Mysql - 帮我优化这个查询
关于系统:
-系统共有8个表 - 用户 - Tutor_Details(导师是一种用户,Tutor_Details表链接到用户) - Learning_packs,(存储由导师创建的包) -learning_packs_tag_relations,(保存用于搜索的标签关系) - coachs_tag_relations 和标签以及 订单(包含导师包的购买详细信息), order_details 链接到订单和 coach_details。
要更清楚地了解所涉及的表格,请查看最后的表格部分。
- 遵循基于标签的搜索方法。当新导师注册和导师创建包时创建标签关系(这使得导师和包可搜索)。有关详细信息,请查看下面的标签在该系统中如何工作?部分。
以下是我试图优化的更复杂查询的更简单表示(不是实际的):-我在查询中使用了诸如部分解释
之类的语句
== =================================================== ===================================================
select
SUM(DISTINCT( t.tag LIKE "%Dictatorship%" )) as key_1_total_matches,
SUM(DISTINCT( t.tag LIKE "%democracy%" )) as key_2_total_matches,
td.*, u.*, count(distinct(od.id_od)), `if (lp.id_lp > 0) then some conditional logic on lp fields else 0 as tutor_popularity`
from Tutor_Details AS td JOIN Users as u on u.id_user = td.id_user
LEFT JOIN Learning_Packs_Tag_Relations AS lptagrels ON td.id_tutor = lptagrels.id_tutor
LEFT JOIN Learning_Packs AS lp ON lptagrels.id_lp = lp.id_lp
LEFT JOIN `some other tables on lp.id_lp - let's call learning pack tables set (including
Learning_Packs table)`
LEFT JOIN Order_Details as od on td.id_tutor = od.id_author LEFT JOIN Orders as o on
od.id_order = o.id_order
LEFT JOIN Tutors_Tag_Relations as ttagrels ON td.id_tutor = ttagrels.id_tutor
JOIN Tags as t on (t.id_tag = ttagrels.id_tag) OR (t.id_tag = lptagrels.id_tag)
where `some condition on Users table's fields`
AND CASE WHEN ((t.id_tag = lptagrels.id_tag) AND (lp.id_lp > 0)) THEN `some
conditions on learning pack tables set` ELSE 1 END
AND CASE WHEN ((t.id_tag = wtagrels.id_tag) AND (wc.id_wc > 0)) THEN `some
conditions on webclasses tables set` ELSE 1 END
AND CASE WHEN (od.id_od>0) THEN od.id_author = td.id_tutor and `some conditions on Orders table's fields` ELSE 1 END
AND ( t.tag LIKE "%Dictatorship%" OR t.tag LIKE "%democracy%")
group by td.id_tutor HAVING key_1_total_matches = 1 AND key_2_total_matches = 1
order by tutor_popularity desc, u.surname asc, u.name asc limit
0,20
=============================================
以上是什么意思查询做什么?
- 是否对搜索关键字进行 AND 逻辑搜索(本例中为 2 - “民主”和“独裁”)。
- 仅返回两个关键字都存在于两个集合的并集中的导师 - 导师详细信息和导师创建的所有包的详细信息。
为了让事情变得清楚 - 假设一个名为“Sandeepan Nath”的导师创建了一个包“我的第一个包”,那么: -
- 搜索“Sandeepan Nath”会返回 Sandeepan Nath。
- 搜索“Sandeepan first”会返回 Sandeepan Nath。
- 搜索“Sandeepan secondary”不会返回 Sandeepan Nath。
=================================================== ======================================
问题
返回的结果上面的查询是正确的(并且逻辑按照预期工作),但是在重负载的数据库上查询所花费的时间约为 25 秒,而正常查询时间为 0.005 - 0.0002 秒,这使得它完全无法使用。
由于所有可能的字段尚未建立索引,可能会导致一些延迟,但我希望有一个更好的查询作为解决方案,尽可能优化,显示相同的结果
======= =================================================== ===================================
标签在这个系统中如何工作?
