按时间戳合并事实记录

Question

我在 Microsoft SQL 服务器数据库中。我想合并员工的开始和结束时间。例如，给定以下简单的 table:

create table dbo.foo (
employee_key bigint,
effective_start_date date,
effective_end_date date,
account_name varchar(100));


insert into foo (employee_key,
effective_start_date,
effective_end_date,
account_name)
values (1
    ,'2017-01-01'
    ,'2017-01-31'
    ,'Google')
,(1
    ,'2017-02-01'
    ,'2017-02-28'
    ,'Apple')
,(1
    ,'2017-03-01'
    ,'2017-03-31'
    ,'Google')
,(1
    ,'2017-04-01'
    ,'9999-12-31'
    ,'Google')

该员工已多次移动帐户。我想在 2 月保持更改并再次返回。但是，我不想在最后看到 2 条记录。相反，我希望从 2017-03-01 到 9999-12-31 Google 到运行的任期。

顺便说一句，这是一个例子。也就是说，"broken" 记录并不总是在员工任期结束时。

预期结果：

employee_key | effective_start_date | effective_end_date | account_name
1            | 2017-01-01           | 2017-01-31         | Google
1            | 2017-02-01           | 2017-02-28         | Apple
1            | 2017-03-01           | 9999-12-31         | Google

Answer 1

你可以试试这个SQL

-- Table to group by
SELECT Row_number() 
         OVER ( 
           ORDER BY employee_key, effective_start_date, effective_end_date) RN, 
       *, 
       Cast(NULL AS BIGINT) 
       GroupBy 
INTO   #tmp 
FROM   #foo 

-- Variables 
DECLARE @GroupBy BIGINT = 0 
DECLARE @lastEmployee_key BIGINT 
DECLARE @lastAccount_Name VARCHAR(100) 


--- Creating the data group 
UPDATE #tmp 
SET    @GroupBy = groupby = CASE 
                              WHEN employee_key = Isnull(@lastEmployee_key, -1) 
                                   AND account_name = Isnull(@lastAccount_Name, 
                                                      '') 
                                   THEN @GroupBy 
                              ELSE @GroupBy + 1 
                            END, 
       @lastEmployee_key = employee_key = employee_key, 
       @lastAccount_Name = account_name = account_name 


--- Output 
SELECT employee_key, 
       Min(effective_start_date) effective_start_date, 
       Max(effective_end_date)   effective_end_date, 
       account_name 
FROM   #tmp 
GROUP  BY groupby, 
          employee_key, 
          account_name

结果是

employee_key | effective_start_date|effective_end_date | account_name
-------------| --------------------|------------------ | -------------
1            | 2017-01-01          |2017-01-31         | Google
1            | 2017-02-01          |2017-02-28         | Apple
1            | 2017-03-01          |9999-12-31         | Google

Answer 2

您可以使用 row_number 并获得此结果...我使用了两个子查询，但没有它也可以解决：

;With Cte as (
select *, Bucket = sum(accn) over(partition by employee_key order by effective_start_date) 
    from (
        Select *,case when account_name <> lag(account_name) over(partition by employee_key order by effective_start_date) 
            then 1 else 0 end as AccN
            from foo ) A
    )
    Select top (1) with ties employee_key, Min(effective_start_date) over(partition by employee_key,Bucket) AS [effective_start_date],      
                Max(effective_end_date) over(partition by employee_key,Bucket) AS [effective_end_date], account_name
    from cte
    order by row_number() over(partition by employee_key, bucket order by effective_start_date)

输出如下：

+--------------+----------------------+--------------------+--------------+
| employee_key | Effective_start_Date | Effective_End_date | account_name |
+--------------+----------------------+--------------------+--------------+
|            1 | 2017-01-01           | 2017-01-31         | Google       |
|            1 | 2017-02-01           | 2017-02-28         | Apple        |
|            1 | 2017-03-01           | 9999-12-31         | Google       |
+--------------+----------------------+--------------------+--------------+

Demo

按时间戳合并事实记录

Consolidating fact records by timestamp

sql-server

dirty-data

fact