CODEWITHSUNDEEP
sqlsql-serverrolling-sum
Lambo
Lambo

Reputation: 897

Rolling Sum when date is continuous

I'm trying to find how many days people have continuously worked in SQL. I'm thinking a rolling sum might be the solution but don't know how to work it out.

My sample data is 

| Employee | work_period |
| 1        | 2019-01-01  |
| 1        | 2019-01-02  |
| 1        | 2019-01-03  |
| 1        | 2019-01-04  |
| 1        | 2019-01-05  |
| 1        | 2019-01-10  |
| 1        | 2019-01-11  |
| 1        | 2019-01-12  |
| 2        | 2019-01-20  |
| 2        | 2019-01-22  |
| 2        | 2019-01-23  |
| 2        | 2019-01-24  |

The designated result should be

| Employee | work_period | Continuous Days |
| 1        | 2019-01-01  | 1               |
| 1        | 2019-01-02  | 2               |
| 1        | 2019-01-03  | 3               |
| 1        | 2019-01-04  | 4               |
| 1        | 2019-01-05  | 5               |
| 1        | 2019-01-10  | 1               |
| 1        | 2019-01-11  | 2               |
| 1        | 2019-01-12  | 3               |
| 2        | 2019-01-20  | 1               |
| 2        | 2019-01-22  | 1               |
| 2        | 2019-01-23  | 2               |
| 2        | 2019-01-24  | 3               |

If the days are not continuous, the continuous counting will re-start from 1.

Upvotes: 1

Views: 228

Answers (3)

Gordon Linoff
Gordon Linoff

Reputation: 1269623

This is similar to John's answer but a bit simpler.

You can identify groups of adjacent rows by subtracting a sequence of numbers -- the difference is constant. So:

select Employee, work_period,
       row_number9) over (partition by employee, grp order by work_period) as day_counter
      ,Cont_Days = row_number() over (partition by Employee,Grp Order by Work_Period)
from (select t.*,
             dateadd(day,
                     - row_number() over (partition by employee order by work_period),
                     work_period
                    ) as grp
      from t
     ) t;

Another interesting way to do this is to identify the rows where the "islands" start and then use datediff():

select t.*,
       datediff(day,
                max(case when island_start_flag = 1 then workperiod end) over (partition by employee order by workperiod),
                workperiod
               ) + 1 as days_counter
from (select t.*,
             (case when lag(workperiod) over (partition by employee order by workperiod) >= dateadd(day, -1, workperiod)
                   then 0 else 1
              end) as island_start_flag
      from t
     ) t;

Upvotes: 2

John Cappelletti
John Cappelletti

Reputation: 81930

Just another option ... Very similar to a Gaps-and-Islands, but without the final aggregation.

Example

Select Employee
      ,work_period
      ,Cont_Days = row_number() over (partition by Employee,Grp Order by Work_Period)
 From  (
        Select *
              ,Grp = datediff(day,'1900-01-01',work_period) - row_number() over (partition by Employee Order by Work_Period) 
          From YourTable
       ) A

Returns

Employee    work_period Cont_Days
1           2019-01-01  1
1           2019-01-02  2
1           2019-01-03  3
1           2019-01-04  4
1           2019-01-05  5
1           2019-01-10  1
1           2019-01-11  2
1           2019-01-12  3
2           2019-01-20  1
2           2019-01-22  1
2           2019-01-23  2
2           2019-01-24  3

Upvotes: 2

sticky bit
sticky bit

Reputation: 37472

You can first use lag() to check if the previous row (as sorted by work_period) per employee has exactly day lees then the current row. Use that in a CASE expression that returns 0 if the condition is true and 0 otherwise. Then use the windowed version of sum() to sum up the 0s and 1s per employee in the order of work_period. That gives you a number per group of continuous days for each employee. You can then use this group number to PARTITION BY additionally to the user in a windowed version of sum() adding 1 for each row in the partition ordered by work_period.

SELECT employee,
       work_period,
       sum(1) OVER (PARTITION BY employee,
                                 g
                    ORDER BY work_period) continuous_days
       FROM (SELECT employee,
                    work_period,
                    sum(c) OVER (PARTITION BY employee
                                 ORDER BY work_period) g
                    FROM (SELECT employee,
                                 work_period,
                                 CASE
                                   WHEN lag(work_period) OVER (PARTITION BY employee
                                                               ORDER BY work_period) = dateadd(day, -1, work_period) THEN
                                     0
                                   ELSE
                                     1
                                 END c
                                 FROM elbat) x) y;

db<>fiddle

Upvotes: 1

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