mysql左加入哪裏用或，掃描整個表（沒有索引）

我試過這幾種不同的方法，結果不好。mysql左加入哪裏用或，掃描整個表（沒有索引）

核心問題是成員搜索掃描所有成員，忽略索引。

主要的原因（從我可以告訴）是該片段

(Member.priv_profile = 3 OR MyFriend.status_id IN (1,2))

的任何一方或者單獨片段，做工精細，得到一個指數，掃描幾排，從而表現良好。

我真的不想把這個查詢分成2個並且做一個UNION，但是我們可能必須這樣做，除非有人能想出一個讓重要OR得到這個選擇「工作」的好方法。

mysql> ALTER TABLE `members` ADD INDEX A (is_active, last_name, first_name); 
Query OK, 140019 rows affected (6.82 sec) 
Records: 140019 Duplicates: 0 Warnings: 0 

mysql> ALTER TABLE `members` ADD INDEX B (is_active, last_name, first_name, priv_profile); 
Query OK, 140019 rows affected (7.70 sec) 
Records: 140019 Duplicates: 0 Warnings: 0 

mysql> explain SELECT COUNT(*) AS `count` FROM `ao_prod`.`members` AS `Member` 
LEFT JOIN `ao_prod`.`member_friends` AS `MyFriend` ON (`MyFriend`.`member_2_id` = `Member`.`id` AND member_1_id = '150365') 
WHERE `Member`.`is_active` = '1' AND NOT(`Member`.`first_name` = '' AND `Member`.`last_name` = '') AND (`Member`.`priv_profile` = 3 OR `MyFriend`.`status_id` IN (1,2)); 
+----+-------------+----------+------+----------------------------------------------+-------------+---------+-------+--------+--------------------------+ 
| id | select_type | table | type | possible_keys        | key   | key_len | ref | rows | Extra     | 
+----+-------------+----------+------+----------------------------------------------+-------------+---------+-------+--------+--------------------------+ 
| 1 | SIMPLE  | Member | ALL | active_delete,scope,member_search_alerts,A,B | NULL  | NULL | NULL | 140019 | Using where    | 
| 1 | SIMPLE  | MyFriend | ref | member_1_id         | member_1_id | 4  | const | 155 | Using where; Using index | 
+----+-------------+----------+------+----------------------------------------------+-------------+---------+-------+--------+--------------------------+ 
2 rows in set (0.00 sec) 

// without the "public profile" part 

mysql> explain SELECT COUNT(*) AS `count` FROM `ao_prod`.`members` AS `Member` 
LEFT JOIN `ao_prod`.`member_friends` AS `MyFriend` ON (`MyFriend`.`member_2_id` = `Member`.`id` AND member_1_id = '150365') 
WHERE `Member`.`is_active` = '1' AND NOT(`Member`.`first_name` = '' AND `Member`.`last_name` = '') AND (`MyFriend`.`status_id` IN (1,2)); 
+----+-------------+----------+--------+------------------------------------------------------+-------------+---------+------------------------------+------+--------------------------+ 
| id | select_type | table | type | possible_keys          | key   | key_len | ref       | rows | Extra     | 
+----+-------------+----------+--------+------------------------------------------------------+-------------+---------+------------------------------+------+--------------------------+ 
| 1 | SIMPLE  | MyFriend | range | member_1_id           | member_1_id | 5  | NULL       | 251 | Using where; Using index | 
| 1 | SIMPLE  | Member | eq_ref | PRIMARY,active_delete,scope,member_search_alerts,A,B | PRIMARY  | 4  | ao_prod.MyFriend.member_2_id | 1 | Using where    | 
+----+-------------+----------+--------+------------------------------------------------------+-------------+---------+------------------------------+------+--------------------------+ 
2 rows in set (0.00 sec) 

// without the "my connection" part 

mysql> explain SELECT COUNT(*) AS `count` FROM `ao_prod`.`members` AS `Member` 
LEFT JOIN `ao_prod`.`member_friends` AS `MyFriend` ON (`MyFriend`.`member_2_id` = `Member`.`id` AND member_1_id = '42983') 
WHERE `Member`.`is_active` = '1' AND (NOT(`Member`.`first_name` = '' AND `Member`.`last_name` = '')) AND (`Member`.`priv_profile` = 3); 
+----+-------------+----------+------+----------------------------------------------+-------------+---------+-------------+------+-------------+ 
| id | select_type | table | type | possible_keys        | key   | key_len | ref   | rows | Extra  | 
+----+-------------+----------+------+----------------------------------------------+-------------+---------+-------------+------+-------------+ 
| 1 | SIMPLE  | Member | ref | active_delete,scope,member_search_alerts,A,B | scope  | 2  | const,const | 2007 | Using where | 
| 1 | SIMPLE  | MyFriend | ref | member_1_id         | member_1_id | 4  | const  | 252 | Using index | 
+----+-------------+----------+------+----------------------------------------------+-------------+---------+-------------+------+-------------+ 
2 rows in set (0.01 sec) 

