ГРУПИРАНЕ ПО последователни дати, разделени с пропуски

create table t ("date" date, "value" int);
insert into t ("date", "value") values
    ('2011-10-31', 2),
    ('2011-11-01', 8),
    ('2011-11-02', 10),
    ('2012-09-13', 1),
    ('2012-09-14', 4),
    ('2012-09-15', 5),
    ('2012-09-16', 20),
    ('2012-10-30', 10);

По-проста и по-евтина версия:

select min("date"), max("date"), sum(value)
from (
    select
        "date", value,
        "date" - (dense_rank() over(order by "date"))::int g
    from t
) s
group by s.g
order by 1

Първият ми опит беше по-сложен и скъп:

create temporary sequence s;
select min("date"), max("date"), sum(value)
from (
    select 
        "date", value, d,
        case 
            when lag("date", 1, null) over(order by s.d) is null and "date" is not null 
                then nextval('s')
            when lag("date", 1, null) over(order by s.d) is not null and "date" is not null 
                then lastval()
            else 0 
        end g
    from 
        t
        right join
        generate_series(
            (select min("date") from t)::date, 
            (select max("date") from t)::date + 1, 
            '1 day'
        ) s(d) on s.d::date = t."date"
) q
where g != 0
group by g
order by 1
;
drop sequence s;

Резултатът:

    min     |    max     | sum 
------------+------------+-----
 2011-10-31 | 2011-11-02 |  20
 2012-09-13 | 2012-09-16 |  30
 2012-10-30 | 2012-10-30 |  10
(3 rows)