Identificar valores no consecutivos siempre es un poco complicado e involucra varias subconsultas anidadas (al menos no puedo encontrar una mejor solución).
El primer paso es identificar valores no consecutivos para el año:
Paso 1) Identifica valores no consecutivos
select company,
profession,
year,
case
when row_number() over (partition by company, profession order by year) = 1 or
year - lag(year,1,year) over (partition by company, profession order by year) > 1 then 1
else 0
end as group_cnt
from qualification
Esto devuelve el siguiente resultado:
company | profession | year | group_cnt ---------+------------+------+----------- Google | Programmer | 2000 | 1 Google | Sales | 2000 | 1 Google | Sales | 2001 | 0 Google | Sales | 2002 | 0 Google | Sales | 2004 | 1 Mozilla | Sales | 2002 | 1
Ahora con el valor group_cnt podemos crear "ID de grupo" para cada grupo que tenga años consecutivos:
Paso 2) Definir ID de grupo
select company,
profession,
year,
sum(group_cnt) over (order by company, profession, year) as group_nr
from (
select company,
profession,
year,
case
when row_number() over (partition by company, profession order by year) = 1 or
year - lag(year,1,year) over (partition by company, profession order by year) > 1 then 1
else 0
end as group_cnt
from qualification
) t1
Esto devuelve el siguiente resultado:
company | profession | year | group_nr ---------+------------+------+---------- Google | Programmer | 2000 | 1 Google | Sales | 2000 | 2 Google | Sales | 2001 | 2 Google | Sales | 2002 | 2 Google | Sales | 2004 | 3 Mozilla | Sales | 2002 | 4 (6 rows)
Como puede ver, cada "grupo" tiene su propio group_nr y esto finalmente podemos usarlo para agregar agregando otra tabla derivada:
Paso 3) Consulta final
select company,
profession,
array_agg(year) as years
from (
select company,
profession,
year,
sum(group_cnt) over (order by company, profession, year) as group_nr
from (
select company,
profession,
year,
case
when row_number() over (partition by company, profession order by year) = 1 or
year - lag(year,1,year) over (partition by company, profession order by year) > 1 then 1
else 0
end as group_cnt
from qualification
) t1
) t2
group by company, profession, group_nr
order by company, profession, group_nr
Esto devuelve el siguiente resultado:
company | profession | years
---------+------------+------------------
Google | Programmer | {2000}
Google | Sales | {2000,2001,2002}
Google | Sales | {2004}
Mozilla | Sales | {2002}
(4 rows)
Que es exactamente lo que querías, si no me equivoco.
