[英]MYSQL delete all rows, but the same query on select return only 3 rows
我无法理解为什么select和delete的相同查询具有不同的行为。
我需要删除除5个最新行之外的所有行。
我知道我的这个任务的解决方案并不好,但我的问题是为什么MySQL没有删除相同的行,返回select为同一个查询子句
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;
我们可以尝试在这里执行删除限制连接:
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;
这种反连接背后的想法是,我们将删除任何与前五个记录之一不匹配的记录,这些记录按created_at
列的顺序递减。
请注意,我们不能WHERE IN
此处使用WHERE IN
查询,因为MySQL将返回此版本尚不支持LIMIT
的可怕错误消息。
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obove值的所有行:
delete tbl
from tbl
where id <= @last_id_to_delete;
您可以将两个查询合并为一个。 使用WHERE子句中的子查询:
delete tbl
from tbl
where id <= (select id from(
select id
from tbl
order by id desc
limit 1
offset 5
)x);
(请注意,您需要将子查询结果包装到派生表中,以避免错误:“您无法在FROM子句中为更新指定目标表'tbl'。”
或者通过加入单行子查询:
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;
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