我有以下数据：
customer
+---------------------------------------------------------------------------------------------+
| id | email | firstname | lastname | no_pub | customer_address |
|---------------------------------------------------------------------------------------------|
| 1 | martin@domain.com | Martin | Scorcese | 0 | 4 |
| 2 | robert@domain.com | Robert | De Niro | 0 | 7 |
| 3 | bruce@domain;com | Bruce | Willis | 0 | 10 |
+---------------------------------------------------------------------------------------------+
address
+------------------------------------------+
| id | city | zipcode |
|------------------------------------------|
| 4 | Paris | 75001 |
| 7 | Marseille | 13000 |
| 10 | Bordeaux | 33000 |
+------------------------------------------+
sf_geo_cities
+------------------------------------------+
| id | region_id | zipcode |
|------------------------------------------|
| 1 | 1 | 75001 |
| 2 | 2 | 13000 |
| 2 | 3 | 33000 |
+------------------------------------------+
sf_geo_regions
+------------------------------------------+
| id | name | zipcode |
|------------------------------------------|
| 1 | Ile-de-France | 75001 |
| 2 | Cote d'Azur | 13000 |
| 2 | Gironde | 33000 |
+------------------------------------------+
这是一个示例，这些表中显然有更多数据（300k +客户，400k +地址） 最后，我想获得以下数据：
+---------------------------------------------------------------------------------------------+
| id | email | firstname | lastname | city | region |
|---------------------------------------------------------------------------------------------|
| 1 | martin@domain.com | Martin | Scorcese | Paris | Ile-de-France |
| 2 | robert@domain.com | Robert | De Niro | Marseille | Cote d'Azur |
| 3 | bruce@domain;com | Bruce | Willis | NULL | NULL |
+---------------------------------------------------------------------------------------------+
我尝试了以下SQL查询：
SELECT c.id, c.email, c.firstname, c.lastname, gc.name, gr.name
FROM customer c
LEFT JOIN address ad ON ad.id = c.customer_address
JOIN sf_geo_cities gc ON gc.zipcode = ad.zipcode
JOIN sf_geo_regions gr ON gr.id = gc.region_id
WHERE no_pub = 0
但这需要很长时间，我必须关闭mysql服务。连接乘法行可能存在问题。
是否有一个简单的查询来获取我期望的数据？
更新＃1： 以下是评论中的EXPLAIN：
id select_type table type possible_keys key key_len ref rows Extra
1 SIMPLE c ref PRIMARY,UNIQ_81398E097D3656A4,UNIQ_81398E09E7927C7... no_pub 1 const 136220 Using where
1 SIMPLE a eq_ref PRIMARY PRIMARY 8 evotest.c.account 1 Using index
1 SIMPLE ad eq_ref PRIMARY PRIMARY 8 evotest.c.customer_address 1 NULL
1 SIMPLE gc ALL zipcode NULL NULL NULL 38194 Range checked for each record (index map: 0x8)
1 SIMPLE gr eq_ref PRIMARY PRIMARY 4 evotest.gc.region_id 1 NULL
更新＃2：数据库示例 这是我的数据库，包含最少的样本数据。
--
-- Structure de la table `account`
--
CREATE TABLE IF NOT EXISTS `account` (
`id` bigint(20) unsigned NOT NULL,
`identifier` varchar(255) COLLATE utf8_bin NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_bin AUTO_INCREMENT=315688 ;
--
-- Structure de la table `customer`
--
CREATE TABLE IF NOT EXISTS `customer` (
`id` bigint(20) unsigned NOT NULL,
`account` bigint(20) unsigned NOT NULL,
`customer_address` bigint(20) unsigned DEFAULT NULL,
`email` varchar(255) COLLATE utf8_bin NOT NULL,
`lastname` varchar(255) COLLATE utf8_bin NOT NULL,
`firstname` varchar(255) COLLATE utf8_bin NOT NULL,
`no_pub` tinyint(1) NOT NULL DEFAULT '0'
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_bin AUTO_INCREMENT=315224 ;
--
-- Structure de la table `address`
--
CREATE TABLE IF NOT EXISTS `address` (
`id` bigint(20) unsigned NOT NULL,
`city` varchar(64) COLLATE utf8_bin DEFAULT NULL,
`street` varchar(255) COLLATE utf8_bin DEFAULT NULL,
`complement` varchar(128) COLLATE utf8_bin DEFAULT NULL,
`zipcode` varchar(16) COLLATE utf8_bin DEFAULT NULL,
`country_id` int(11) DEFAULT NULL,
`cedex` tinyint(1) NOT NULL DEFAULT '0',
`abroad` tinyint(1) NOT NULL DEFAULT '0'
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_bin AUTO_INCREMENT=442743 ;
--
-- Structure de la table `sf_geo_cities`
--
CREATE TABLE IF NOT EXISTS `sf_geo_cities` (
`id` int(11) NOT NULL,
`region_id` int(11) NOT NULL,
`name` varchar(255) NOT NULL,
`slug` varchar(255) NOT NULL,
`zipcode` varchar(5) NOT NULL,
`insee_code` int(11) NOT NULL,
`latitude` float NOT NULL,
`longitude` float NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8 AUTO_INCREMENT=38106 ;
CREATE TABLE IF NOT EXISTS `sf_geo_regions` (
`id` int(11) NOT NULL,
`name` varchar(255) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8 AUTO_INCREMENT=25 ;
CREATE TABLE IF NOT EXISTS `sf_geo_countries` (
`id` int(11) NOT NULL,
`code` int(11) NOT NULL,
`alpha2` varchar(2) NOT NULL,
`alpha3` varchar(3) NOT NULL,
`name_en` varchar(45) NOT NULL,
`name_fr` varchar(45) NOT NULL,
`is_default` tinyint(1) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8 AUTO_INCREMENT=242 ;
CREATE TABLE IF NOT EXISTS `sf_user_data` (
`id` int(11) NOT NULL,
`user_id` bigint(20) unsigned NOT NULL,
`main_activity_type_id` int(11) DEFAULT NULL,
`main_activity_id` int(11) DEFAULT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci AUTO_INCREMENT=19001 ;
CREATE TABLE IF NOT EXISTS `sf_activity_types` (
`id` int(11) NOT NULL,
`name` varchar(255) COLLATE utf8_unicode_ci NOT NULL,
`identifier` varchar(255) COLLATE utf8_unicode_ci NOT NULL COMMENT 'Ne doit pas être modifié, il s agit de la clé dans le tableau de constantes $constants stockant les taux dans l entité Calculator'
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci AUTO_INCREMENT=3 ;
CREATE TABLE IF NOT EXISTS `sf_activities` (
`id` int(11) NOT NULL,
`name` varchar(255) COLLATE utf8_unicode_ci NOT NULL,
`asks_for_custom` tinyint(1) NOT NULL DEFAULT '0'
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci AUTO_INCREMENT=27 ;
--
-- Index pour les tables exportées
--
ALTER TABLE `account`
ADD PRIMARY KEY (`id`), ADD UNIQUE KEY `identifier_UNIQUE` (`identifier`);
ALTER TABLE `customer`
ADD PRIMARY KEY (`id`), ADD UNIQUE KEY `UNIQ_81398E097D3656A4` (`account`), ADD UNIQUE KEY `UNIQ_81398E09E7927C74` (`email`), ADD UNIQUE KEY `UNIQ_81398E091193CB3F` (`customer_address`), ADD KEY `no_pub` (`no_pub`);
ALTER TABLE `address`
ADD PRIMARY KEY (`id`), ADD KEY `IDX_D4E6F81F92F3E70` (`country_id`), ADD KEY `zipcode` (`zipcode`);
ALTER TABLE `sf_geo_cities`
ADD PRIMARY KEY (`id`), ADD KEY `IDX_B56556A198260155` (`region_id`), ADD KEY `zipcode` (`zipcode`);
ALTER TABLE `sf_geo_regions`
ADD PRIMARY KEY (`id`);
ALTER TABLE `sf_geo_countries`
ADD PRIMARY KEY (`id`), ADD KEY `IDX_F86325E277153098` (`code`), ADD KEY `IDX_F86325E2B762D672` (`alpha2`), ADD KEY `IDX_F86325E2C065E6E4` (`alpha3`);
ALTER TABLE `sf_user_data`
ADD PRIMARY KEY (`id`), ADD UNIQUE KEY `UNIQ_E904BFD1A76ED395` (`user_id`), ADD KEY `IDX_E904BFD12E864BE8` (`main_activity_type_id`), ADD KEY `IDX_E904BFD15543A800` (`main_activity_id`);
ALTER TABLE `sf_activity_types`
ADD PRIMARY KEY (`id`);
ALTER TABLE `sf_activities`
ADD PRIMARY KEY (`id`);
--
-- Contraintes pour les tables exportées
--
ALTER TABLE `customer`
ADD CONSTRAINT `FK_81398E091193CB3F` FOREIGN KEY (`customer_address`) REFERENCES `address` (`id`),
ADD CONSTRAINT `FK_81398E097D3656A4` FOREIGN KEY (`account`) REFERENCES `account` (`id`);
ALTER TABLE `address`
ADD CONSTRAINT `FK_D4E6F81F92F3E70` FOREIGN KEY (`country_id`) REFERENCES `sf_geo_countries` (`id`);
ALTER TABLE `sf_geo_cities`
ADD CONSTRAINT `FK_B56556A198260155` FOREIGN KEY (`region_id`) REFERENCES `sf_geo_regions` (`id`);
ALTER TABLE `sf_user_data`
ADD CONSTRAINT `FK_E904BFD12E864BE8` FOREIGN KEY (`main_activity_type_id`) REFERENCES `sf_activity_types` (`id`),
ADD CONSTRAINT `FK_E904BFD15543A800` FOREIGN KEY (`main_activity_id`) REFERENCES `sf_activities` (`id`),
ADD CONSTRAINT `FK_E904BFD1A76ED395` FOREIGN KEY (`user_id`) REFERENCES `account` (`id`);
INSERT INTO `account` (`id`, `identifier`) VALUES ('1', 'martin@domain.com'), ('2', 'robert@domain.com'), ('3', 'bruce@domain.com');
INSERT INTO `sf_geo_countries` (`id`, `code`, `alpha2`, `alpha3`, `name_en`, `name_fr`, `is_default`) VALUES ('1', '1', 'FR', 'FRA', 'France', 'France', '1');
INSERT INTO `address` (`id`, `city`, `street`, `complement`, `zipcode`, `country_id`, `cedex`, `abroad`) VALUES ('1', 'Paris', '1 rue de Paris', NULL, '75001', '1', '0', '0'), ('2', 'Marseille', '1 rue de Marseille', NULL, '13000', '1', '0', '0');
INSERT INTO `customer` (`id`, `account`, `customer_address`, `email`, `lastname`, `firstname`, `no_pub`) VALUES ('1', '1', '1', 'martin@domain.com', 'Scorcese', 'Martin', '0'), ('2', '2', '2', 'robert@domain.com', 'De Niro', 'Robert', '0'), ('3', '3', NULL, 'bruce@domain.com', 'Willis', 'Bruce', '0');
INSERT INTO `sf_activities` (`id`, `name`, `asks_for_custom`) VALUES ('1', 'Activity #1', '0'), ('2', 'Activity #2', '0');
INSERT INTO `sf_activity_types` (`id`, `name`, `identifier`) VALUES ('1', 'Activity Type #1', 'activity-type-1'), ('2', 'Activity Type #2', 'activity-type-2');
INSERT INTO `sf_geo_regions` (`id`, `name`) VALUES ('1', 'Ile-de-France'), ('2', 'Cote d''Azur');
