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.任何指导将不胜感激.
我正在尝试获取特定会议的所有项目并加入其最大会议日期<X并加入最大日期的委员会首字母缩写词.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.
您可以通过转到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;
审查/审查答复(修订):*
我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给出一些反馈的.总的来说,我非常感谢所有人的帮助,谢谢.
为了测试,我查看了以下查询:
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.
在最初的回答工作,似乎是最有效的选择提出了,比我多得多.
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)
添加DISTINCT到select语句时,此查询还会返回正确的结果.此查询的执行效果不如第一次(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
此查询确实使用返回预期结果DISTINCT.当使用EXPLAIN和拉动行数时,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
正如评论中所指出的,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.
这个有点疯狂..让我们一步一步来做:
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(id = 2 - am2.meeting_id <> am1.meeting_id除外)相结合.我们只希望在实际会议(m2.date < m1.date)之前举行会议.在所有这些会议中,我们只想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;
这样我们得到了第四列(max_date).
最后一步是选择acronym具有最后日期(max_date)的会议.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)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)
最后的查询是:
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,我很惊讶你可以max(m2.date)在子查询中使用.
另一种解决方案 - 在子查询(派生表)中使用第二个查询.使用会议和联接表加入委员会max_date.只保留带有首字母缩略词的行和没有的行max_date.
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
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