Question

SELECT `apps`.*,
       COUNT(all_users.id) AS total_users,
       COUNT(active_users.id) AS active_users
FROM `apps`
INNER JOIN `social_users` AS `all_users` ON `all_users`.`app_id` = `apps`.`id`
INNER JOIN `social_users` AS `active_users` ON `active_users`.`app_id` = `apps`.`id`
WHERE `active_users`.`is_active` = 'true'
GROUP BY `id`
ORDER BY `total_users` ASC LIMIT 30
OFFSET 0

我有2张桌子 apps -> id, name
social_user -> id, app_id, is_active

我想在apps表上运行查询，并显示所有用户和所有活跃用户我两次加入社交用户问题是，如果我加入它一次（无论哪个）它运行得非常快。第二次连接会导致性能下降，而且数字根本不准确，我得到180000个用户而不是750个。

我怎样才能更好更快地完成？

谢谢

Answer 1

SELECT `apps`.*,
       COUNT(all_users.id) AS total_users,
       SUM(IF(all_users.is_active='true',1,0)) AS active_users
  FROM `apps`
 INNER JOIN `social_users` AS `all_users` ON `all_users`.`app_id` = `apps`.`id`
 GROUP BY `id`
 ORDER BY `total_users` ASC LIMIT 30
OFFSET 0

Answer 2

速度将在很大程度上取决于您的索引，但是我看到的第一个问题是您需要将WHERE子句移动到ON表的active_users子句中：

SELECT `apps`.*,
       COUNT(all_users.id) AS total_users,
       COUNT(active_users.id) AS active_users
FROM `apps`
INNER JOIN `social_users` AS `all_users` ON `all_users`.`app_id` = `apps`.`id`
INNER JOIN `social_users` AS `active_users` ON `active_users`.`app_id` = `apps`.`id`
       AND `active_users`.`is_active` = 'true'
GROUP BY `id`
ORDER BY `total_users` ASC LIMIT 30
OFFSET 0

这至少可以为您提供正确数量的结果。为了避免链接两次，你可以使你的第二个COUNT（）条件，考虑：

SELECT `apps`.*,
       COUNT(all_users.id) AS total_users,
       COUNT(IF(all_users.is_active = 'true',all_users.id,NULL)) AS active_users
FROM `apps`
INNER JOIN `social_users` AS `all_users` ON `all_users`.`app_id` = `apps`.`id`
GROUP BY `id`
ORDER BY `total_users` ASC LIMIT 30
OFFSET 0

通过这种方式，您只能链接到该表一次，并且只有在您的活动计数处于活动状态时才会有条件地计数。

祝你好运！

mysql加入同一个表两次运行速度很慢，结果错误

2 个答案: