如何使用JOIN而不是子查询（NOT IN）

Question

我想列出$_SESSION['userid']关注的所有人，而不是

以下查询适用于跟随演员，它可以正常工作

IN （慢）

SELECT user.* FROM user 
WHERE user.userid 
     IN (SELECT follow.followtoid FROM follow 
WHERE follow.followerid = $_SESSION['userid'])

加入（快速）

SELECT p.* 
FROM user p
    JOIN follow f ON p.userid = f.followtoid
WHERE f.followerid = $_SESSION['userid']

对于那些仍然没有跟随演员的人来说，它有效但似乎很慢 请建议我如何使用JOIN代替NOT IN

NOT IN （慢）

SELECT user.* FROM user 
WHERE user.userid 
     NOT IN (SELECT follow.followtoid FROM follow 
WHERE follow.followerid = $_SESSION['userid'])

Answer 1

您可以使用LEFT JOIN并删除所有匹配的行;

SELECT p.* 
FROM user p
LEFT JOIN follow f 
  ON p.userid = f.followtoid
 AND f.followerid = $_SESSION['userid']
WHERE f.followtoid IS NULL

Answer 2

试试这个：

SELECT u.* 
FROM user u
LEFT JOIN follow f ON u.userid = f.followtoid AND f.followerid = $_SESSION['userid']
WHERE f.followtoid IS NULL;

或

SELECT u.* 
FROM user u
WHERE NOT EXISTS (SELECT 1 FROM follow f 
                  WHERE u.userid = f.followtoid AND f.followerid = $_SESSION['userid']
                 );