- 当导师注册、输入标签并根据导师的详细信息(如姓名等)创建标签关系。
- 当导师创建包时,再次输入标签并根据包的详细信息(如包名称、描述等)创建标签关系。
- 标签关系对于存储在tutorials_tag_relations中的导师和存储在learning_packs_tag_relations中的包。所有单独的标签都存储在标签表中。
=================================================== ==================
表格
下面的大多数表格都包含许多其他字段,我在这里省略了这些字段。
CREATE TABLE IF NOT EXISTS `users` (
`id_user` int(10) unsigned NOT NULL AUTO_INCREMENT,
`name` varchar(100) NOT NULL DEFAULT '',
`surname` varchar(155) NOT NULL DEFAULT '',
PRIMARY KEY (`id_user`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8 AUTO_INCREMENT=636 ;
CREATE TABLE IF NOT EXISTS `tutor_details` (
`id_tutor` int(10) NOT NULL AUTO_INCREMENT,
`id_user` int(10) NOT NULL DEFAULT '0',
PRIMARY KEY (`id_tutor`),
KEY `Users_FKIndex1` (`id_user`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=51 ;
CREATE TABLE IF NOT EXISTS `orders` (
`id_order` int(10) unsigned NOT NULL AUTO_INCREMENT,
PRIMARY KEY (`id_order`),
KEY `Orders_FKIndex1` (`id_user`),
) ENGINE=InnoDB DEFAULT CHARSET=utf8 AUTO_INCREMENT=275 ;
ALTER TABLE `orders`
ADD CONSTRAINT `Orders_ibfk_1` FOREIGN KEY (`id_user`) REFERENCES `users`
(`id_user`) ON DELETE NO ACTION ON UPDATE NO ACTION;
CREATE TABLE IF NOT EXISTS `order_details` (
`id_od` int(10) unsigned NOT NULL AUTO_INCREMENT,
`id_order` int(10) unsigned NOT NULL DEFAULT '0',
`id_author` int(10) NOT NULL DEFAULT '0',
PRIMARY KEY (`id_od`),
KEY `Order_Details_FKIndex1` (`id_order`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8 AUTO_INCREMENT=284 ;
ALTER TABLE `order_details`
ADD CONSTRAINT `Order_Details_ibfk_1` FOREIGN KEY (`id_order`) REFERENCES `orders`
(`id_order`) ON DELETE NO ACTION ON UPDATE NO ACTION;
CREATE TABLE IF NOT EXISTS `learning_packs` (
`id_lp` int(10) unsigned NOT NULL AUTO_INCREMENT,
`id_author` int(10) unsigned NOT NULL DEFAULT '0',
PRIMARY KEY (`id_lp`),
KEY `Learning_Packs_FKIndex2` (`id_author`),
KEY `id_lp` (`id_lp`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8 AUTO_INCREMENT=23 ;
CREATE TABLE IF NOT EXISTS `tags` (
`id_tag` int(10) unsigned NOT NULL AUTO_INCREMENT,
`tag` varchar(255) DEFAULT NULL,
PRIMARY KEY (`id_tag`),
UNIQUE KEY `tag` (`tag`),
KEY `id_tag` (`id_tag`),
KEY `tag_2` (`tag`),
KEY `tag_3` (`tag`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=3419 ;
CREATE TABLE IF NOT EXISTS `tutors_tag_relations` (
`id_tag` int(10) unsigned NOT NULL DEFAULT '0',
`id_tutor` int(10) DEFAULT NULL,
KEY `Tutors_Tag_Relations` (`id_tag`),
KEY `id_tutor` (`id_tutor`),
KEY `id_tag` (`id_tag`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
ALTER TABLE `tutors_tag_relations`
ADD CONSTRAINT `Tutors_Tag_Relations_ibfk_1` FOREIGN KEY (`id_tag`) REFERENCES
`tags` (`id_tag`) ON DELETE NO ACTION ON UPDATE NO ACTION;
CREATE TABLE IF NOT EXISTS `learning_packs_tag_relations` (
`id_tag` int(10) unsigned NOT NULL DEFAULT '0',
`id_tutor` int(10) DEFAULT NULL,
`id_lp` int(10) unsigned DEFAULT NULL,
KEY `Learning_Packs_Tag_Relations_FKIndex1` (`id_tag`),
KEY `id_lp` (`id_lp`),
KEY `id_tag` (`id_tag`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
ALTER TABLE `learning_packs_tag_relations`
ADD CONSTRAINT `Learning_Packs_Tag_Relations_ibfk_1` FOREIGN KEY (`id_tag`)
REFERENCES `tags` (`id_tag`) ON DELETE NO ACTION ON UPDATE NO ACTION;
