Count consecutive Workdays where employee has taken SICK leave

Question

I am trying to identify a way within sql to count the number of consecutive workdays (Monday - Friday) an employee has taken SICK leave. My problem in the past has been trying to count days whereas we cross the weekend. Currently I have 4 queries identifying patterns around the weekend that I run and export into an Excel Spreadsheet and manipulate from there. My goal is to be able to list an EMPLID (employee id), EMPL_RCD ( employee record), Min(DUR) (Date of absence), Max(DUR), and count of consecutive workdays.
Any help or guidance would will be appreciated. I have attached my current query.
I run my current query 4 times using the patterns below one at a time.
SELECT DISTINCT A.EMPLID,A.EMPL_RCD, A.DUR,b.dur,c.dur,d.dur, A.TL_QUANTITY, A.TRC
FROM PS_TL_RPTD_TIME A, PS_TL_RPTD_TIME B, PS_TL_RPTD_TIME C,PS_TL_RPTD_TIME D
WHERE A.EMPLID = B.EMPLID
AND A.EMPL_RCD = B.EMPL_RCD
AND A.EMPLID = C.EMPLID
AND A.EMPL_RCD = C.EMPL_RCD
AND A.EMPLID = D.EMPLID
AND A.EMPL_RCD = D.EMPL_RCD
AND B.EMPLID = C.EMPLID
AND B.EMPL_RCD = C.EMPL_RCD
AND B.TRC = C.TRC
AND A.TRC = B.TRC
AND A.TRC = C.TRC
AND A.TRC = D.TRC
AND (B.DUR = A.DUR+3 AND C.DUR = A.DUR+4 AND D.DUR = A.DUR+5) Friday, Monday, Tuesday, Wednesday
AND (B.DUR = A.DUR+1 AND C.DUR = A.DUR+4 AND D.DUR = A.DUR+5) Thursday, Friday, Monday, Tuesday
AND (B.DUR = A.DUR+1 AND C.DUR = A.DUR+2 AND D.DUR = A.DUR+5) Wednesday, Thursday, Friday, Monday
AND (B.DUR = A.DUR+1 AND C.DUR = A.DUR+2 AND D.DUR = A.DUR+3) -- Same workweek
AND A.TRC LIKE 'SICK%'
AND NOT EXISTS ( SELECT 'x' FROM PS_JOB J
WHERE J.EMPLID = A.EMPLID
AND J.EMPL_RCD = A.EMPL_RCD
AND J.EMPL_STATUS IN ('P','L')
AND J.EFFDT = ( SELECT MAX(J1.EFFDT) FROM PS_JOB J1
WHERE J1.EMPLID = J.EMPLID
AND J1.EMPL_RCD = J.EMPL_RCD
AND J1.EFFDT <= D.DUR))

Answer 1

You should consider a technique from data warehousing where you use a table to describe dates. To keep things simple, let's create a table that holds all your valid business days for the past 2 weeks...
create table business_dates (business_day date);
begin
insert into business_dates values (to_date('16-oct-2006'));
insert into business_dates values (to_date('17-oct-2006'));
insert into business_dates values (to_date('18-oct-2006'));
insert into business_dates values (to_date('19-oct-2006'));
insert into business_dates values (to_date('20-oct-2006'));
insert into business_dates values (to_date('23-oct-2006'));
insert into business_dates values (to_date('24-oct-2006'));
insert into business_dates values (to_date('25-oct-2006'));
insert into business_dates values (to_date('26-oct-2006'));
insert into business_dates values (to_date('27-oct-2006'));
insert into business_dates values (to_date('30-oct-2006'));
insert into business_dates values (to_date('31-oct-2006'));
insert into business_dates values (to_date('01-nov-2006'));
insert into business_dates values (to_date('02-nov-2006'));
insert into business_dates values (to_date('03-nov-2006'));
end;
now we create a table that shows whether each employee was sick or not for a given work day...
create table attendance (empid number, business_day date, sick char(1));
insert into attendance values (100, to_date('16-oct-2006'), 'N');
insert into attendance values (100, to_date('17-oct-2006'), 'Y');
insert into attendance values (100, to_date('18-oct-2006'), 'Y');
insert into attendance values (100, to_date('19-oct-2006'), 'N');
insert into attendance values (100, to_date('20-oct-2006'), 'N');
insert into attendance values (100, to_date('23-oct-2006'), 'Y');
insert into attendance values (100, to_date('24-oct-2006'), 'Y');
insert into attendance values (100, to_date('25-oct-2006'), 'Y');
insert into attendance values (100, to_date('26-oct-2006'), 'N');
insert into attendance values (100, to_date('27-oct-2006'), 'N');
insert into attendance values (100, to_date('30-oct-2006'), 'N');
insert into attendance values (100, to_date('31-oct-2006'), 'Y');
insert into attendance values (100, to_date('01-nov-2006'), 'N');
insert into attendance values (100, to_date('02-nov-2006'), 'N');
insert into attendance values (100, to_date('03-nov-2006'), 'N');
insert into attendance values (105, to_date('16-oct-2006'), 'Y');
insert into attendance values (105, to_date('17-oct-2006'), 'Y');
insert into attendance values (105, to_date('18-oct-2006'), 'N');
insert into attendance values (105, to_date('19-oct-2006'), 'N');
insert into attendance values (105, to_date('20-oct-2006'), 'Y');
insert into attendance values (105, to_date('23-oct-2006'), 'N');
insert into attendance values (105, to_date('24-oct-2006'), 'N');
insert into attendance values (105, to_date('25-oct-2006'), 'Y');
insert into attendance values (105, to_date('26-oct-2006'), 'Y');
insert into attendance values (105, to_date('27-oct-2006'), 'Y');
insert into attendance values (105, to_date('30-oct-2006'), 'Y');
insert into attendance values (105, to_date('31-oct-2006'), 'Y');
insert into attendance values (105, to_date('01-nov-2006'), 'N');
insert into attendance values (105, to_date('02-nov-2006'), 'N');
insert into attendance values (105, to_date('03-nov-2006'), 'Y');
Now the query to get each sick occurrence and the number of consecutive days for each employee is...
select empid, first_sick_day, sick_count from
(select empid,
first_value(business_day) over (partition by empid, groupno order by business_day) as first_sick_day,
     row_number() over (partition by empid, groupno order by business_day) as rn,
count(*) over (partition by empid, groupno) as sick_count
from
(select empid, business_day, daynum-rownum groupno
     from
          (SELECT a.empid, a.business_day, d.day_num as daynum
               FROM attendance a,
                    (select rownum as day_num, business_day
                    from (select business_day from business_dates order by business_day)) d
               WHERE sick = 'Y' AND a.business_day = d.business_day
               ORDER BY 1,2 )
where rn = 1;
The above query can be modified slightly to only give you the sick occurrence with the maximum number of consecutive days for each employee.
Having a separate date table is nice because you can take in account weekends, holidays or any other nonwork day by just removing that date from the table. Generating this table is easy as date dimension examples can be found on on the web, and the amount of rows is small (250 rows per year approx).
JR