Question

好的，所以我是派生表的新手。当我运行这个查询时，它给了我一个MYSQL错误1248.任何人都知道为此工作？这可能是我做过的最复杂的查询，我只想让它工作。谢谢！

delete from table_1
    where thing_id in (
        select tid from(
            select thing_id as tid
            FROM table_1
            inner join table_2
            on table_1.thing_id = table_2.thing_id
            where stuff = 'stuff'
            )
        )

Answer 1

MySQL通常不允许您在语句的其余部分引用要删除（或更新）的表。你可以使用嵌套的子查询解决这个问题（你似乎正在尝试这样做）。但是，我认为最好使用显式join来进行更新：

delete t1
    from table_1 t1 join
         (select t1.thing_id as tid
          from table_1 t1 inner join
               table_2 t2 
               on t1.thing_id = t2.thing_id
          where stuff = 'stuff'
        ) tt
        on t1.thing_id = tt.tid;

那就是说，我认为这相当于只在join上执行table2：

delete t1
    from table_1 t1 join
         table_2 t2 
         on t1.thing_id = t2.thing_id;
   where stuff = 'stuff';

后一种方法也应具有更好的性能。

Answer 2

您忘记了表格别名

delete from table_1
where thing_id in (
    select tid from(
        select table_1.thing_id as tid
        FROM table_1
        inner join table_2
        on table_1.thing_id = table_2.thing_id
        where table_1.stuff = 'stuff'
        ) tmp
    )

MYSQL错误1248我被卡住了

2 个答案: