Question

有此表

CREATE TABLE BOOKINGS
    ([RES_ID] varchar(4), [ATTENDANCE_DATE] datetime, [AUTOID] int);

INSERT INTO BOOKINGS
    ([RES_ID], [ATTENDANCE_DATE], [AUTOID])
VALUES
    ('A001', '2018-01-01 00:00:00', 1),
    ('A002', '2018-01-01 00:00:00', 2),
    ('A003', '2018-01-01 00:00:00', 3),
    ('A001', '2018-01-02 00:00:00', 4),
    ('A002', '2018-01-02 00:00:00', 5),
    ('A003', '2018-01-02 00:00:00', 6),
    ('A002', '2018-01-03 00:00:00', 7),
    ('A003', '2018-01-03 00:00:00', 8);

我想用id='A001'删除所有行，因为在最近的日期，没有保留（即已被取消）。

我已经尝试过了：

with cte as
(
  select *,
    row_number() over(partition by [res_id]
                      order by  [ATTENDANCE_DATE] desc) rn
  from BOOKINGS
)
DELETE FROM cte where rn > 1;

但这保留了'A001'（id = 4）的最新行，我不希望这样。

预期输出为：

A002    2018-01-03 00:00:00.000 7
A003    2018-01-03 00:00:00.000 8

Answer 1

一种方法是将每个resid的最新日期与总体上的最新日期进行比较。您可以使用窗口功能来做到这一点：

with todelete as (
      select b.*,
             max(attendance_date) over (partition by res_id) as max_ad_resid,
             max(attendance_date) over () as max_ad
      from bookings b
     )
delete from todelete
    where max_ad_resid < max_ad;

Answer 2

您可以像这样简单地删除

delete
from BOOKINGS
where ATTENDANCE_DATE != (
  select max(ATTENDANCE_DATE) from BOOKINGS
)

Answer 3

这是您要寻找的吗？

        DELETE FROM BOOKINGS WHERE
    [ATTENDANCE_DATE] <= (SELECT MAX([ATTENDANCE_DATE]) FROM BOOKINGS where RES_ID = 'A001')

这将获得RES_ID A001的最大日期（无保留），并删除该日期之前或之前的所有条目。

按日期仅保留最近的行

3 个答案: