使用 LEAD 函数筛选为连续值的一个实例的连续行的日期差异

Question

基本上我需要的结果是每个项目的价格变化范围，我需要提取项目价格和交易日期，结束日期设置为下一次项目价格变化。

鉴于此 table

create table myTable(id int, Price decimal(10,6), StartDate datetime)
insert into myTable
select 1, 92.576842, '2018-04-06 23:00:00.000' union all
select 1, 92.700000, '2018-04-12 21:39:00.000' union all
select 1, 92.700000, '2018-04-26 00:01:00.000' union all
select 1, 92.700000, '2018-06-04 23:20:00.000' union all
select 1, 116.700000, '2018-07-04 21:38:00.000' union all
select 1, 116.700000, '2018-08-01 22:31:31.000' union all
select 1, 118.610597, '2018-08-13 23:34:22.000' union all
select 1, 116.700000, '2018-09-05 23:52:28.000'

和这个查询

select id, Price, StartDate, LEAD(StartDate) over (partition by id order by id, StartDate) endDate
from myTable
order by id, StartDate

查询结果：

id  Price       StartDate               endDate
1   92.576842   2018-04-06 23:00:00.000 2018-04-12 21:39:00.000
1   92.700000   2018-04-12 21:39:00.000 2018-04-26 00:01:00.000
1   92.700000   2018-04-26 00:01:00.000 2018-06-04 23:20:00.000
1   92.700000   2018-06-04 23:20:00.000 2018-07-04 21:38:00.000
1   116.700000  2018-07-04 21:38:00.000 2018-08-01 22:31:31.000
1   116.700000  2018-08-01 22:31:31.000 2018-08-13 23:34:22.000
1   118.610597  2018-08-13 23:34:22.000 2018-09-05 23:52:28.000
1   116.700000  2018-09-05 23:52:28.000 NULL

如何只获取连续 Price 个实例的最早日期

结果应如下所示 - 请注意最后一行是重复的 Price 但它是必需的

id  Price       StartDate               EndDate
1   92.576842   2018-04-06 23:00:00.000 2018-04-12 21:39:00.000
1   92.700000   2018-04-12 21:39:00.000 2018-07-04 21:38:00.000
1   116.700000  2018-07-04 21:38:00.000 2018-08-13 23:34:22.000
1   118.610597  2018-08-13 23:34:22.000 2018-09-05 23:52:28.000
1   116.700000  2018-09-05 23:52:28.000 NULL

Answer 1

这是 "islands" 的一种方法，它采用 2 row_number() 计算，为 "island" 中的所有行提供一个共享 ID，然后可以按该 ID 进行分组：

select
    id, Price, group_id, min(StartDate) StartDate, max(enddate) enddate
from (
        select id, Price, StartDate
            , LEAD(StartDate,1,DATEADD(YEAR,1,StartDate)) over (partition by id order by id,StartDate) endDate
            , row_number() over(partition by id order by StartDate ASC) 
              - row_number() over(partition by id, price order by StartDate ASC) AS group_id
        from myTable
      ) d
group by
    id, Price, group_id
order by
    id,StartDate
;

| id |      Price | group_id |            StartDate |              enddate |
|----|------------|----------|----------------------|----------------------|
|  1 |  92.576842 |        0 | 2018-04-06T23:00:00Z | 2018-04-12T21:39:00Z |
|  1 |       92.7 |        1 | 2018-04-12T21:39:00Z | 2018-07-04T21:38:00Z |
|  1 |      116.7 |        4 | 2018-07-04T21:38:00Z | 2018-08-13T23:34:22Z |
|  1 | 118.610597 |        6 | 2018-08-13T23:34:22Z | 2018-09-05T23:52:28Z |
|  1 |      116.7 |        5 | 2018-09-05T23:52:28Z | 2019-09-05T23:52:28Z |

http://sqlfiddle.com/#!18/293d1/6

如果您希望 NULL 作为最后一个结束日期，请更改 lead() 使其不提供默认值

LEAD(StartDate,1)) over (partition by id order by id,StartDate) endDate

使用 LEAD 函数筛选为连续值的一个实例的连续行的日期差异

Date difference over consecutive rows filtered to one instance of consecutive values using LEAD function

lead

data-partitioning

sql-server-2014