// as a subquery vs. join (no workie) 

mysql> explain SELECT COUNT(*) AS `count` FROM `ao_prod`.`members` AS `Member` 
WHERE `Member`.`is_active` = '1' AND NOT(`Member`.`first_name` = '' AND `Member`.`last_name` = '') AND (`Member`.`id` IN (
SELECT member_2_id FROM member_friends WHERE member_1_id = 150365 AND status_id IN (1,2) 
)); 
+----+--------------------+----------------+-------+----------------------------------------------+-------------+---------+------+--------+--------------------------+ 
| id | select_type  | table   | type | possible_keys        | key   | key_len | ref | rows | Extra     | 
+----+--------------------+----------------+-------+----------------------------------------------+-------------+---------+------+--------+--------------------------+ 
| 1 | PRIMARY   | Member   | ALL | active_delete,scope,member_search_alerts,A,B | NULL  | NULL | NULL | 140019 | Using where    | 
| 2 | DEPENDENT SUBQUERY | member_friends | range | member_1_id         | member_1_id | 5  | NULL | 155 | Using where; Using index | 
+----+--------------------+----------------+-------+----------------------------------------------+-------------+---------+------+--------+--------------------------+ 
2 rows in set (0.01 sec) 

// sketch of the possible, ugly UNION 

mysql> explain SELECT COUNT(*) AS `count` FROM `ao_prod`.`members` AS `Member` LEFT JOIN `ao_prod`.`member_friends` AS `MyFriend` ON (`MyFriend`.`member_2_id` = `Member`.`id` AND member_1_id = '42983') WHERE `Member`.`is_active` = '1' AND (NOT(`Member`.`first_name` = '' AND `Member`.`last_name` = '')) AND (`MyFriend`.`status_id` IN (1,2)) 
    -> UNION 
    -> SELECT COUNT(*) AS `count` FROM `ao_prod`.`members` AS `Member` WHERE `Member`.`is_active` = '1' AND (NOT(`Member`.`first_name` = '' AND `Member`.`last_name` = '')) AND (`Member`.`priv_profile` = 3) 
    -> GROUP BY Member.id 
    -> ; 
+----+--------------+------------+--------+------------------------------------------------------+-------------+---------+------------------------------+------+----------------------------------------------+ 
| id | select_type | table  | type | possible_keys          | key   | key_len | ref       | rows | Extra          | 
+----+--------------+------------+--------+------------------------------------------------------+-------------+---------+------------------------------+------+----------------------------------------------+ 
| 1 | PRIMARY  | MyFriend | range | member_1_id           | member_1_id | 5  | NULL       | 251 | Using where; Using index      | 
| 1 | PRIMARY  | Member  | eq_ref | PRIMARY,active_delete,scope,member_search_alerts,A,B | PRIMARY  | 4  | ao_prod.MyFriend.member_2_id | 1 | Using where         | 
| 2 | UNION  | Member  | ref | active_delete,scope,member_search_alerts,A,B   | scope  | 2  | const,const     | 2007 | Using where; Using temporary; Using filesort | 
| NULL | UNION RESULT | <union1,2> | ALL | NULL             | NULL  | NULL | NULL       | NULL |            | 
+----+--------------+------------+--------+------------------------------------------------------+-------------+---------+------------------------------+------+----------------------------------------------+ 
4 rows in set (0.02 sec) 

// using index hinting to no avail 

mysql> explain SELECT COUNT(*) AS `count` 
FROM `ao_prod`.`members` AS `Member` 
USE INDEX (A) 
LEFT JOIN `ao_prod`.`member_friends` AS `MyFriend` ON (`MyFriend`.`member_2_id` = `Member`.`id` AND member_1_id = '150365') 
WHERE `Member`.`is_active` = '1' AND NOT(`Member`.`first_name` = '' AND `Member`.`last_name` = '') AND (`Member`.`priv_profile` = 3 OR `MyFriend`.`status_id` IN (1,2)); 
+----+-------------+----------+------+---------------+-------------+---------+-------+--------+--------------------------+ 
| id | select_type | table | type | possible_keys | key   | key_len | ref | rows | Extra     | 
+----+-------------+----------+------+---------------+-------------+---------+-------+--------+--------------------------+ 
| 1 | SIMPLE  | Member | ALL | A    | NULL  | NULL | NULL | 140245 | Using where    | 
| 1 | SIMPLE  | MyFriend | ref | member_1_id | member_1_id | 4  | const | 181 | Using where; Using index | 
+----+-------------+----------+------+---------------+-------------+---------+-------+--------+--------------------------+ 
2 rows in set (0.01 sec)

下面是創建（顯示全，醜表和所有其他指標）對涉及的表聲明

CREATE TABLE IF NOT EXISTS `member_friends` (
    `id` varchar(36) NOT NULL, 
    `created` datetime DEFAULT NULL, 
    `modified` datetime DEFAULT NULL, 
    `member_1_id` int(11) NOT NULL DEFAULT '0', 
    `member_2_id` int(11) NOT NULL DEFAULT '0', 
    `status_id` tinyint(3) NOT NULL DEFAULT '0', 
    `requested_by` tinyint(3) NOT NULL DEFAULT '0', 
    `requested` datetime DEFAULT NULL, 
    `accepted` datetime DEFAULT NULL, 
    PRIMARY KEY (`id`), 
    KEY `member_1_id` (`member_1_id`,`status_id`,`member_2_id`) 
) ENGINE=InnoDB DEFAULT CHARSET=utf8; 

ALTER TABLE `members_fields` 
ADD PRIMARY KEY (`id`), ADD KEY `key` (`key`), ADD KEY `member_key` (`member_id`,`key`); 

CREATE TABLE IF NOT EXISTS `members` (
`id` int(11) NOT NULL, 
    `created` datetime DEFAULT NULL, 
    `modified` datetime DEFAULT NULL, 
    `profile_updated` datetime NOT NULL, 
    `last_login` datetime DEFAULT NULL, 
    `is_active` tinyint(1) NOT NULL, 
    `email` varchar(256) NOT NULL DEFAULT '', 
    `password` varchar(40) NOT NULL, 
    `first_name` varchar(128) NOT NULL DEFAULT '', 
    `middle_name` varchar(128) NOT NULL, 
    `last_name` varchar(128) NOT NULL DEFAULT '', 
    `suffix` varchar(32) NOT NULL, 
    `company` varchar(128) NOT NULL, 
    `address` varchar(128) NOT NULL, 
    `address_2` varchar(128) NOT NULL, 
    `city` varchar(128) NOT NULL, 
    `state` varchar(5) NOT NULL, 
    `zip` varchar(16) NOT NULL, 
    `location_name` varchar(128) NOT NULL, 
    `image_url` varchar(256) NOT NULL, 
    `slug` varchar(64) NOT NULL, 
    `headline` varchar(256) NOT NULL, 
    `experience_level` varchar(64) NOT NULL, 
    `apply_job_states` varchar(256) NOT NULL COMMENT 'CSV list', 
    `apply_job_us` tinyint(1) NOT NULL DEFAULT '0', 
    `apply_job_ca` tinyint(1) NOT NULL DEFAULT '0', 
    `apply_job_traveling` tinyint(1) NOT NULL DEFAULT '0', 
    `apply_job_international` tinyint(1) NOT NULL DEFAULT '0', 
    `apply_job_fulltime` tinyint(1) NOT NULL DEFAULT '0', 
    `apply_job_parttime` tinyint(1) NOT NULL DEFAULT '0', 
    `apply_job_perdiem` tinyint(1) NOT NULL DEFAULT '0', 
    `contact_for_professional_opportunities` tinyint(1) NOT NULL DEFAULT '0', 
    `contact_for_job_inquiries` tinyint(1) NOT NULL DEFAULT '0', 
    `contact_for_new_ventures` tinyint(1) NOT NULL DEFAULT '0', 
    `contact_for_expertise_requests` tinyint(1) NOT NULL DEFAULT '0', 
    `country` varchar(2) NOT NULL, 
    `timezone` varchar(32) NOT NULL, 
    `phone` varchar(16) NOT NULL, 
    `fax` varchar(16) NOT NULL, 
    `birthday` varchar(5) NOT NULL COMMENT 'MM/DD (required)', 
    `birth_year` varchar(4) DEFAULT NULL COMMENT 'YYYY (optional)', 
    `corp_id` int(11) NOT NULL DEFAULT '0', 
    `is_deleted` tinyint(1) NOT NULL, 
    `url` varchar(256) DEFAULT NULL, 
    `emails` varchar(512) NOT NULL COMMENT 'JSON list of alternate emails', 
    `phones` varchar(512) NOT NULL COMMENT 'JSON list of alternate phones', 
    `lat` float NOT NULL, 
    `lon` float NOT NULL, 
    `facebook_id` varchar(32) NOT NULL, 
    `connect_id` int(11) NOT NULL, 
    `is_student` tinyint(1) NOT NULL DEFAULT '0', 
    `is_career_center_recruiter` tinyint(1) NOT NULL DEFAULT '0', 
    `is_continuing_education_portal_manager` tinyint(1) NOT NULL DEFAULT '0', 
    `is_manually_approved` tinyint(1) NOT NULL DEFAULT '0', 
    `is_employer` tinyint(1) NOT NULL DEFAULT '0', 
    `is_jobseeker` tinyint(1) NOT NULL DEFAULT '0', 
    `is_jobseeker_badge` tinyint(1) NOT NULL DEFAULT '0', 
    `is_contributor` tinyint(1) NOT NULL DEFAULT '0', 
    `priv_profile` tinyint(3) NOT NULL DEFAULT '1', 
    `priv_email` tinyint(3) NOT NULL DEFAULT '0', 
    `priv_phone` tinyint(3) NOT NULL DEFAULT '0', 
    `has_certification` tinyint(1) DEFAULT NULL, 
    `has_state_license` tinyint(1) DEFAULT NULL, 
    `job_title` varchar(64) NOT NULL, 
    `occupation_id` int(11) NOT NULL, 
    `occupation_other` varchar(64) NOT NULL, 
    `work_setting_id` int(11) NOT NULL, 
    `work_setting_other` varchar(64) NOT NULL, 
    `memberships_honors_awards` text NOT NULL 
) ENGINE=MyISAM DEFAULT CHARSET=utf8 AUTO_INCREMENT=1093688 ; 

