MYSQL 删除所有行，但 select return 上的相同查询只有 3 行

Question

我不明白为什么 select 和删除的相同查询有不同的行为。
我需要删除除 5 个最新行之外的所有行。
我知道我对这个任务的解决方案不好，但我的问题是为什么 MySQL 不删除相同的行，即 return select 对于相同的查询子句

查看代码

drop table if exists tbl;
create table tbl
(
    id         serial,
    cal        date COMMENT 'some column',
    created_at datetime default NOW()
);

insert into tbl
values
       (default, '2018-07-15', '2018-07-15 12:00'),
       (default, '2018-07-16', '2018-07-16 12:00'),
       (default, '2018-07-17', '2018-07-17 12:00'),
       (default, '2018-07-18', '2018-07-18 12:00'),
       (default, '2018-08-01', '2018-08-01 12:00'),
       (default, '2018-08-04', '2018-08-04 12:00'),
       (default, '2018-08-16', '2018-08-16 12:00'),
       (default, '2018-08-17', '2018-08-17 12:00');

  select *
  from tbl;

    #     +----+------------+---------------------+
    #     | id | cal        | created_at          |
    #     +----+------------+---------------------+
    #     | 1  | 2018-07-15 | 2018-07-15 12:00:00 |
    #     | 2  | 2018-07-16 | 2018-07-16 12:00:00 |
    #     | 3  | 2018-07-17 | 2018-07-17 12:00:00 |
    #     | 4  | 2018-07-18 | 2018-07-18 12:00:00 |
    #     | 5  | 2018-08-01 | 2018-08-01 12:00:00 |
    #     | 6  | 2018-08-04 | 2018-08-04 12:00:00 |
    #     | 7  | 2018-08-16 | 2018-08-16 12:00:00 |
    #     | 8  | 2018-08-17 | 2018-08-17 12:00:00 |
    #     +----+------------+---------------------+

现在我需要删除 ID 为 1,2,3 的行

SET @row_number = 0;
select *
from tbl
where tbl.id in (
    select T.id
    from (SELECT (@row_number := @row_number + 1) as num, tbl.id
          from tbl
          order by created_at desc
         ) as T
    where T.num > 5);


# +----+------------+---------------------+
# | id | cal        | created_at          |
# +----+------------+---------------------+
# | 3  | 2018-07-17 | 2018-07-17 12:00:00 |
# | 2  | 2018-07-16 | 2018-07-16 12:00:00 |
# | 1  | 2018-07-15 | 2018-07-15 12:00:00 |
# +----+------------+---------------------+

现在我使用删除操作

SET @row_number = 0;
delete
from tbl
where tbl.id in (
    select T.id
    from (SELECT (@row_number := @row_number + 1) as num, tbl.id
          from tbl
          order by created_at desc
         ) as T
    where T.num > 5);

select * from tbl; # <-- result empty
# +----+-----+------------+
# | id | cal | created_at |
# +----+-----+------------+

我哭了；

Answer 1

我们可以尝试在此处进行删除限制连接：

DELETE t1
FROM tbl t1
LEFT JOIN
(
    SELECT id
    FROM tbl
    ORDER BY created_at DESC
    LIMIT 5
) t2
    ON t1.id = t2.id
WHERE
    t2.id IS NULL;

这个 anti-join 背后的想法是，我们将删除不匹配前五个记录之一的任何记录，按 [= 降序排列11=]列。

请注意，我们不能在此处使用 WHERE IN 查询，因为 MySQL 会 return 可怕的错误消息，即此版本尚不支持 LIMIT .

Answer 2

使用LIMIT和OFFSET获取要删除的最高ID:

set @last_id_to_delete = (
  select id
  from tbl
  order by id desc
  limit 1
  offset 5
);

然后删除ID等于或小于以上值的所有行：

delete tbl
from tbl
where id <= @last_id_to_delete;

db-fiddle

您可以将两个查询合二为一。在 WHERE 子句中使用子查询：

delete tbl
from tbl
where id <= (select id from(
  select id
  from tbl
  order by id desc
  limit 1
  offset 5
)x);

(注意需要将子查询结果包装成派生的table，避免报错："You can't specify target table 'tbl' for update in FROM clause"。)

db-fidle

或通过加入 single-row 子查询：

delete t
from tbl t
join (
  select id as last_id_to_delete
  from tbl
  order by id desc
  limit 1
  offset 5
) x on t.id <= x.last_id_to_delete;

db-fidle

MYSQL 删除所有行，但 select return 上的相同查询只有 3 行

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