Question

我需要帮助限制内部联接。

表格

用户：

UID
左
活性

文件：

裂

UID = FID

SELECT uid
FROM users AS u
INNER JOIN files AS x ON u.uid = x.fid
WHERE u.left = '0'
AND u.active = '1'
ORDER BY `u`.`uid` ASC

结果：

我需要将INNER JOIN限制为5。所以“uid”不会出现超过5次。像：

更新

这是php代码

 $res = do_sqlquery("SELECT uid FROM users as u INNER JOIN files as x ON u.fid=x.fid WHERE u.left = '0'  AND u.active='1'");
   if (mysql_num_rows($res) > 0)
   {
       while ($arr = mysql_fetch_assoc($res))
       {
       $x=$arr["uid"];
       quickQuery("UPDATE users SET pots = pots+".$GLOBALS["bui"]."*".$cleaint."/900 WHERE id = '$x'");
       }
   }

Answer 1

仅限MySQL的解决方案，不使用PHP

SELECT uid
from (
    SELECT u.uid, @r:= if(@u=u.uid,@r+1,1) r, @u:=u.uid
    FROM users AS u
    CROSS JOIN (select @u:=null) g
    INNER JOIN files AS x ON u.uid = x.fid
    WHERE u.`left` = '0'
    AND u.active = '1'
    ORDER BY u.uid ASC
) U
WHERE U.r <= 5
ORDER BY uid ASC

或更一般地说，只是没有MySQL变量的SQL

SELECT u.uid
from (
    SELECT u.uid, count(*) C
    FROM users AS u
    INNER JOIN files AS x ON u.uid = x.fid
    WHERE u.`left` = '0'
    AND u.active = '1'
    GROUP BY u.uid
) U inner join (
   select 1 n union all select 2 union all select 3
    union all select 4 union all select 5) V ON U.C>=V.n
ORDER BY U.uid ASC

Answer 2

在MySQL中：

SELECT CONCAT(`uid`, ":", MAX(5, COUNT(`fid`))) `uid_count`
FROM ...
GROUP BY `uid`

给你的结果如下：

uid_count
3:4
7:1
47:5

例如，在PHP中，您可以在获取后进行拆分：

while (FALSE !== ($row = mysql_fetch_assoc($query))) {
    ($uid, $count) = split(':', $row['uid_count'], 2);
}

帮助限制内部联接

2 个答案: