以下是其他方法：
1） 使用 NOT IN 您在上添加了一个条件 usertable.user_id 不应该在id列表中：这些id是表的所有id friendrequests 以及 friends 那些有 to_id 或者 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

根据rick james评论编辑，与无工会相同：

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 我看到了这个答案，试着用同样的逻辑：你做一些 LEFT JOIN 并添加一个条件，其中这些连接返回空值

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，由于没有样本数据或sql fiddle来测试这些查询，我无法告诉您它是否返回您想要的结果，或者它是否真的更快。。。
编辑：这个网页是从2009年，所以结果有变化，因为！我不会删除它，因为您仍然具有我以前使用的查询结构：
我建议您查看本页以了解这些查询之间的区别，并为您提供一些实现所需的想法：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，这并不有效：
然而，这个查询的效率比前两个稍微低一点：它需要0.92秒。
这并没有造成太大的性能下降，但是，查询所需的时间要多27%。