SQL getting status of a period - sql

Question

I'm looking for a SQL solution for the following problem.
I want a list of employees who are more then 14 days sick in a row.
I've a sql table with the following:
First_name, Last_Name, INDIRECT_ID, SHIFT_DATE
John, Doe, Sick, 2016-01-01
John, Doe, Sick, 2016-01-02
John, Doe, working, 2016-01-03
John, Doe, Sick, 2016-01-04
John, Doe, Sick, 2016-01-05
etc.
I thought to do this by seeing if they are sick for 10x (2x 5 working days) in two weeks. But maybe there is a much simpler solution for it. But Now I'm also getting duplicate answers.
select FIRST_NAME, LAST_NAME
from (select t.*
,(select count(*)
from LABOR_TICKET t2
where t2.EMPLOYEE_ID = t.EMPLOYEE_ID and
t2.INDIRECT_ID = t.INDIRECT_ID and
t2.SHIFT_DATE >= t.SHIFT_DATE and
t2.SHIFT_DATE < DATEADD(day, 14, t.SHIFT_DATE)) NumWithin14Days
from LABOR_TICKET t
where SHIFT_DATE between '2016-01-01' and '2016-04-01'
) LABOR_TICKET
INNER JOIN
EMPLOYEE ON LABOR_TICKET.EMPLOYEE_ID = EMPLOYEE.ID
where NumWithin14Days >= 10 AND INDIRECT_ID = 'SICK'

Answer 1

Try this,
First create all the 14 days intervals in between the From Date and To Date.
Then check the count of the 'Sick' is 14 in each interval for every employee.
DECLARE #ST_DATE DATE='2016-01-01'
,#ED_DATE DATE='2016-04-01'
;WITH CTE_DATE AS (
SELECT #ST_DATE AS ST_DATE,DATEADD(DAY,13,#ST_DATE) AS ED_DATE
UNION ALL
SELECT DATEADD(DAY,1,ED_DATE),DATEADD(DAY,14,ED_DATE)
FROM CTE_DATE
WHERE DATEADD(DAY,14,ED_DATE) <= #ED_DATE
)
SELECT FIRST_NAME, LAST_NAME
FROM CTE_DATE
INNER JOIN LABOR_TICKET ON SHIFT_DATE BETWEEN ST_DATE AND ED_DATE
WHERE INDIRECT_ID = 'Sick'
GROUP BY FIRST_NAME, LAST_NAME
HAVING COUNT(*) >= 14

Answer 2

Pseudo code to give you idea for all employees
if you have a calendar table like below
create table dates
(
datetime date
)
insert into dates
select '2016-01-01'
union all
select '2016-01-02'
Now you can left join this with your main table like
select
mt.firstname,dt.date,count(indirect_id)
from
datestable dt
left join
maintable mt
on mt.date=dt.date
and mt.indirect_id='sick'
group by mt.firstname,dt.date
having count(indirect_id)>=14
order by dt.date

Answer 3

you should have thrown more sample data.
try this,(I am sure it will work with other sample data)few thing are just there to filter data.
declare #t table(First_name varchar(50), Last_Name varchar(50), INDIRECT_ID varchar(50), SHIFT_DATE date)
insert into #t values
('John', 'Doe', 'Sick', '2016-01-01')
,('John', 'Doe', 'Sick', '2016-01-02')
,('John','Doe','working','2016-01-03')
,('John', 'Doe', 'Sick', '2016-01-04')
,('John', 'Doe', 'Sick', '2016-01-05')
declare #name varchar(50)='John'
declare #month int=1
;With CTE as
(
select top 1 First_name,Last_Name,SHIFT_DATE,1 rn from #T where First_name=#name
and INDIRECT_ID='Sick' order by SHIFT_DATE
union all
select t.First_name,t.Last_Name,t.SHIFT_DATE, rn+1 from #T t
inner join cte c on t.First_name=c.First_name
where INDIRECT_ID='Sick'
and t.SHIFT_DATE=DATEADD(day,1,c.SHIFT_DATE)
and t.SHIFT_DATE<='2016-01-31'
)
select * from CTE where rn>=14

Answer 4

declare #t table(First_name varchar(50), Last_Name varchar(50), INDIRECT_ID varchar(50), SHIFT_DATE date)
insert into #t values
('John', 'Doe', 'Sick', '2016-01-01')
,('John', 'Doe', 'Sick', '2016-01-02')
,('John','Doe','working','2016-01-03')
,('John', 'Doe', 'Sick', '2016-04-04')
,('John', 'Doe', 'Sick', '2016-05-05')
select s.*
,u.*
,Sickdays =
case
when s.indirect_id = 'Sick' and u.indirect_id = 'Sick' then datediff(dd,u.shift_date,s.shift_date)
else 0
end
from
(
select t.*,
row_number() over(partition by last_name,first_name order by shift_date desc) rn
from #t t
) s
join
(select t.*,
row_number() over(partition by last_name,first_name order by shift_date desc) rn
from #t t
) u on s.last_name = u.last_name and s.first_name = u.first_name and s.rn = u.rn - 1
where
case
when s.indirect_id = 'Sick' and u.indirect_id = 'Sick' then datediff(dd,u.shift_date,s.shift_date)
else 0
end > 13