SQL Consecutive Date Cumulative Count - sql

Question

I am doing some roster analysis and need to identify when an employee has worked for 5 or more consecutive days. In my table, I can extract data something like the below (note, there are lot more columns, this is just a cut down example):
Emp
Start
First_Entry
1234
23/06/2016
1
1234
24/06/2016
1
1234
24/06/2016
0
1234
25/06/2016
1
1234
26/06/2016
1
1234
27/06/2016
1
1234
28/06/2016
1
1234
29/06/2016
1
1234
29/06/2016
0
1234
30/06/2016
1
1234
2/07/2016
1
1234
3/07/2016
1
1234
3/07/2016
0
1234
4/07/2016
1
1234
4/07/2016
0
1234
5/07/2016
1
1234
6/07/2016
1
1234
9/07/2016
1
1234
10/07/2016
1
1234
11/07/2016
1
1234
12/07/2016
1
And what I am after is something like this:
Emp
Start
First_Entry
Consecutive_Days
Over_5
Status
1234
23/06/2016
1
1
0
Worked < 5
1234
24/06/2016
1
2
0
Worked < 5
1234
24/06/2016
0
2
0
Worked < 5
1234
25/06/2016
1
3
0
Worked < 5
1234
26/06/2016
1
4
0
Worked < 5
1234
27/06/2016
1
5
1
Worked >= 5
1234
28/06/2016
1
6
1
Worked >= 5
1234
29/06/2016
1
7
1
Worked >= 5
1234
29/06/2016
0
7
1
Worked >= 5
1234
30/06/2016
1
8
1
Worked >= 5
1234
02/07/2016
1
1
0
Worked < 5
1234
03/07/2016
1
2
0
Worked < 5
1234
03/07/2016
0
2
0
Worked < 5
1234
04/07/2016
1
3
0
Worked < 5
1234
04/07/2016
0
3
0
Worked < 5
1234
05/07/2016
1
4
0
Worked < 5
1234
06/07/2016
1
5
1
Worked >= 5
1234
09/07/2016
1
1
0
Worked < 5
1234
10/07/2016
1
2
0
Worked < 5
1234
11/07/2016
1
3
0
Worked < 5
1234
12/07/2016
1
4
0
Worked < 5
I'm really not sure how to go about getting the cumulative count for consecutive days, so any help you can give will be amazing

Answer 1

Probably someone would come up with a brilliant solution but this would do. Your problem looks like an "Gaps and Islands" problem. Finding islands of date ranges we can find out the rest easily. In the below SQL, #mindate is not a must, but makes it easier.
CREATE TABLE #temptable
(
[Emp] CHAR(4),
[startDate] DATE,
[First_Entry] BIT
);
INSERT INTO #temptable
(
[Emp],
[startDate],
[First_Entry]
)
VALUES
('1234', N'2016-06-23', 1),
('1234', N'2016-06-24', 1),
('1234', N'2016-06-24', 0),
('1234', N'2016-06-25', 1),
('1234', N'2016-06-26', 1),
('1234', N'2016-06-27', 1),
('1234', N'2016-06-28', 1),
('1234', N'2016-06-29', 1),
('1234', N'2016-06-29', 0),
('1234', N'2016-06-30', 1),
('1234', N'2016-07-02', 1),
('1234', N'2016-07-03', 1),
('1234', N'2016-07-03', 0),
('1234', N'2016-07-04', 1),
('1234', N'2016-07-04', 0),
('1234', N'2016-07-05', 1),
('1234', N'2016-07-06', 1),
('1234', N'2016-07-09', 1),
('1234', N'2016-07-10', 1),
('1234', N'2016-07-11', 1),
('1234', N'2016-07-12', 1);
DECLARE #minDate DATE;
SELECT #minDate = DATEADD(d, -1, MIN(startDate))
FROM #temptable;
WITH firstOnly
AS (SELECT *
FROM #temptable
WHERE First_Entry = 1),
grouper (emp, startDate, grp)
AS (SELECT Emp,
startDate,
DATEDIFF(d, #minDate, startDate) - ROW_NUMBER() OVER (PARTITION BY Emp ORDER BY startDate)
FROM firstOnly),
islands (emp, START, [end])
AS (SELECT emp,
MIN(startDate),
MAX(startDate)
FROM grouper
GROUP BY emp,
grp),
consecutives (emp, startDate, consecutive_days)
AS (SELECT f.Emp,
f.startDate,
-- i.START,
-- i.[end],
ROW_NUMBER() OVER (PARTITION BY f.Emp, i.START ORDER BY i.START)
FROM firstOnly f
INNER JOIN islands i
ON f.startDate
BETWEEN i.START AND i.[end])
SELECT t.Emp,
t.startDate,
t.First_Entry,
c.consecutive_days,
CAST(CASE
WHEN c.consecutive_days < 5 THEN
0
ELSE
1
END AS BIT) Over_5,
CASE
WHEN c.consecutive_days < 5 THEN
'Worked < 5'
ELSE
'Worked >= 5'
END [Status]
FROM consecutives c
INNER JOIN #temptable t
ON t.Emp = c.emp
AND t.startDate = c.startDate;
DROP TABLE #temptable;

Answer 2

This is a island and gap problem, You can try to use LAG window function to get the previous startDate row for each Emp, ten use SUM window function to calculate which days are continuous.
Finally, We can use CASE WHEN expression to judge whether the day is greater than 5.
;WITH CTE AS (
SELECT [Emp],
[startDate],
[First_Entry],
SUM(CASE WHEN DATEDIFF(dd,f_Dt,startDate) <= 1 THEN 0 ELSE 1 END) OVER(PARTITION BY Emp ORDER BY startDate) grp
FROM (
SELECT *,
LAG(startDate,1,startDate) OVER(PARTITION BY Emp ORDER BY startDate) f_Dt
FROM T
) t1
)
SELECT [Emp],
[startDate],
[First_Entry],
SUM(CASE WHEN First_Entry = 1 THEN 1 ELSE 0 END) OVER(PARTITION BY Emp,grp ORDER BY startDate) Consecutive_Days,
(CASE WHEN SUM(CASE WHEN First_Entry = 1 THEN 1 ELSE 0 END) OVER(PARTITION BY Emp,grp ORDER BY startDate) >= 5 THEN 1 ELSE 0 END) Over_5,
(CASE WHEN SUM(CASE WHEN First_Entry = 1 THEN 1 ELSE 0 END) OVER(PARTITION BY Emp,grp ORDER BY startDate) >= 5 THEN 'Worked >= 5' ELSE 'Worked < 5' END) Status
FROM CTE
sqlfiddle