Question

我正在尝试从我的数据库中获取最新记录。该表如下所示。

Ticket_id|  date|   comments
EWU-752-84170|  4/10/2015 13:26|bla
HCX-943-86125|  4/10/2015 13:39|    ola
IKW-626-96314|  4/10/2015 13:42|    jkj
EWU-752-84170|  4/10/2015 13:28|    blo
EWU-752-84170|  4/10/2015 13:37|    ala
HCX-943-86125|  4/10/2015 15:11|    kbdkj
EWU-752-84170|  4/10/2015 13:43|    cla

并且输出应该发送最近的记录，如下所示。

Ticket_id|  date|   comments
EWU-752-84170|  4/10/2015 13:43|    cla
HCX-943-86125|  4/10/2015 15:11|    kbdkj
IKW-626-96314|  4/10/2015 13:42|    jkj

Answer 1

ORDER BY date DESC limit 1

这应该可以正常工作。

Answer 2

SELECT t1.*
FROM
  mytable t1 LEFT JOIN mytable t2
  ON
    t1.Ticket_id=t2.Ticket_id
    AND t1.date<t2.date
WHERE
  t2.date IS NULL

Answer 3

这将选择所有票证，其中不存在具有较晚日期的另一张票证（因此它仅选择最新票证）

select * from mytable t1
where not exists (
    select 1 from mytable t2
    where t2.ticket_id = t1.ticket_id
    and t2.date > t1.date
)

Answer 4

    select * from dailyhandover t1
where not exists (
    select * from dailyhandover t2
    where t2.ticket_id = t1.ticket_id
    and t2.date > t1.date
) and customer_name='bata'

从MySQL获取最新记录 - 这有许多重复记录

4 个答案: