我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做到这一点。任何指导都将不胜感激。
我正在尝试获取特定会议的所有项目并加入他们的最大会议日期＆lt; X并加入最大日期的委员会首字母缩略词。 X是当前的会议日期。
我尝试过几个不同的查询，但除了下面的查询之外，没有任何查询一直返回预期的结果。
您可以转到rextester来查看此查询。
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;
审核/测试答案（已修订）：*
我已根据所做的评论修改了测试。
由于我对这个问题给予了赏金，我觉得我应该展示我是如何评估答案并提供一些反馈的。总的来说，我非常感谢所有帮助，谢谢。
为了进行测试，我查看了以下问题：
我的原始查询与EXPLAIN
+----+-------------+---------------------------+------+----------------------------------------------+
| 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的答案。
初始答案有效，似乎是最有效的选项，远远超过我的。
Paul Spiegel的第一个查询提取的行数最少，比我的更短，更易读。它也不需要引用一个更好的日期来编写它。
+----+--------------------+-------+------+--------------------------+
| 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)
将DISTINCT添加到select语句时，此查询也会返回正确的结果。这个查询的效果不如第一个（但它很接近）。
+----+-------------+------------++------+-------------------------+
| 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
此查询确实使用DISTINCT返回预期结果。当使用EXPLAIN和被拉动的行数时，与原始查询相比，此查询更有效，但Paul Spiegel的更好一点。
+----+-------------+------------+------+---------------------------------+
| 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'回答
如评论中所述，如果会议是连续的，这个答案是有效的，但不幸的是，它们可能并不存在。
答案 0 :(得分：5)
这个有点疯狂..让我们一步一步地做：
第一步是基本联接
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;
我们选择会议（id = 2）和相应的agenda_item_ids。这将返回前三列所需的行。
下一步是获取每个议程项目的最后一次会议日期。我们需要将第一个查询与连接表和相应的会议连接起来（id = 2 - am2.meeting_id <> am1.meeting_id除外）。我们只希望会议在实际会议之前有一个日期（m2.date < m1.date）。在所有这些会议中，我们只想要每个议程项目的最新日期。因此，我们按议程项目进行分组，然后选择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）。
最后一步是选择会议的acronym和最后一个日期（max_date）。这是一个疯狂的部分 - 我们可以在SELECT子句中使用相关的子查询。我们可以使用max(m2.date)进行关联：
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;
说实话，我很惊讶您可以在子查询中使用max(m2.date)。
另一种解决方案 - 在子查询（派生表）中使用第二个查询。使用max_date加入会议和联接表的委员会。仅保留带有首字母缩写词和行而没有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;
答案 1 :(得分：3)
Using your schema I used the below query, assuming that all meetings entries are sequential:
set @mymeeting = 2;
select j.agenda_item_id, m.id, m.date, mp.date, c.acronym
from meetings m
left join join_agenda_items_meetings j on j.meeting_id = m.id
left join join_agenda_items_meetings jp on jp.meeting_id = m.id -1 and jp.agenda_item_id = j.agenda_item_id
left join meetings mp on mp.id = jp.meeting_id
left join committees c on mp.committee_id = c.id
where m.id = @mymeeting;
I create a variable just to make it easy to change meetings on the fly.
Here is a functional example in Rextester
Thanks for making your schema so easy to reproduce!
答案 2 :(得分：3)
我发现这个问题非常具有挑战性，我所取得的成果并不令人惊讶，但我设法摆脱了其中一个子查询，可能还有一些联接，这就是结果：
select distinct me.ID, me.DATE, rel.AGENDA_ITEM_ID, sub.MAX_DATE, co.ACRONYM
from MEETINGS me
join JOIN_AGENDA_ITEMS_MEETINGS rel /* Note 1*/
on me.ID = rel.MEETING_ID
left join (
select t1.AGENDA_ITEM_ID, max(t2.DATE) MAX_DATE
from JOIN_AGENDA_ITEMS_MEETINGS t1
join MEETINGS t2
on t2.ID = t1.MEETING_ID
where t2.DATE < '2017-02-02'
group by t1.AGENDA_ITEM_ID
) sub
on rel.AGENDA_ITEM_ID = sub.AGENDA_ITEM_ID /* Note 2 */
left join JOIN_AGENDA_ITEMS_MEETINGS rel2
on rel2.AGENDA_ITEM_ID = rel.AGENDA_ITEM_ID /* Note 3 */
left join MEETINGS me2
on rel2.MEETING_ID = me2.ID and
sub.MAX_DATE = me2.DATE /* Note 4 */
left join COMMITTEES co
on co.ID = me2.COMMITTEE_ID
where me.ID = 2 and
(sub.MAX_DATE is null or me2.DATE is not null) /* Note 5 */
order by rel.AGENDA_ITEM_ID, rel2.MEETING_ID;
备注
您不需要加入AGENDA_ITEMS，因为ID已在关系表中提供
到目前为止，我们有当前的会议，议程项目和他们的＆＃34;计算＆＃34;最长日期
我们会收到每个议程项目的所有会议......
...以便我们可以选择与我们之前计算的最长日期相匹配的会议
此条件是必需的，因为必须保留rel2所有的联接（因为某些议程项目可能没有以前的会议，因此MAX_DATE = null），但这样{{1}会给一些议程项目提出不受欢迎的会议。