php - 删除最近 80 之后的 MYSQL 值

Question

我目前正在使用：

SELECT * FROM users ORDER BY date_time DESC LIMIT 0, 80

要显示最近的 80 个用户名，我如何进行查询以删除这 80 个之后的任何其他条目？

我想我需要一个 cron 作业，但它需要执行什么样的查询？

Answer 1

这里有不同的方法：

delete users from users
LEFT JOIN
(
  select id from users order by date_time DESC LIMIT 0, 80) t1 ON users.id=t1.id
where t1.id is null

SQL Fiddle 演示

并且没有限制：

delete users from users
LEFT JOIN
(
  select id from
(
SELECT  id, 
        @curRow := @curRow + 1 AS row_number
FROM    users,(SELECT @curRow := 0) r
ORDER BY date_time DESC) t where row_number<=80
) t1 ON users.id=t1.id
where t1.id is null

SQLFiddle 演示

Answer 2

您需要使用子查询

DELETE FROM USERS where ID NOT IN
(SELECT ID FROM users ORDER BY date_time DESC LIMIT 0, 80)

因为删除时不能使用限制

Answer 3

尝试使用LEFT JOIN.

DELETE  a
FROM    users a
        LEFT JOIN
        (
            SELECT  ID
            FROM    users
            ORDER   BY date_time DESC 
            LIMIT   0, 80
        ) b ON a.ID = b.ID
WHERE   b.ID IS NULL

Answer 4

未经测试，但可能使用用户变量来分配序列号：-

DELETE users
FROM users
INNER JOIN
(
    SELECT ID, @Sequence:=@Sequence + 1 AS aSequence
    FROM (SELECT ID, date_time FROM users ORDER BY date_time) a
    CROSS JOIN (SELECT @Sequence:=0) b
) c
ON users.ID = c.ID
AND c.aSequence > 80

Answer 5

使用中间子查询怎么样？只是一个想法。

DELETE FROM `users`
   WHERE id NOT IN (
      SELECT id
      FROM (
         SELECT id
         FROM `users`
         ORDER BY date_time DESC
         LIMIT 80 
   ) AS q
);