INSERT INTO `sf_geo_cities` (`id`, `region_id`, `name`, `slug`, `zipcode`, `insee_code`, `latitude`, `longitude`) VALUES ('1', '1', 'Paris', 'paris', '75001', '1', '0', '0'), ('2', '2', 'Marseille', 'marseille', '13000', '2', '0', '0');
INSERT INTO `sf_user_data` (`id`, `user_id`, `main_activity_type_id`, `main_activity_id`) VALUES ('1', '1', '1', '1'), ('2', '3', '2', '2');
使用这么少量的数据，我可以运行以下查询，返回我想要的一切。但是这个查询在我的真实数据库中运行了很长时间，拥有300k +客户和400k +地址
SELECT c.id, c.email, c.firstname, c.lastname, acttypes.name AS activity_type, act.name AS activity, gc.name AS city, gr.name AS region
FROM customer c
JOIN account a ON a.id = c.account
LEFT JOIN sf_user_data ud ON ud.user_id = a.id
LEFT JOIN sf_activity_types acttypes ON acttypes.id = ud.main_activity_type_id
LEFT JOIN sf_activities act ON act.id = ud.main_activity_id
LEFT JOIN address ad ON ad.id = c.customer_address
LEFT JOIN sf_geo_cities gc ON gc.zipcode = ad.zipcode
LEFT JOIN sf_geo_regions gr ON gr.id = gc.region_id
WHERE no_pub = 0
更新＃3：跟踪加入问题
此查询运行速度很快：
SELECT
c.id
,c.email
,c.firstname
,c.lastname
,acttypes.name AS activity_type
,act.name AS activity
,ad.zipcode AS address_zipcode
-- ,gc.name AS city
-- ,gr.name AS region
FROM
customer c
JOIN account a ON a.id = c.account
LEFT JOIN sf_user_data ud ON ud.user_id = a.id
LEFT JOIN sf_activity_types acttypes ON acttypes.id = ud.main_activity_type_id
LEFT JOIN sf_activities act ON act.id = ud.main_activity_id
LEFT JOIN address ad ON ad.id = c.customer_address
-- LEFT JOIN sf_geo_cities gc ON gc.zipcode = ad.zipcode
-- LEFT JOIN sf_geo_regions gr ON gr.id = gc.region_id
WHERE
no_pub = 0
此查询也可以快速运行：
SELECT
c.id
,c.email
,c.firstname
,c.lastname
-- ,acttypes.name AS activity_type
-- ,act.name AS activity
,ad.zipcode AS address_zipcode
,gc.name AS city
-- ,gr.name AS region
FROM
customer c
JOIN account a ON a.id = c.account
-- LEFT JOIN sf_user_data ud ON ud.user_id = a.id
-- LEFT JOIN sf_activity_types acttypes ON acttypes.id = ud.main_activity_type_id
-- LEFT JOIN sf_activities act ON act.id = ud.main_activity_id
LEFT JOIN address ad ON ad.id = c.customer_address
LEFT JOIN sf_geo_cities gc ON gc.zipcode = ad.zipcode
-- LEFT JOIN sf_geo_regions gr ON gr.id = gc.region_id
WHERE
no_pub = 0
此查询永远运行：
SELECT
c.id
,c.email
,c.firstname
,c.lastname
,acttypes.name AS activity_type
,act.name AS activity
,ad.zipcode AS address_zipcode
,gc.name AS city
-- ,gr.name AS region
FROM
customer c
JOIN account a ON a.id = c.account
LEFT JOIN sf_user_data ud ON ud.user_id = a.id
LEFT JOIN sf_activity_types acttypes ON acttypes.id = ud.main_activity_type_id
LEFT JOIN sf_activities act ON act.id = ud.main_activity_id
LEFT JOIN address ad ON ad.id = c.customer_address
LEFT JOIN sf_geo_cities gc ON gc.zipcode = ad.zipcode
-- LEFT JOIN sf_geo_regions gr ON gr.id = gc.region_id
WHERE
no_pub = 0
加入活动是可以的，加入地址是可以的，但尝试使用所有这些联接并不起作用。
答案 0 :(得分：1)
正如我的评论中提到的，非常重要的一点是返回了多少条记录。一个语句中的记录太多可能导致问题，如果您对结果进行分页，一次选择n条记录，则可能会更好。
要对查询进行疑难解答，请通过联接构建它以尝试确定性能何时落入悬崖。从：
开始go build然后一次取消注释一个连接，直到您的查询执行速度非常慢。分析该查询并查看是否可以使用索引进行改进。然后继续取消注释连接，直到您有完整的查询。