从mysql获取数据，耗时近436秒

Question

我是 android 开发人员，在 mysql 方面没有太多经验。我在 mysql 数据库中有 3 个表。

第一个表：-UserTable

+-------------+-------------+----------+----------------------+
| user_id     | name      |   image    |   joining_date       |
+-------------+-------------+----------+----------------------+
|         100 | King      | defualt.jpg | 2018-05-23 20:09:27 |
|         101 | Kochhar   | defualt.jpg | 2018-05-23 20:09:27 |
| - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - |
| - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - |

第二张表：- friendRequests

+-------------+-------------+----------+
| unique_id   | from_id     |   to_id  |
+-------------+-------------+----------+
|         1  | 100          | 101      |
|         2  | 200          | 110      |
| - - - - - - - - - - - - - - - - - - -|
| - - - - - - - - - - - - - - - - - - -|

第三张桌子：- 朋友

+-------------+-------------+----------+
| unique_id   | from_id     |   to_id  |
+-------------+-------------+----------+
|         1  | 104          | 101      |
|         2  | 206          | 110      |
| - - - - - - - - - - - - - - - - - - -|
| - - - - - - - - - - - - - - - - - - -|

我想获取不是我朋友的用户列表，而且我们还没有待处理的朋友请求。所以我尝试了以下查询：-

SELECT usertable.name,
       usertable.user_id,
       usertable.image
FROM   `UserTable` AS usertable
WHERE  usertable.user_id <> '100'
       AND (SELECT Count(*)
            FROM   friendrequests
            WHERE  ( to_id = '100'
                     AND from_id = usertable.user_id )
                    OR ( to_id = usertable.user_id
                         AND from_id = '100' )) = 0
       AND (SELECT Count(*)
            FROM   friends
            WHERE  to_id = usertable.user_id
                   AND from_id = '100'
                    OR to_id = '100'
                       AND from_id = usertable.user_id) = 0
ORDER  BY usertable.joining_date DESC
LIMIT  10  

它正在运行，但需要 436 秒。编写此查询的正确方法是什么？

Answer 1

这里有一些其他的方法：

1) 使用NOT IN

您在 usertable.user_id 上添加了一个不应出现在 id 列表中的条件：这些 id 是表 friendrequests 和 friends 的所有 id 具有 to_id 或 from_id = 100。

SELECT 
  usertable.name,
  usertable.user_id,
  usertable.image
FROM `UserTable` AS usertable
WHERE usertable.user_id <> '100'
  AND usertable.user_id NOT IN 
    (SELECT from_id as user_id FROM friendrequests WHERE to_id = '100'
     UNION
     SELECT to_id as user_id FROM friendrequests WHERE from_id = '100'
     UNION
     SELECT from_id as user_id FROM friends WHERE to_id = '100'
     UNION
     SELECT to_id as user_id FROM friends WHERE from_id = '100')               
ORDER  BY usertable.joining_date DESC
LIMIT  10  

EDIT 根据 Rick James 的评论，同样没有联合：

SELECT 
  usertable.name,
  usertable.user_id,
  usertable.image
FROM `UserTable` AS usertable
WHERE usertable.user_id <> '100'
  AND usertable.user_id NOT IN (SELECT from_id as user_id FROM friendrequests WHERE to_id = '100')
  AND usertable.user_id NOT IN (SELECT to_id as user_id FROM friendrequests WHERE from_id = '100')
  AND usertable.user_id NOT IN (SELECT from_id as user_id FROM friends WHERE to_id = '100')
  AND usertable.user_id NOT IN (SELECT to_id as user_id FROM friends WHERE from_id = '100')               
ORDER  BY usertable.joining_date DESC
LIMIT  10  

2) 使用LEFT JOIN

我看到 this answer 并尝试使用相同的逻辑：您使用另一个表创建一些 LEFT JOIN 并添加一个条件，其中这些连接返回 NULL 值

SELECT 
  usertable.name,
  usertable.user_id,
  usertable.image
FROM `UserTable` AS usertable
LEFT JOIN friendrequests as FRa on FRa.from_id = usertable_user_id
LEFT JOIN friendrequests as FRb on FRb.to_id   = usertable_user_id
LEFT JOIN friends        as Fa  on Fa.from_id  = usertable_user_id
LEFT JOIN friends        as Fb  on Fb.to_id    = usertable_user_id
WHERE usertable.user_id <> '100' 
  AND FRa.from_id IS NULL
  AND FRb.to_id IS NULL
  AND Fa.from_id IS NULL
  AND Fb.to_id IS NULL

3) 使用NOT EXISTS

SELECT 
  usertable.name,
  usertable.user_id,
  usertable.image
FROM `UserTable` AS usertable
WHERE usertable.user_id <> '100'
  AND NOT EXISTS (SELECT from_id as user_id FROM friendrequests WHERE to_id = '100')
  AND NOT EXISTS (SELECT to_id as user_id FROM friendrequests WHERE from_id = '100')
  AND NOT EXISTS (SELECT from_id as user_id FROM friends WHERE to_id = '100')
  AND NOT EXISTS (SELECT to_id as user_id FROM friends WHERE from_id = '100')               
ORDER  BY usertable.joining_date DESC
LIMIT  10  

抱歉，没有示例数据或 SQL Fiddle 来测试这些查询，我无法告诉您它是否返回您想要的内容，或者它是否真的更快...

---

编辑：这个页面是从 2009 年开始的，所以结果已经改变了！我不会删除它，因为您仍然拥有我之前使用的查询结构：

我建议您查看此页面以了解这些查询之间的区别，并为您提供一些想法来实现您想要的：https://explainextended.com/2009/09/18/not-in-vs-not-exists-vs-left-join-is-null-mysql/

本页提出了第三种方法（使用NOT EXISTS），但根据测试它效率不高：

然而，这个查询的效率比前两个略低：它需要 0.92 秒。

这并没有造成太大的性能下降，但是查询需要多花 27% 的时间。