首先从一个表中选择id然后删除那些不在第二个表中的ID

Question

What i'm trying to do is delete the rows in lobby table that don't have the ids in useronline table. that way i will be able to eliminate the ones who aren't "online". (the actual script is not about who's not online but its the same logic) Is there a way that I can first select the ids from useronline , then search in lobby for the ones that aren't those i've just selected, and delete them with a while loop?

This is my non-working script to show you what i've got for the idea so far:

$sql = mysql_query("SELECT DISTINCT `id` FROM `useronline` WHERE 1");
while($row = mysql_fetch_array( $sql )) {
mysql_query("DELETE * 
FROM  `lobby` 
WHERE  `tableid` NOT IN ('$row') <-- Can't figure out how to make this part
LIMIT 0 , 30");
}

Answer 1

You can do this with one query.

DELETE FROM  `lobby` 
WHERE  `tableid` NOT IN (SELECT DISTINCT `id` FROM `useronline`)

Answer 2

为了使代码保持原样，您希望将$ row更改为$ row [“id”]，如下所示：

WHERE  `tableid` NOT IN ('" . $row["id"] . "')