EDITAR Aquí hay una versión de SQL Server de la consulta:
with LowerBound as (select second_day.EmployeeId
, second_day."DATE" as LowerDate
, row_number() over (partition by second_day.EmployeeId
order by second_day."DATE") as RN
from T second_day
left outer join T first_day
on first_day.EmployeeId = second_day.EmployeeId
and first_day."DATE" = dateadd(day, -1, second_day."DATE")
and first_day.IsPresent = 1
where first_day.EmployeeId is null
and second_day.IsPresent = 1)
, UpperBound as (select first_day.EmployeeId
, first_day."DATE" as UpperDate
, row_number() over (partition by first_day.EmployeeId
order by first_day."DATE") as RN
from T first_day
left outer join T second_day
on first_day.EmployeeId = second_day.EmployeeId
and first_day."DATE" = dateadd(day, -1, second_day."DATE")
and second_day.IsPresent = 1
where second_day.EmployeeId is null
and first_day.IsPresent = 1)
select LB.EmployeeID, max(datediff(day, LowerDate, UpperDate) + 1) as LongestStreak
from LowerBound LB
inner join UpperBound UB
on LB.EmployeeId = UB.EmployeeId
and LB.RN = UB.RN
group by LB.EmployeeId
SQL Server versión de los datos de prueba:
create table T (EmployeeId int
, "DATE" date not null
, IsPresent bit not null
, constraint T_PK primary key (EmployeeId, "DATE")
)
insert into T values (1, '2000-01-01', 1);
insert into T values (2, '2000-01-01', 0);
insert into T values (3, '2000-01-01', 0);
insert into T values (3, '2000-01-02', 1);
insert into T values (3, '2000-01-03', 1);
insert into T values (3, '2000-01-04', 0);
insert into T values (3, '2000-01-05', 1);
insert into T values (3, '2000-01-06', 1);
insert into T values (3, '2000-01-07', 0);
insert into T values (4, '2000-01-01', 0);
insert into T values (4, '2000-01-02', 1);
insert into T values (4, '2000-01-03', 1);
insert into T values (4, '2000-01-04', 1);
insert into T values (4, '2000-01-05', 1);
insert into T values (4, '2000-01-06', 1);
insert into T values (4, '2000-01-07', 0);
insert into T values (5, '2000-01-01', 0);
insert into T values (5, '2000-01-02', 1);
insert into T values (5, '2000-01-03', 0);
insert into T values (5, '2000-01-04', 1);
insert into T values (5, '2000-01-05', 1);
insert into T values (5, '2000-01-06', 1);
insert into T values (5, '2000-01-07', 0);
Lo sentimos, este se escribe en Oracle, para sustituir la adecuada Aritmética de fechas de SQL Server.
Supuestos:
- fecha es o bien un valor de fecha o DateTime con el componente de tiempo de 00:00:00.
- La clave principal es
(EmployeeId, Date)
- Todos los campos son
not null
Si una fecha es que falta para el empleado, que eran no presente. (Se utiliza para manejar el comienzo y el final de la serie de datos, sino que también significa que faltan fechas en el medio se romperá rayas Podría ser un problema dependiendo de los requerimientos
with LowerBound as (select second_day.EmployeeId
, second_day."DATE" as LowerDate
, row_number() over (partition by second_day.EmployeeId
order by second_day."DATE") as RN
from T second_day
left outer join T first_day
on first_day.EmployeeId = second_day.EmployeeId
and first_day."DATE" = second_day."DATE" - 1
and first_day.IsPresent = 1
where first_day.EmployeeId is null
and second_day.IsPresent = 1)
, UpperBound as (select first_day.EmployeeId
, first_day."DATE" as UpperDate
, row_number() over (partition by first_day.EmployeeId
order by first_day."DATE") as RN
from T first_day
left outer join T second_day
on first_day.EmployeeId = second_day.EmployeeId
and first_day."DATE" = second_day."DATE" - 1
and second_day.IsPresent = 1
where second_day.EmployeeId is null
and first_day.IsPresent = 1)
select LB.EmployeeID, max(UpperDate - LowerDate + 1) as LongestStreak
from LowerBound LB
inner join UpperBound UB
on LB.EmployeeId = UB.EmployeeId
and LB.RN = UB.RN
group by LB.EmployeeId
de datos de prueba:..
create table T (EmployeeId number(38)
, "DATE" date not null check ("DATE" = trunc("DATE"))
, IsPresent number not null check (IsPresent in (0, 1))
, constraint T_PK primary key (EmployeeId, "DATE")
)
/
insert into T values (1, to_date('2000-01-01', 'YYYY-MM-DD'), 1);
insert into T values (2, to_date('2000-01-01', 'YYYY-MM-DD'), 0);
insert into T values (3, to_date('2000-01-01', 'YYYY-MM-DD'), 0);
insert into T values (3, to_date('2000-01-02', 'YYYY-MM-DD'), 1);
insert into T values (3, to_date('2000-01-03', 'YYYY-MM-DD'), 1);
insert into T values (3, to_date('2000-01-04', 'YYYY-MM-DD'), 0);
insert into T values (3, to_date('2000-01-05', 'YYYY-MM-DD'), 1);
insert into T values (3, to_date('2000-01-06', 'YYYY-MM-DD'), 1);
insert into T values (3, to_date('2000-01-07', 'YYYY-MM-DD'), 0);
insert into T values (4, to_date('2000-01-01', 'YYYY-MM-DD'), 0);
insert into T values (4, to_date('2000-01-02', 'YYYY-MM-DD'), 1);
insert into T values (4, to_date('2000-01-03', 'YYYY-MM-DD'), 1);
insert into T values (4, to_date('2000-01-04', 'YYYY-MM-DD'), 1);
insert into T values (4, to_date('2000-01-05', 'YYYY-MM-DD'), 1);
insert into T values (4, to_date('2000-01-06', 'YYYY-MM-DD'), 1);
insert into T values (4, to_date('2000-01-07', 'YYYY-MM-DD'), 0);
insert into T values (5, to_date('2000-01-01', 'YYYY-MM-DD'), 0);
insert into T values (5, to_date('2000-01-02', 'YYYY-MM-DD'), 1);
insert into T values (5, to_date('2000-01-03', 'YYYY-MM-DD'), 0);
insert into T values (5, to_date('2000-01-04', 'YYYY-MM-DD'), 1);
insert into T values (5, to_date('2000-01-05', 'YYYY-MM-DD'), 1);
insert into T values (5, to_date('2000-01-06', 'YYYY-MM-DD'), 1);
insert into T values (5, to_date('2000-01-07', 'YYYY-MM-DD'), 0);
Por alguna razón, no pude crear la tabla temporal ... me muestra el error de un objeto desconocido ... así que utilicé una declaración select into ... – Vishal