SQL solves the problem of continuous login deformation holiday filtering

One 、 explain

Let's first talk about the effects that need to be achieved , It is to judge holidays and working days , If it's a weekday , Return to the current day , If it is a holiday or rest , Return to the next working date
06-01 Working day 06-01
06-02 Weekend 06-04 * because The next working day is 06-04
06-03 Weekend 06-04
06-07、06-08、06-09 The holiday season , return 06-10, That is, the next working day

There are two solutions explained before the continuous login days , It can solve the problem of continuous login , Then it is often used in actual production , But for the deformation of some such problems , Whether it can be easily solved ？

Two 、 Requirement specification

There's a demand recently , It is not demand , It is a small function implementation . The need to do event warning before , It is necessary to feed back the alarm events , Judge whether it is the feedback of the day .
At first, the idea was simple , Only considering the weekend , Set the alarm event if it is a weekend , Feedback on Monday is considered as feedback on the day , It was achieved in this way ：
If it's Saturday , Feedback date increased 2, If it's Sunday , Feedback date increased 1

case dayofweek(alarm_date)
	when 7 then date_add(fb_date,2)
	when 0 then date_add(fb_date,1)
else fb_date end as fb_date

But in practice , Found a checkpoint with holidays , Especially in the first half of the year, there are many holidays , Therefore, not filtering holidays has a great impact on the results , in addition , For some weekends , It may also be working days off .

3、 ... and 、 Filter holidays

Why is it said that this problem is a deformation of the continuous login problem ？ Let's think about it , Are holidays continuous ？ Whether it's weekends or holidays , Are consecutive dates . Then the question becomes , We need to distinguish which are holidays ？ Which are working days ？
I think of two ways to realize this

Train of thought

To write UDF, stay jar Maintain a file in the package , Documents can record the date of holidays , Write a custom one UDF Implement this function
This method can be implemented , But the process will be a little cumbersome , But the advantage is that it can be reused , Write once , Use everywhere

Train of thought two

Maintain a holiday dimension table , Through some sql Logic implementation , The advantage is that the implementation process is not cumbersome , The disadvantage is that reusability is not strong . For shortcomings , In fact, you can output the results to a dimension table for use , At the same time, some other dimensions can be added , Because there are many time and holiday dimensions in the production environment .

Four 、 Function realization

Adopt idea 2 to realize , Maintain a dimension table , Whether the date is a holiday is recorded in the dimension table , There are also two ways to implement

The way 1： Equal difference solution

Recall the solution of continuous login problem , Calculate equal difference , Find consecutive days , Sort in positive order , Returns the next day of consecutive days , Is the result of the need
The flow chart is as follows ：

date	Whether it's a holiday or not	Return value
06-01	1	06-03
06-02	1	06-03
06-03	0	06-03
06-04	1	06-05
06-05	0	06-05

1	select 
2	  log_date, 
3	  date_add(log_date, rn) next_work_day 
4	from ( 
5	select 
6	  log_date, 
7	  row_number() 
8	  over(partition by start_day order by log_date desc) rn 
9	from ( 
10	select 
11	  log_date, 
12	  date_sub(log_date, rn) start_day 
13	from( 
14	select 
15	  log_date, 
16	  row_number() 
17	  over(order by log_date) rn 
18	from ( 
19	select 
20	  log_date 
21	from 
22	  tmp_bdp.tmp_log_date 
23	) t1 
24	) t2 
25	) t3 
26	) t4 

The way 2： Subtraction before and after

It is also a way to simulate continuous login , Find the next day of consecutive days , Is the value to return

1	select 
2	  log_date, 
3	  max(log_date) 
4	  over(partition by flag order by log_date) next_day 
5	from ( 
6		select 
7		  log_date, 
8		  sum(if(diff_date > 1, 1, 0)) 
9		  over(order by log_date) flag 
10		from ( 
11			select  
12			  log_date, 
13			  datediff(log_date, lag_date) diff_date 
14			from ( 
15				select 
16				  `date` as log_date, 
17				  lag(`date`, 1, '1970-01-01')  
18				  over(order by `date`) lag_date 
19				from 
20				  bili_dim.dim_date_info_d 
21				where year(`date`) = 2022 and holiday_type <> 0 
22			) t1 
23		) t2 
24	) t3