Question

不确定如何使用滞后功能获取数据。

我有此数据：

date                        value   id
---------------------------------------
2019-03-01 00:00:00.000     1       a
2019-02-28 00:00:00.000     2       a
2019-02-27 00:00:00.000     3       a
2019-02-26 00:00:00.000     4       a
2019-03-01 00:00:00.000     4       b
2019-02-28 00:00:00.000     3       b
2019-02-27 00:00:00.000     2       b
2019-02-26 00:00:00.000     1       b

我需要lag 1和lag2值，并且每个id仅需要1行

 id  value lag1 lag2
--------------------
 a    1     2    3
 b    4     3    2

我的查询是

select 
    id, date, value, 
    lag(value, 1) over (partition by id order by date),
    lag(value, 2) over (partition by id order by date) 
from 
    data;

但是我得到的行多于1行，因此每行的滞后都会被计算出来。我知道函数的行为方式如此，但无法获得所需的输出。使用了另一个查询

select top 1 
from 
    (select 
         id, date, value, 
         lag(value, 1) over (partition by id order by date),
         lag(value, 2) over (partition by id order by date)
     from 
         data)

这仅返回1行，但我有多个ID。我需要为每个ID限制1行。感谢您的帮助

Answer 1

您可以进行聚合：

select t.id, 
       max(case when seq = 1 then value end) as value1,
       max(case when seq = 2 then value end) as value2,
       max(case when seq = 3 then value end) as value3,
       max(case when seq = 4 then value end) as value4,
       max(case when seq = 5 then value end) as value5
from (select t.*,
             row_number() over (partition by id order by date desc) as seq
      from table t
     ) t
group by t.id;

Answer 2

lag()是一个窗口函数，因此它不会更改行数。您需要进行一些过滤。

怎么样？

select id, date, value, value_1, value_2
from (select id, date, value,
             lag(value, 1) over (partition by id order by 
date) as value_1,
             lag(value, 2) over (partition by id order by date) as value_2,
             row_number() over (partition by id order by date desc) as seqnum
      from data
     ) d
where seqnum = 1

无法在SQL Server中使用LAG功能检索数据

2 个答案: