index_merge全表扫描-2秒mysql选择

Question

我有这个选择：

  SELECT MAX(id) FROM chat
  WHERE (`to` = 1 and `del_to_status` = '0') or (`from` = 1 and `del_from_status` = '0')
  GROUP BY CASE WHEN 1 = `to` THEN `from` ELSE `to` END

聊天：

`chat` (
  `id` int(11) UNSIGNED NOT NULL AUTO_INCREMENT,
  `from` int(11) UNSIGNED NOT NULL,
  `to` int(11) UNSIGNED NOT NULL,
  `message` text NOT NULL,
  `del_from_status` tinyint(1) NOT NULL DEFAULT '0',
  `del_to_status` tinyint(1) NOT NULL DEFAULT '0',
  PRIMARY KEY (`id`),
  KEY `from` (`from`),
  KEY `to` (`to`),
);

问题在于它正在使用全表扫描：

这需要很多时间。有什么想法可以更快获得结果？

Answer 1

您如何看待该解决方案：

select grouped_by_to.user, greatest(grouped_by_to.id, grouped_by_from.id ) from 
(
    select c1.to as user, max(id) as id from chat c1
    group by c1.to 
) grouped_by_to

join
(
    select c1.from as user, max(id) as id from chat c1
    group by c1.from
) grouped_by_from on grouped_by_from.user = grouped_by_to.user

请注意，我忽略了del_to_status列，您可以轻松地添加它们。

但是实际上我认为您的整个数据库架构是错误的，我认为您需要更多类似的东西：

`messages` (
  `message_id` int(11) UNSIGNED NOT NULL AUTO_INCREMENT,
  `user_id` int(11) UNSIGNED NOT NULL,
  `message` text NOT NULL,
  `message_date` timestamp NOT NULL,
  PRIMARY KEY (`message_id`),
);

`conversatinos` (
  `conversation_id` int(11) UNSIGNED NOT NULL AUTO_INCREMENT,
  `message_id` int(11) UNSIGNED NOT NULL,
  PRIMARY KEY (`conversation_id`),
);

`users` (
  `user_id` int(11) UNSIGNED NOT NULL AUTO_INCREMENT,
  `user_name` int(11) UNSIGNED NOT NULL,
  PRIMARY KEY (`user_id`),
);

还有可能，如果您需要：

`chat` (
  `id` int(11) UNSIGNED NOT NULL AUTO_INCREMENT,
  `message_id` int(11) UNSIGNED NOT NULL,
  PRIMARY KEY (`id`),
);