Question

我有一张桌子，我只按ID排序显示最新的30行。

我正在尝试使用下面的查询删除30个最新行之后的所有行。

DELETE FROM table WHERE type = 'test' ORDER BY id DESC LIMIT 30, 60

我一直收到以下错误

#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ' 60' at line 1

我做错了什么？

Answer 1

试试这个，

DELETE FROM table
WHERE ID IN
        (
        SELECT ID
        FROM
            (
                SELECT ID
                FROM table
                WHERE Type = 'TEST'
                ORDER BY ID
                LIMIT 30,60
            ) a
        )

Answer 2

第二次编辑：虽然MySQL支持删除语句中的LIMIT，但它不允许使用OFFSET。这意味着您不能跳过前30行。

对id（或任何其他主键）进行子选择：

DELETE FROM table WHERE id IN (SELECT id FROM table WHERE type = 'test' ORDER BY id DESC LIMIT 30, 60)

Answer 3

这种方式不可能。您可以使用嵌套的select语句尝试它，有点像这样：

DELETE FROM table
WHERE type = 'test'
AND ID IN (SELECT id from table where type = 'test' order by id desc limit 30 )

Answer 4

试试这个

DELETE FROM table WHERE id in(SELECT id FROM table WHERE type = "test" order by id desc limit 30, 60)

Answer 5

我无法在子查询中使用limit子句，因此我使用的解决方案有点凌乱： -

select group_concat(id) into @idList from
( 
select id from  table order by id desc limit 0,30
) as saveIds;
delete from table where not find_in_set(id,@idList)

或者，

select group_concat(id) into @idList from
( 
select id from  table order by id desc limit 30
) as saveIds;
delete from table where find_in_set(id,@idList)

Mysql删除顺序依次

5 个答案: