mysql 删除日期大于特定日期的行

mysql delete row that have a date greater than a specific date

我有一个table这样的

----------------------
| idDoc | date       | 
----------------------
| 1     | 2018-01-20 |
| 1     | 2018-07-15 |
| 1     | 2017-07-31 |
| 1     | 2019-01-17 |
| 1     | 2019-07-30 |
| 1     | 2020-01-11 |
| 1     | 2020-07-31 |
| 1     | 2021-01-20 |
| 15    | 2018-11-31 |
| 15    | 2019-03-17 |
| 15    | 2018-05-31 |
| 15    | 2017-05-29 |
| 15    | 2019-09-20 |
| 15    | 2020-12-31 |
| 5     | 2018-01-31 |
| 5     | 2017-07-31 |
| 5     | 2018-04-23 |
| 5     | 2019-11-31 |
| 5     | 2019-12-08 |
----------------------

我希望(通过单个查询)变成这样:

----------------------
| idDoc | date       | 
----------------------
| 1     | 2017-07-31 |
| 15    | 2017-05-29 |
| 5     | 2017-07-31 |
----------------------

要获取的日期总是较旧的日期,要删除的字段将始终(具有相同的 ID)所有大于它的日期。

有什么建议吗?

如果你想删除数据库中的数据,试试这个:

delete t1
from demo t1
join demo t2
on t1.idDoc = t2.idDoc
and t1.`date` > t2.`date`;

请参阅此处 demo

如果你想select记录样例数据,试试这个:

select t1.*
from demo t1
join (select idDoc, min(`date`) minDate from demo group by idDoc) t2
on t1.idDoc = t2.idDoc
and t1.`date` = t2.`minDate`;

我会这样处理:

delete t
    from t join
         (select t2.idDoc, min(t2.date) as mindate
          from t t2
          group by t2.idDoc
         ) t2
         on t.idDoc = t2.idDoc
    where t.date > t2.mindate;

如果你可以指定更多,它并不完全清楚你想要什么,但你可以试试这个查询

select *
FROM mytable
WHERE 1 order by date asc group by idDoc

您可以使用 join 和 date greater than 来实现此目的,请参见 select 的示例:

select * from docData d
LEFT JOIN docMaster m on d.idDoc=m.idDoc
where d.dateDoc>'2018-01-20';

我在这里举例:http://rextester.com/VAOB18193