`id`, `name`)
VALUES
    (1, 'Item 1'),
    (2, 'Item 2'),
    (3, 'Item 3')
;

DROP TABLE IF EXISTS `join_agenda_items_meetings`;
CREATE TABLE join_agenda_items_meetings
    (`id` int, `agenda_item_id` int, `meeting_id` int)
;

INSERT INTO join_agenda_items_meetings
    (`id`, `agenda_item_id`, `meeting_id`)
VALUES
    (1, 1, 1),
    (2, 1, 2),
    (3, 2, 1),
    (4, 3, 2),
    (5, 2, 1),
    (6, 1, 3)
;




SELECT agenda_items.id, 
       meetings.id, 
       meetings.date, 
       sub_one.max_date, 
       sub_two.acronym 
FROM   agenda_items 
       LEFT JOIN (SELECT ai.id                AS ai_id, 
                         me.id                AS me_id, 
                         Max(me.date) AS max_date 
                  FROM   agenda_items AS ai 
                         JOIN join_agenda_items_meetings AS jaim 
                           ON jaim.agenda_item_id = ai.id 
                         JOIN meetings AS me 
                           ON me.id = jaim.meeting_id 
                  WHERE  me.date < '2017-02-02' 
                  GROUP  BY ai_id) sub_one 
              ON sub_one.ai_id = agenda_items.id 
       LEFT JOIN (SELECT agenda_items.id       AS age_id, 
                         meetings.date AS meet_date, 
                         committees.acronym    AS acronym 
                  FROM   agenda_items 
                         JOIN join_agenda_items_meetings 
                           ON join_agenda_items_meetings.agenda_item_id = agenda_items.id 
                         JOIN meetings 
                           ON meetings.id = join_agenda_items_meetings.meeting_id 
                         JOIN committees 
                           ON committees.id = meetings.committee_id 
                  WHERE  meetings.date) sub_two 
              ON sub_two.age_id = agenda_items.id 
                 AND sub_one.max_date = sub_two.meet_date 
       JOIN join_agenda_items_meetings 
         ON agenda_items.id = join_agenda_items_meetings.agenda_item_id 
       JOIN meetings 
         ON meetings.id = join_agenda_items_meetings.meeting_id 
WHERE  meetings.id = 2;

审查/测试答案(修订):*

我已经根据评论修改了测试。

由于我对这个问题悬赏，我觉得我应该展示我如何评估答案并提供一些反馈。总的来说，我非常感谢所有提供帮助的人，谢谢。

为了测试，我审查了针对以下方面的查询:

initial rextester
一个modified version of the initial rextester对 2 个独立数据集的所有 4 个查询
来 self 实际数据库的更大数据集

我使用 EXPLAIN 的原始查询

+----+-------------+---------------------------+------+----------------------------------------------+
| id | select_type | table                     | rows | Extra                                        |
+----+-------------+---------------------------+------+----------------------------------------------+
|  1 | PRIMARY     | meetings                  |    1 |                                              |
|  1 | PRIMARY     | join_agenda_item_meetings | 1976 | Using where; Using index                     |
|  1 | PRIMARY     | agenda_items              |    1 | Using index                                  |
|  1 | PRIMARY     | <derived2>                | 1087 |                                              |
|  1 | PRIMARY     | <derived3>                | 2202 |                                              |
|  3 | DERIVED     | join_agenda_item_meetings | 1976 | Using index                                  |
|  3 | DERIVED     | meetings                  |    1 | Using where                                  |
|  3 | DERIVED     | committees                |    1 |                                              |
|  3 | DERIVED     | agenda_items              |    1 | Using index                                  |
|  2 | DERIVED     | jaim                      | 1976 | Using index; Using temporary; Using filesort |
|  2 | DERIVED     | me                        |    1 | Using where                                  |
|  2 | DERIVED     | ai                        |    1 | Using index                                  |
+----+-------------+---------------------------+------+----------------------------------------------+
12 rows in set (0.02 sec)

Paul Spiegel 的回答。

最初的答案有效并且似乎是最有效的选项，比我的要多得多。

Paul Spiegel 的第一个查询提取的行最少，比我的更短且可读性更好。它也不需要引用日期，这在编写时也会更好。

+----+--------------------+-------+------+--------------------------+
| id | select_type        | table | rows | Extra                    |
+----+--------------------+-------+------+--------------------------+
|  1 | PRIMARY            | m1    |    1 |                          |
|  1 | PRIMARY            | am1   | 1976 | Using where; Using index |
|  1 | PRIMARY            | am2   |    1 | Using index              |
|  1 | PRIMARY            | m2    |    1 |                          |
|  2 | DEPENDENT SUBQUERY | am3   |    1 | Using index              |
|  2 | DEPENDENT SUBQUERY | m3    |    1 | Using where              |
|  2 | DEPENDENT SUBQUERY | c3    |    1 | Using where              |
+----+--------------------+-------+------+--------------------------+
7 rows in set (0.00 sec)

在将 DISTINCT 添加到 select 语句时，此查询也会返回正确的结果。虽然这个查询的性能不如第一个(但很接近)。

+----+-------------+------------++------+-------------------------+
| id | select_type | table      | rows | Extra                    |
+----+-------------+------------++------+-------------------------+
|  1 | PRIMARY     | <derived2> |    5 | Using temporary          |
|  1 | PRIMARY     | am         |    1 | Using index              |
|  1 | PRIMARY     | m          |    1 |                          |
|  1 | PRIMARY     | c          |    1 | Using where              |
|  2 | DERIVED     | m1         |    1 |                          |
|  2 | DERIVED     | am1        | 1787 | Using where; Using index |
|  2 | DERIVED     | am2        |    1 | Using index              |
|  2 | DERIVED     | m2         |    1 |                          |
+----+-------------+------------+------+--------------------------+
8 rows in set (0.00 sec)

Stefano Zanini 的回答

此查询使用 DISTINCT 返回预期结果。当使用 EXPLAIN 和被拉取的行数时，与我原来的查询相比，这个查询更有效，但 Paul Spiegel 的稍微好一点。

+----+-------------+------------+------+---------------------------------+
| id | select_type | table      | rows | Extra                           |
+----+-------------+------------+------+---------------------------------+
|  1 | PRIMARY     | me         |    1 | Using temporary; Using filesort |
|  1 | PRIMARY     | rel        | 1787 | Using where; Using index        |
|  1 | PRIMARY     | <derived2> | 1087 |                                 |
|  1 | PRIMARY     | rel2       |    1 | Using index                     |
|  1 | PRIMARY     | me2        |    1 | Using where                     |
|  1 | PRIMARY     | co         |    1 |                                 |
|  2 | DERIVED     | t1         | 1787 | Using index                     |
|  2 | DERIVED     | t2         |    1 | Using where                     |
+----+-------------+------------+------+---------------------------------+
8 rows in set (0.00 sec)

EoinS 的回答

如评论中所述，如果 session 是连续的，则此答案有效，但不幸的是，它们可能并非如此。

最佳答案

这个有点疯狂..让我们一步一步来:

第一步是基本连接

set @meeting_id = 2;

select am1.meeting_id,
       am1.agenda_item_id,
       m1.date as meeting_date
from meetings m1
join join_agenda_items_meetings am1 on am1.meeting_id = m1.id
where m1.id = @meeting_id;

我们选择 session (id = 2) 和相应的 agenda_item_ids。这将返回我们需要的前三列行。

下一步是获取每个议程项目的最后 session 日期。我们需要将第一个查询与连接表和相应的 session 连接起来(id = 2 - am2.meeting_id <> am1.meeting_id 除外)。我们只希望 session 的日期早于实际 session ( m2.date < m1.date )。在所有这些 session 中，我们只需要每个议程项目的最新日期。所以我们按议程项目分组并选择 max(m2.date) :

select am1.meeting_id,
       am1.agenda_item_id,
       m1.date as meeting_date,
       max(m2.date) as max_date
from meetings m1
join join_agenda_items_meetings am1 on am1.meeting_id = m1.id
left join join_agenda_items_meetings am2 
    on  am2.agenda_item_id = am1.agenda_item_id
    and am2.meeting_id <> am1.meeting_id
left join meetings m2 
    on  m2.id = am2.meeting_id
    and m2.date < m1.date
where m1.id = @meeting_id
group by m1.id, am1.agenda_item_id;

这样我们得到第四列(max_date)。

最后一步是选择 acronym session 的最后日期 ( max_date )。这是最疯狂的部分——我们可以在 SELECT 子句中使用相关子查询。我们可以使用 max(m2.date)对于相关性:

select c3.acronym
from meetings m3
join join_agenda_items_meetings am3 on am3.meeting_id = m3.id
join committees c3 on c3.id = m3.committee_id
where am3.agenda_item_id = am2.agenda_item_id
  and m3.date = max(m2.date)

最终查询将是:

select am1.meeting_id,
       am1.agenda_item_id,
       m1.date as meeting_date,
       max(m2.date) as max_date,
       (   select c3.acronym
           from meetings m3
           join join_agenda_items_meetings am3 on am3.meeting_id = m3.id
           join committees c3 on c3.id = m3.committee_id
           where am3.agenda_item_id = am2.agenda_item_id
             and m3.date = max(m2.date)
       ) as acronym
from meetings m1
join join_agenda_items_meetings am1 on am1.meeting_id = m1.id
left join join_agenda_items_meetings am2 
    on  am2.agenda_item_id = am1.agenda_item_id
    and am2.meeting_id <> am1.meeting_id
left join meetings m2 
    on  m2.id = am2.meeting_id
    and m2.date < m1.date
where m1.id = @meeting_id
group by m1.id, am1.agenda_item_id;

http://rextester.com/JKK60222

说实话，我很惊讶你可以使用 max(m2.date)在子查询中。

另一个解决方案 - 在子查询(派生表)中使用第二个查询。使用 max_date 通过 session 和加入表加入委员会.只保留带有首字母缩写词的行和没有 max_date 的行.

select t.*, c.acronym
from (
    select am1.meeting_id,
           am1.agenda_item_id,
           m1.date as meeting_date,
           max(m2.date) as max_date
    from meetings m1
    join join_agenda_items_meetings am1 on am1.meeting_id = m1.id
    left join join_agenda_items_meetings am2 
        on  am2.agenda_item_id = am1.agenda_item_id
        and am2.meeting_id <> am1.meeting_id
    left join meetings m2 
        on  m2.id = am2.meeting_id
        and m2.date < m1.date
    where m1.id = @meeting_id
    group by m1.id, am1.agenda_item_id
) t
left join join_agenda_items_meetings am
    on  am.agenda_item_id = t.agenda_item_id
    and t.max_date is not null
left join meetings m
    on  m.id   = am.meeting_id
    and m.date = t.max_date
left join committees c on c.id = m.committee_id
where t.max_date is null or c.acronym is not null;

http://rextester.com/BBMDFL23101

关于mysql - 查询是否可以优化 : Get a records max date then join the max date's values，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/42564643/

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mysql - 查询是否可以优化 : Get a records max date then join the max date's values