ALTER TABLE `members` 
ADD PRIMARY KEY (`id`), ADD KEY `is_cc` (`is_career_center_recruiter`,`corp_id`), ADD KEY `is_ce` (`is_continuing_education_portal_manager`,`corp_id`), ADD KEY `corp_id` (`corp_id`), ADD KEY `active_delete` (`is_active`,`is_deleted`), ADD KEY `delete` (`is_deleted`), ADD KEY `email_pass` (`email`,`password`), ADD KEY `apply_job_states` (`apply_job_states`,`apply_job_us`,`apply_job_ca`), ADD KEY `experience_level` (`experience_level`), ADD KEY `latlon` (`lat`,`lon`), ADD KEY `location` (`state`,`zip`), ADD KEY `slug` (`slug`,`is_active`,`priv_profile`), ADD KEY `scope` (`is_active`,`priv_profile`,`state`), ADD KEY `member_search_alerts` (`is_active`,`is_jobseeker`,`profile_updated`,`priv_profile`,`apply_job_us`,`apply_job_ca`);

更新：根據要求，這裏的優化設置

mysql> SELECT @@optimizer_switch\G 
*************************** 1. row *************************** 
@@optimizer_switch: index_merge=on,index_merge_union=on,index_merge_sort_union=on,index_merge_intersection=on 
1 row in set (0.00 sec)

注意：這已經過測試

服務器版本：5.6.20-68.0-56日誌 - Percona的XtraDB集羣（GPL），釋放25.7
Server版本：5.5.29-0ubuntu0.12.04.1
Server版本：72年5月1日 - 源代碼分發

來源

2014-11-14 zeroasterisk

爲什麼要爲'聲明members'只顯示一個指數'active_delete'（旁邊的主鍵），同時解釋計劃表現出更多的可能密鑰此表：'active_delete，範圍，member_search_alerts，A，B'？ – krokodilko 2014-11-14 20:20:05

對不起 - 這個例子是從一個真實的情況中拉出來的，因此還有一些其他「無關」的索引我沒有清理......我們在這張表上還有其他類型的查詢。 – zeroasterisk 2014-11-17 13:57:08

您可以運行'SELECT @@ optimizer_switch;'並將此命令的結果追加到問題中嗎？ – krokodilko 2014-11-17 23:45:20

在這種情況下，表1是MyISAM，另一個是InnoDB

當我切換既InnoDB它奇蹟般地從ALL變爲ref和從掃描所有行的一個子集。

mysql> explain SELECT COUNT(*) AS `count` FROM `ao_prod`.`members` AS `Member` LEFT JOIN `ao_prod`.`member_friends` AS `MyFriend` ON (`MyFriend`.`member_2_id` = `Member`.`id` AND member_1_id = '150365') WHERE `Member`.`is_active` = '1' AND NOT(`Member`.`first_name` = '' AND `Member`.`last_name` = '') AND (`Member`.`priv_profile` = 3 OR `MyFriend`.`status_id` IN (1,2)); 
+----+-------------+----------+------+---------------+-------------+---------+-------+--------+--------------------------+ 
| id | select_type | table | type | possible_keys | key   | key_len | ref | rows | Extra     | 
+----+-------------+----------+------+---------------+-------------+---------+-------+--------+--------------------------+ 
| 1 | SIMPLE  | Member | ALL | A    | NULL  | NULL | NULL | 140245 | Using where    | 
| 1 | SIMPLE  | MyFriend | ref | member_1_id | member_1_id | 4  | const | 181 | Using where; Using index | 
+----+-------------+----------+------+---------------+-------------+---------+-------+--------+--------------------------+ 
2 rows in set (0.00 sec) 

mysql> ALTER TABLE `members` ENGINE = InnoDB; 
Query OK, 140245 rows affected (1 min 8.10 sec) 
Records: 140245 Duplicates: 0 Warnings: 0 

mysql> explain SELECT COUNT(*) AS `count` FROM `ao_prod`.`members` AS `Member` LEFT JOIN `ao_prod`.`member_friends` AS `MyFriend` ON (`MyFriend`.`member_2_id` = `Member`.`id` AND member_1_id = '150365') WHERE `Member`.`is_active` = '1' AND NOT(`Member`.`first_name` = '' AND `Member`.`last_name` = '') AND (`Member`.`priv_profile` = 3 OR `MyFriend`.`status_id` IN (1,2)); 
+----+-------------+----------+------+---------------+-------------+---------+-------+-------+--------------------------+ 
| id | select_type | table | type | possible_keys | key   | key_len | ref | rows | Extra     | 
+----+-------------+----------+------+---------------+-------------+---------+-------+-------+--------------------------+ 
| 1 | SIMPLE  | Member | ref | A    | A   | 1  | const | 53916 | Using where    | 
| 1 | SIMPLE  | MyFriend | ref | member_1_id | member_1_id | 4  | const | 181 | Using where; Using index | 
+----+-------------+----------+------+---------------+-------------+---------+-------+-------+--------------------------+

來源

2014-11-21 02:22:03 zeroasterisk

創建與這兩個指數沿着這兩張表，並運行你的第一個查詢實際使用index A爲memebers表：

+----+-------------+----------+------+---------------+-------------+---------+-------+------+--------------------------+ 
| id | select_type | table | type | possible_keys | key   | key_len | ref | rows | Extra     | 
+----+-------------+----------+------+---------------+-------------+---------+-------+------+--------------------------+ 
| 1 | SIMPLE  | Member | ref | A,B   | A   | 1  | const | 3199 | Using index condition | 
| 1 | SIMPLE  | MyFriend | ref | member_1_id | member_1_id | 4  | const | 2 | Using where; Using index | 
+----+-------------+----------+------+---------------+-------------+---------+-------+------+--------------------------+

測試上：5.6.19-0ubuntu0.14.04.1

同樣在SQLFiddle

來源

2014-11-19 20:23:43

是的 - 如果我們只有幾條記錄，它似乎使用索引...但不與'成員'100k記錄。 – zeroasterisk 2014-11-20 00:22:07

@zeroasterisk在'members'中測試了200K，在'member_friends'中測試了500K，並且仍在使用索引A. – 2014-11-20 10:10:37

請提供您在這兩個表上的其他索引，可能它們會阻止優化器選擇正確的索引。 – 2014-11-20 10:15:10

mysql左加入哪裏用或，掃描整個表（沒有索引）

回答

相關問題