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Question

使用group by时，我需要帮助查询运行缓慢：

SELECT           `customers`.`id` 
                ,`customers`.`firstname`
                ,`customers`.`lastname`
                ,`customers`.`address`
                ,`customers`.`address_co`
                ,`customers`.`country`
                ,`customers`.`zipcode`
                ,`customers`.`city`
                ,`customers`.`phone`
                ,`customers`.`mobilephone`
                ,`customers`.`ssn`
                ,`users`.`email`
                ,`customer_lists`.`name`        AS `customerList`
                ,COUNT( transactions.id )       AS transactions_count
                ,SUM( transactions.sum_incl )   AS sum_incl
                ,SUM( transactions.sum_excl )   AS sum_excl
                ,`customers`.`created_at` 
FROM            `customers` 
INNER JOIN      `users` 
ON              `customers`.`user_id` = `users`.`id` 
LEFT JOIN       `transactions` 
ON              `customers`.`id` = `transactions`.`customer_id` 
INNER JOIN      `customer_lists`
ON              `customers`.`customer_list_id` = `customer_lists`.`id` 
WHERE           `customer_lists`.`club_id` =  '1'
GROUP BY        `customers`.`id` 
ORDER BY        `customers`.`created_at` DESC

当我运行EXPLAIN时，它会显示Using temporary; Using filesort。

**id**  **select_type** **table**       **type**    **possible_keys**                                       **key**                             **key_len** **ref**                             **rows**    **Extra**
1       SIMPLE          customer_lists  ref         PRIMARY,club_id_index                                   club_id_index                       8           const                               1           Using temporary; Using filesort
1       SIMPLE          customers       ref         customers_user_id_foreign,customers_customer_list_...   customers_customer_list_id_index    8           kund_workbox_nu.customer_lists.id   867  
1       SIMPLE          users           eq_ref      PRIMARY                                                 PRIMARY                             8           kund_workbox_nu.customers.user_id   1    
1       SIMPLE          transactions    ref         customer_id_index                                       customer_id_index                   8           kund_workbox_nu.customers.id        3108

我该如何解决？

更新

表：

客户

CREATE TABLE `customers` (
  `id` bigint(20) unsigned NOT NULL AUTO_INCREMENT,
  `user_id` bigint(20) unsigned NOT NULL,
  `customer_list_id` bigint(20) unsigned NOT NULL,
  `firstname` varchar(100) COLLATE utf8_unicode_ci NOT NULL,
  `lastname` varchar(100) COLLATE utf8_unicode_ci NOT NULL,
  `address` varchar(100) COLLATE utf8_unicode_ci NOT NULL,
  `address_co` varchar(100) COLLATE utf8_unicode_ci NOT NULL,
  `country` varchar(100) COLLATE utf8_unicode_ci NOT NULL,
  `zipcode` varchar(10) COLLATE utf8_unicode_ci NOT NULL,
  `city` varchar(100) COLLATE utf8_unicode_ci NOT NULL,
  `phone` varchar(20) COLLATE utf8_unicode_ci NOT NULL,
  `mobilephone` varchar(20) COLLATE utf8_unicode_ci NOT NULL,
  `ssn` varchar(15) COLLATE utf8_unicode_ci NOT NULL,
  `last_login` datetime NOT NULL,
  `added_information` tinyint(1) NOT NULL DEFAULT '0',
  `created_at` timestamp NOT NULL DEFAULT '0000-00-00 00:00:00',
  `updated_at` timestamp NOT NULL DEFAULT '0000-00-00 00:00:00',
  PRIMARY KEY (`id`),
  KEY `customers_user_id_foreign` (`user_id`),
  KEY `customers_customer_list_id_index` (`customer_list_id`),
  CONSTRAINT `customers_customer_list_id_foreign` FOREIGN KEY (`customer_list_id`) REFERENCES `customer_lists` (`id`) ON DELETE CASCADE ON UPDATE CASCADE,
  CONSTRAINT `customers_user_id_foreign` FOREIGN KEY (`user_id`) REFERENCES `users` (`id`) ON DELETE CASCADE ON UPDATE CASCADE
) ENGINE=InnoDB AUTO_INCREMENT=29739 DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;

交易

CREATE TABLE `transactions` (
  `id` bigint(20) unsigned NOT NULL AUTO_INCREMENT,
  `transaction_type_id` bigint(20) unsigned NOT NULL,
  `customer_id` bigint(20) unsigned NOT NULL,
  `created_at` timestamp NOT NULL DEFAULT '0000-00-00 00:00:00',
  `updated_at` timestamp NOT NULL DEFAULT '0000-00-00 00:00:00',
  `sum_incl` decimal(15,4) NOT NULL,
  `sum_excl` decimal(15,4) NOT NULL,
  PRIMARY KEY (`id`),
  KEY `transactions_customer_id_foreign` (`customer_id`),
  KEY `transactions_transaction_type_id_foreign` (`transaction_type_id`),
  CONSTRAINT `transactions_customer_id_foreign` FOREIGN KEY (`customer_id`) REFERENCES `customers` (`id`) ON DELETE CASCADE ON UPDATE CASCADE,
  CONSTRAINT `transactions_transaction_type_id_foreign` FOREIGN KEY (`transaction_type_id`) REFERENCES `transaction_types` (`id`) ON DELETE CASCADE ON UPDATE CASCADE
) ENGINE=InnoDB AUTO_INCREMENT=580116 DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;

客户名单

CREATE TABLE `customer_lists` (
  `id` bigint(20) unsigned NOT NULL AUTO_INCREMENT,
  `club_id` bigint(20) unsigned NOT NULL,
  `name` varchar(90) COLLATE utf8_unicode_ci NOT NULL,
  `created_at` timestamp NOT NULL DEFAULT '0000-00-00 00:00:00',
  `updated_at` timestamp NOT NULL DEFAULT '0000-00-00 00:00:00',
  PRIMARY KEY (`id`),
  KEY `customer_lists_club_id_foreign` (`club_id`),
  CONSTRAINT `customer_lists_club_id_foreign` FOREIGN KEY (`club_id`) REFERENCES `clubs` (`id`) ON DELETE CASCADE ON UPDATE CASCADE
) ENGINE=InnoDB AUTO_INCREMENT=11 DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;

Answer 1

也许尝试将一个表分组作为派生表进行分组？

SELECT
      `customers`.`id`
    , `customers`.`firstname`
    , `customers`.`lastname`
    , `customers`.`address`
    , `customers`.`address_co`
    , `customers`.`country`
    , `customers`.`zipcode`
    , `customers`.`city`
    , `customers`.`phone`
    , `customers`.`mobilephone`
    , `customers`.`ssn`
    , `users`.`email`
    , `customer_lists`.`name` AS `customerList`
    , `trans`.transactions_count
    , `trans`.sum_incl
    , `trans`.sum_excl
    , `customers`.`created_at`
FROM `customers`
INNER JOIN `users` ON `customers`.`user_id` = `users`.`id`
INNER JOIN `customer_lists` ON `customers`.`customer_list_id` = `customer_lists`.`id`
LEFT JOIN (
    SELECT
          `transactions`.`customer_id`
        , COUNT(transactions.id) AS transactions_count
        , SUM(transactions.sum_incl) AS sum_incl
        , SUM(transactions.sum_excl) AS sum_excl
    FROM `transactions`
    GROUP BY `transactions`.`customer_id`
    ) `trans` ON `customers`.`id` = `trans`.`customer_id`
WHERE `customer_lists`.`club_id` = '1'
ORDER BY `customers`.`created_at` DESC

该子查询中的where子句也可以提供帮助。

Answer 2

您是否尝试在所有JOIN列上创建索引，即customers。user_id，users。id，transactions。{{1} ，...？

使用GROUP BY优化SQL查询

更新

表：

客户

交易

客户名单

2 个答案: