我怎样才能 return 满足特定条件的行列表与它之前的行？

Question

precise_value_date          local_limit_amount
2020-12-10 18:45:37.650     1800000.000000
2020-12-18 19:01:09.530     3225000.000000
2020-12-31 00:00:00.000     1800000.000000
2021-01-18 00:00:00.000     2300000.000000
2021-04-27 00:00:00.000     1500000.000000
2021-07-22 00:00:00.000     500000.000000
2021-08-31 00:00:00.000     500000.000000
2021-12-31 00:00:00.000     0.000000

我想return今天日期之后的所有行以及最后一行意思是如果今天是 2020-12-24 00:00:00.000 我将收到该日期之后的所有数据以及该日期之前的一行所以获取的数据如下：

precise_value_date          local_limit_amount
2020-12-18 19:01:09.530     3225000.000000
2020-12-31 00:00:00.000     1800000.000000
2021-01-18 00:00:00.000     2300000.000000
2021-04-27 00:00:00.000     1500000.000000
2021-07-22 00:00:00.000     500000.000000
2021-08-31 00:00:00.000     500000.000000
2021-12-31 00:00:00.000     0.000000

如果今天的日期恰好在其中一个获取的数据中，我只会 return 那些 >= 它而不是最后一行意思是如果今天是 2021-04-27 00:00:00.000 则 table 将如下所示：

precise_value_date          local_limit_amount
2021-04-27 00:00:00.000     1500000.000000
2021-07-22 00:00:00.000     500000.000000
2021-08-31 00:00:00.000     500000.000000
2021-12-31 00:00:00.000     0.000000

谢谢！

Answer 1

您可以使用 lead():

select t.*
from (select t.*,
             lead(precise_value_date) over (order by precise_value_date) as next_precise_value_date
      from t
     ) t
where next_precise_value_date is null or
      next_precise_value_date >= curdate()
order by precise_value_date;

或者，您可以使用 union all:

select x.*
from ((select top (1) t.*
       from t
       where t.precise_value_date < convert(date, getdate())
       order by t.precise_value_date desc
      ) union all
      (select t.*
       from t
       where t.precise_value_date >= convert(date, getdate())
      )
     ) x;

Answer 2

这看起来像 union all:

select *
from (
    select * from mytable where precise_value_date < current_date
    union all
    (select * from mytable where precise_value_date >= current_date order by precise_value_date limit 1)
) t
order by precise_value_date

在MySQL中，甚至不需要外层的select：

select * from mytable where precise_value_date < current_date
union all
(select * from mytable where precise_value_date >= current_date order by precise_value_date limit 1)
order by precise_value_date

在 SQL 服务器中：

select *
from (
    select * from mytable where precise_value_date < convert(date, getdate())
    union all
    (select top (1) * from mytable where precise_value_date >= convert(date, getdate()) order by precise_value_date)
) t
order by precise_value_date

我怎样才能 return 满足特定条件的行列表与它之前的行？

How can I return the list of rows meeting a certain condition with the row just before it?

sql

sql-server

datetime

subquery