=================================================== =================================
以下是确切的查询(这也包括课程 - 导师可以创建课程和搜索词与导师创建的课程相匹配):-
SELECT SUM(DISTINCT( t.tag LIKE "%Dictatorship%" )) AS key_1_total_matches,
SUM(DISTINCT( t.tag LIKE "%democracy%" )) AS key_2_total_matches,
COUNT(DISTINCT( od.id_od )) AS tutor_popularity,
CASE
WHEN ( IF(( wc.id_wc > 0 ), ( wc.wc_api_status = 1
AND wc.wc_type = 0
AND wc.class_date > '2010-06-01 22:00:56'
AND wccp.status = 1
AND ( wccp.country_code = 'IE'
OR wccp.country_code IN ( 'INT' )
) ), 0)
) THEN 1
ELSE 0
END AS 'classes_published',
CASE
WHEN ( IF(( lp.id_lp > 0 ), ( lp.id_status = 1
AND lp.published = 1
AND lpcp.status = 1
AND ( lpcp.country_code = 'IE'
OR lpcp.country_code IN ( 'INT' )
) ), 0)
) THEN 1
ELSE 0
END AS 'packs_published',
td . *,
u . *
FROM tutor_details AS td
JOIN users AS u
ON u.id_user = td.id_user
LEFT JOIN learning_packs_tag_relations AS lptagrels
ON td.id_tutor = lptagrels.id_tutor
LEFT JOIN learning_packs AS lp
ON lptagrels.id_lp = lp.id_lp
LEFT JOIN learning_packs_categories AS lpc
ON lpc.id_lp_cat = lp.id_lp_cat
LEFT JOIN learning_packs_categories AS lpcp
ON lpcp.id_lp_cat = lpc.id_parent
LEFT JOIN learning_pack_content AS lpct
ON ( lp.id_lp = lpct.id_lp )
LEFT JOIN webclasses_tag_relations AS wtagrels
ON td.id_tutor = wtagrels.id_tutor
LEFT JOIN webclasses AS wc
ON wtagrels.id_wc = wc.id_wc
LEFT JOIN learning_packs_categories AS wcc
ON wcc.id_lp_cat = wc.id_wp_cat
LEFT JOIN learning_packs_categories AS wccp
ON wccp.id_lp_cat = wcc.id_parent
LEFT JOIN order_details AS od
ON td.id_tutor = od.id_author
LEFT JOIN orders AS o
ON od.id_order = o.id_order
LEFT JOIN tutors_tag_relations AS ttagrels
ON td.id_tutor = ttagrels.id_tutor
JOIN tags AS t
ON ( t.id_tag = ttagrels.id_tag )
OR ( t.id_tag = lptagrels.id_tag )
OR ( t.id_tag = wtagrels.id_tag )
WHERE ( u.country = 'IE'
OR u.country IN ( 'INT' ) )
AND CASE
WHEN ( ( t.id_tag = lptagrels.id_tag )
AND ( lp.id_lp > 0 ) ) THEN lp.id_status = 1
AND lp.published = 1
AND lpcp.status = 1
AND ( lpcp.country_code = 'IE'
OR lpcp.country_code IN (
'INT'
) )
ELSE 1
END
AND CASE
WHEN ( ( t.id_tag = wtagrels.id_tag )
AND ( wc.id_wc > 0 ) ) THEN wc.wc_api_status = 1
AND wc.wc_type = 0
AND
wc.class_date > '2010-06-01 22:00:56'
AND wccp.status = 1
AND ( wccp.country_code = 'IE'
OR wccp.country_code IN (
'INT'
) )
ELSE 1
END
AND CASE
WHEN ( od.id_od > 0 ) THEN od.id_author = td.id_tutor
AND o.order_status = 'paid'
AND CASE
WHEN ( od.id_wc > 0 ) THEN od.can_attend_class = 1
ELSE 1
END
ELSE 1
END
GROUP BY td.id_tutor
HAVING key_1_total_matches = 1
AND key_2_total_matches = 1
ORDER BY tutor_popularity DESC,
u.surname ASC,
u.name ASC
LIMIT 0, 20
请注意 - 提供的数据库结构未显示此查询中的所有字段和表格
============= =================================================== ======================
解释查询输出:- 请看这个截图 http://www.test.examvillage.com/Explain_query.jpg
About the system:
-The system has a total of 8 tables
- Users
- Tutor_Details (Tutors are a type of User,Tutor_Details table is linked to Users)
- learning_packs, (stores packs created by tutors)
- learning_packs_tag_relations, (holds tag relations meant for search)
- tutors_tag_relations and tags and
orders (containing purchase details of tutor's packs),
order_details linked to orders and tutor_details.
For a more clear idea about the tables involved please check the The tables section in the end.
-A tags based search approach is being followed.Tag relations are created when new tutors register and when tutors create packs (this makes tutors and packs searcheable). For details please check the section How tags work in this system? below.
Following is a simpler representation (not the actual) of the more complex query which I am trying to optimize:- I have used statements like explanation of parts
in the query
============================================================================
select
SUM(DISTINCT( t.tag LIKE "%Dictatorship%" )) as key_1_total_matches,
SUM(DISTINCT( t.tag LIKE "%democracy%" )) as key_2_total_matches,
td.*, u.*, count(distinct(od.id_od)), `if (lp.id_lp > 0) then some conditional logic on lp fields else 0 as tutor_popularity`
from Tutor_Details AS td JOIN Users as u on u.id_user = td.id_user
LEFT JOIN Learning_Packs_Tag_Relations AS lptagrels ON td.id_tutor = lptagrels.id_tutor
LEFT JOIN Learning_Packs AS lp ON lptagrels.id_lp = lp.id_lp
LEFT JOIN `some other tables on lp.id_lp - let's call learning pack tables set (including
Learning_Packs table)`
LEFT JOIN Order_Details as od on td.id_tutor = od.id_author LEFT JOIN Orders as o on
od.id_order = o.id_order
LEFT JOIN Tutors_Tag_Relations as ttagrels ON td.id_tutor = ttagrels.id_tutor
JOIN Tags as t on (t.id_tag = ttagrels.id_tag) OR (t.id_tag = lptagrels.id_tag)
where `some condition on Users table's fields`
AND CASE WHEN ((t.id_tag = lptagrels.id_tag) AND (lp.id_lp > 0)) THEN `some
conditions on learning pack tables set` ELSE 1 END
AND CASE WHEN ((t.id_tag = wtagrels.id_tag) AND (wc.id_wc > 0)) THEN `some
conditions on webclasses tables set` ELSE 1 END
AND CASE WHEN (od.id_od>0) THEN od.id_author = td.id_tutor and `some conditions on Orders table's fields` ELSE 1 END
AND ( t.tag LIKE "%Dictatorship%" OR t.tag LIKE "%democracy%")
group by td.id_tutor HAVING key_1_total_matches = 1 AND key_2_total_matches = 1
order by tutor_popularity desc, u.surname asc, u.name asc limit
0,20
=====================================================================
What does the above query do?
- Does AND logic search on the search keywords (2 in this example - "Democracy" and "Dictatorship").
- Returns only those tutors for which both the keywords are present in the union of the two sets - tutors details and details of all the packs created by a tutor.
To make things clear - Suppose a Tutor name "Sandeepan Nath" has created a pack "My first pack", then:-
- Searching "Sandeepan Nath" returns Sandeepan Nath.
- Searching "Sandeepan first" returns Sandeepan Nath.
- Searching "Sandeepan second" does not return Sandeepan Nath.
======================================================================================
The problem
The results returned by the above query are correct (AND logic working as per expectation), but the time taken by the query on heavily loaded databases is like 25 seconds as against normal query timings of the order of 0.005 - 0.0002 seconds, which makes it totally unusable.
It is possible that some of the delay is being caused because all the possible fields have not yet been indexed, but I would appreciate a better query as a solution, optimized as much as possible, displaying the same results
==========================================================================================
How tags work in this system?
- When a tutor registers, tags are entered and tag relations are created with respect to tutor's details like name, surname etc.
- When a Tutors create packs, again tags are entered and tag relations are created with respect to pack's details like pack name, description etc.
- tag relations for tutors stored in tutors_tag_relations and those for packs stored in learning_packs_tag_relations. All individual tags are stored in tags table.
====================================================================
The tables
Most of the following tables contain many other fields which I have omitted here.
CREATE TABLE IF NOT EXISTS `users` (
`id_user` int(10) unsigned NOT NULL AUTO_INCREMENT,
`name` varchar(100) NOT NULL DEFAULT '',
`surname` varchar(155) NOT NULL DEFAULT '',
PRIMARY KEY (`id_user`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8 AUTO_INCREMENT=636 ;
CREATE TABLE IF NOT EXISTS `tutor_details` (
`id_tutor` int(10) NOT NULL AUTO_INCREMENT,
`id_user` int(10) NOT NULL DEFAULT '0',
PRIMARY KEY (`id_tutor`),
KEY `Users_FKIndex1` (`id_user`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=51 ;
CREATE TABLE IF NOT EXISTS `orders` (
`id_order` int(10) unsigned NOT NULL AUTO_INCREMENT,
PRIMARY KEY (`id_order`),
KEY `Orders_FKIndex1` (`id_user`),
) ENGINE=InnoDB DEFAULT CHARSET=utf8 AUTO_INCREMENT=275 ;
ALTER TABLE `orders`
ADD CONSTRAINT `Orders_ibfk_1` FOREIGN KEY (`id_user`) REFERENCES `users`
(`id_user`) ON DELETE NO ACTION ON UPDATE NO ACTION;
CREATE TABLE IF NOT EXISTS `order_details` (
`id_od` int(10) unsigned NOT NULL AUTO_INCREMENT,
`id_order` int(10) unsigned NOT NULL DEFAULT '0',
`id_author` int(10) NOT NULL DEFAULT '0',
PRIMARY KEY (`id_od`),
KEY `Order_Details_FKIndex1` (`id_order`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8 AUTO_INCREMENT=284 ;
ALTER TABLE `order_details`
ADD CONSTRAINT `Order_Details_ibfk_1` FOREIGN KEY (`id_order`) REFERENCES `orders`
(`id_order`) ON DELETE NO ACTION ON UPDATE NO ACTION;
CREATE TABLE IF NOT EXISTS `learning_packs` (
`id_lp` int(10) unsigned NOT NULL AUTO_INCREMENT,
`id_author` int(10) unsigned NOT NULL DEFAULT '0',
PRIMARY KEY (`id_lp`),
KEY `Learning_Packs_FKIndex2` (`id_author`),
KEY `id_lp` (`id_lp`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8 AUTO_INCREMENT=23 ;
CREATE TABLE IF NOT EXISTS `tags` (
`id_tag` int(10) unsigned NOT NULL AUTO_INCREMENT,
`tag` varchar(255) DEFAULT NULL,
PRIMARY KEY (`id_tag`),
UNIQUE KEY `tag` (`tag`),
KEY `id_tag` (`id_tag`),
KEY `tag_2` (`tag`),
KEY `tag_3` (`tag`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=3419 ;
CREATE TABLE IF NOT EXISTS `tutors_tag_relations` (
`id_tag` int(10) unsigned NOT NULL DEFAULT '0',
`id_tutor` int(10) DEFAULT NULL,
KEY `Tutors_Tag_Relations` (`id_tag`),
KEY `id_tutor` (`id_tutor`),
KEY `id_tag` (`id_tag`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
ALTER TABLE `tutors_tag_relations`
ADD CONSTRAINT `Tutors_Tag_Relations_ibfk_1` FOREIGN KEY (`id_tag`) REFERENCES
`tags` (`id_tag`) ON DELETE NO ACTION ON UPDATE NO ACTION;
CREATE TABLE IF NOT EXISTS `learning_packs_tag_relations` (
`id_tag` int(10) unsigned NOT NULL DEFAULT '0',
`id_tutor` int(10) DEFAULT NULL,
`id_lp` int(10) unsigned DEFAULT NULL,
KEY `Learning_Packs_Tag_Relations_FKIndex1` (`id_tag`),
KEY `id_lp` (`id_lp`),
KEY `id_tag` (`id_tag`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
ALTER TABLE `learning_packs_tag_relations`
ADD CONSTRAINT `Learning_Packs_Tag_Relations_ibfk_1` FOREIGN KEY (`id_tag`)
REFERENCES `tags` (`id_tag`) ON DELETE NO ACTION ON UPDATE NO ACTION;
===================================================================================
Following is the exact query (this includes classes also - tutors can create classes and search terms are matched with classes created by tutors):-
SELECT SUM(DISTINCT( t.tag LIKE "%Dictatorship%" )) AS key_1_total_matches,
SUM(DISTINCT( t.tag LIKE "%democracy%" )) AS key_2_total_matches,
COUNT(DISTINCT( od.id_od )) AS tutor_popularity,
CASE
WHEN ( IF(( wc.id_wc > 0 ), ( wc.wc_api_status = 1
AND wc.wc_type = 0
AND wc.class_date > '2010-06-01 22:00:56'
AND wccp.status = 1
AND ( wccp.country_code = 'IE'
OR wccp.country_code IN ( 'INT' )
) ), 0)
) THEN 1
ELSE 0
END AS 'classes_published',
CASE
WHEN ( IF(( lp.id_lp > 0 ), ( lp.id_status = 1
AND lp.published = 1
AND lpcp.status = 1
AND ( lpcp.country_code = 'IE'
OR lpcp.country_code IN ( 'INT' )
) ), 0)
) THEN 1
ELSE 0
END AS 'packs_published',
td . *,
u . *
FROM tutor_details AS td
JOIN users AS u
ON u.id_user = td.id_user
LEFT JOIN learning_packs_tag_relations AS lptagrels
ON td.id_tutor = lptagrels.id_tutor
LEFT JOIN learning_packs AS lp
ON lptagrels.id_lp = lp.id_lp
LEFT JOIN learning_packs_categories AS lpc
ON lpc.id_lp_cat = lp.id_lp_cat
LEFT JOIN learning_packs_categories AS lpcp
ON lpcp.id_lp_cat = lpc.id_parent
LEFT JOIN learning_pack_content AS lpct
ON ( lp.id_lp = lpct.id_lp )
LEFT JOIN webclasses_tag_relations AS wtagrels
ON td.id_tutor = wtagrels.id_tutor
LEFT JOIN webclasses AS wc
ON wtagrels.id_wc = wc.id_wc
LEFT JOIN learning_packs_categories AS wcc
ON wcc.id_lp_cat = wc.id_wp_cat
LEFT JOIN learning_packs_categories AS wccp
ON wccp.id_lp_cat = wcc.id_parent
LEFT JOIN order_details AS od
ON td.id_tutor = od.id_author
LEFT JOIN orders AS o
ON od.id_order = o.id_order
LEFT JOIN tutors_tag_relations AS ttagrels
ON td.id_tutor = ttagrels.id_tutor
JOIN tags AS t
ON ( t.id_tag = ttagrels.id_tag )
OR ( t.id_tag = lptagrels.id_tag )
OR ( t.id_tag = wtagrels.id_tag )
WHERE ( u.country = 'IE'
OR u.country IN ( 'INT' ) )
AND CASE
WHEN ( ( t.id_tag = lptagrels.id_tag )
AND ( lp.id_lp > 0 ) ) THEN lp.id_status = 1
AND lp.published = 1
AND lpcp.status = 1
AND ( lpcp.country_code = 'IE'
OR lpcp.country_code IN (
'INT'
) )
ELSE 1
END
AND CASE
WHEN ( ( t.id_tag = wtagrels.id_tag )
AND ( wc.id_wc > 0 ) ) THEN wc.wc_api_status = 1
AND wc.wc_type = 0
AND
wc.class_date > '2010-06-01 22:00:56'
AND wccp.status = 1
AND ( wccp.country_code = 'IE'
OR wccp.country_code IN (
'INT'
) )
ELSE 1
END
AND CASE
WHEN ( od.id_od > 0 ) THEN od.id_author = td.id_tutor
AND o.order_status = 'paid'
AND CASE
WHEN ( od.id_wc > 0 ) THEN od.can_attend_class = 1
ELSE 1
END
ELSE 1
END
GROUP BY td.id_tutor
HAVING key_1_total_matches = 1
AND key_2_total_matches = 1
ORDER BY tutor_popularity DESC,
u.surname ASC,
u.name ASC
LIMIT 0, 20
Please note - The provided database structure does not show all the fields and tables as in this query
=====================================================================================
The explain query output:-
Please see this screenshot
http://www.test.examvillage.com/Explain_query.jpg
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有关行计数、值分布、索引、数据库大小、内存大小、磁盘布局的信息 - raid 0、5 等 - 当查询缓慢时有多少用户正在访问数据库 - 其他查询正在运行。所有这些因素都会影响性能。
如果只是查询/索引问题,那么解释计划输出的打印可能会揭示一些原因。还需要确切的查询。
Information on row counts, value distributions, indexes, size of the database, size of memory, disk layout - raid 0, 5, etc - how many users are hitting your database when queries are slow - what other queries are running. All these things factor into performance.
Also a print out of the explain plan output may shed some light on the cause if it's simply a query / index issue. The exact query would be needed as well.
您确实应该为查询使用一些更好的格式。
只需在每行的开头添加至少 4 个空格即可获得良好的代码格式。
或者看看这里:https://stackoverflow.com/editing-help
你能提供mysql的执行计划吗?您需要将“EXPLAIN”添加到查询中并复制结果。
会给你一些有用的提示(执行顺序,返回的行,可用/使用的索引)
You really should use some better formatting for the query.
Just add at least 4 spaces to the beginning of each row to get this nice code formatting.
Or have a look here: https://stackoverflow.com/editing-help
Could you provide the execution plan from mysql? You need to add "EXPLAIN" to the query and copy the result.
will give you some useful hints (execution order, returned rows, available/used indexes)
您的问题是,“我怎样才能找到匹配某些标签的导师?”这不是一个很难的问题,所以回答这个问题的查询也不难。
类似的东西:
如果您修改设计以在“导师”表中包含“标签”字段,其中放置适用于该导师的所有标签,那么这将起作用。它将消除连接层和表层。
所有这些连接和表层是否提供了真正的功能,或者只是更多的编程难题?考虑您的应用程序真正需要的功能,然后简化您的数据库设计!
Your question is, "how can I find tutors that match certain tags?" That's not a hard question, so the query to answer it shouldn't be hard either.
Something like:
That will work, if you modify your design to have a "tags" field in your "tutors" table, in which you put all the tags that apply to that tutor. It will eliminate layers of joins and tables.
Are all those layers of joins and tables providing real functionality, or just more programming headaches? Think about the functionality that your app REALLY needs, and then simplify your database design!!
回答我自己的问题。
这种方法的主要问题是在单个查询中连接了太多表。其中一些表,例如
Tags
(具有大量记录 - 将来可以容纳词汇表中的所有英语单词),当与如此多的表连接时,会导致这种乘法效应,这种效应在任何情况下都不会发生。方式予以反击。解决方案基本上是确保在单个查询中不会进行太多连接。将一个大型连接查询分成多个步骤,将一个查询(涉及某些表上的连接)的结果用于下一个连接查询(涉及其他表上的连接)可以减少乘法效应。
稍后我将尝试对此提供更好的解释。
Answering my own question.
The main problem with this approach was that too many tables were joined in a single query. Some of those tables like
Tags
(having large number of records - which can in future hold as many as all the English words in the vocabulary) when joined with so many tables cause this multiplication effect which can in no way be countered.The solution is basically to make sure too many joins are not made in a single query. Breaking one large join query into steps, using the results of the one query (involving joins on some of the tables) for the next join query (involving joins on the other tables) reduces the multiplication effect.
I will try to provide better explanation to this later.