Question

我有这样的疑问

SET @n=0;
DELETE t3 FROM (
         SELECT  id, project_id, task_id, user_id,grouper
                            FROM (  
                                    SELECT  id, project_id, task_id, user_id,
                                    @n:=if(status=55,@n+1,@n),
                                    if(status=55,@n-1,@n) as grouper FROM timelog
                                    WHERE user_id='5' ORDER BY id ASC
                                 ) as t
                where grouper>-1
                group by grouper) as t3 WHERE grouper=1

我收到The target table t3 of the DELETE is not updatable

这个错误有什么解决方案吗？
基本上我正在尝试的是使用select in delete删除标有grouper的表行组。我也很高兴其他解决方案或想法与此不同。

sql fiddle：http://sqlfiddle.com/#!2/33820/2/0

编辑：感谢这里的答案是工作代码（如果有人需要类似的东西）：

SET @n=0;
delete from timelog where id in  ((SELECT  id
                    FROM (  
                            SELECT  id, project_id, task_id, user_id,
                            @n:=if(status=55,@n+1,@n),
                            if(status=55,@n-1,@n) as grouper FROM timelog
                            WHERE user_id='5' ORDER BY id ASC
                         ) as t
        where grouper>-1 and grouper=1
        group by grouper))

Answer 1

希望我有更多时间......但是快速的伪代码......

delete from timelog where id in  ((SELECT  id
                        FROM (  
                                SELECT  id, project_id, task_id, user_id,
                                @n:=if(status=55,@n+1,@n),
                                if(status=55,@n-1,@n) as grouper FROM timelog
                                WHERE user_id='5' ORDER BY id ASC
                             ) as t
            where grouper>-1
            group by grouper) as t3 WHERE grouper=1)

我所做的就是将subselect语句更改为where子句，该子句只返回原始子查询中列出的ID。

编辑 - 括号有点偏，我想我现在有了。说实话，这可以真正清理到一个选择语句，而不是这里的嵌套版本。

DELETE的目标表...不可更新

1 个答案: