Ini adalah variasi dari masalah celah-dan-pulau, dengan komplikasi tambahan dari jumlah baris maksimum di setiap pulau. Ini agak bertele-tele tetapi Anda bisa mulai dengan mengidentifikasi grup yang disebabkan oleh urutan urutan:
select t.*,
row_number() over (partition by "Description" order by "Start") as rn,
case when lag("SequentialOrder")
over (partition by "Description" order by "Start") < "SequentialOrder"
then 1 else 0 end as newblock
from test t
order by "Start";
Start Description MaximunRow SequentialOrder RN NEWBLOCK
--------- ----------- ---------- --------------- --- ----------
12-JUN-15 A 3 3 1 0
13-JUN-15 A 3 4 2 1
14-JUN-15 A 3 5 3 1
01-JUL-15 A 3 4 4 0
02-JUL-15 A 3 3 5 0
04-JUL-15 A 3 4 6 1
01-AUG-15 B 2 5 1 0
16-AUG-15 B 2 7 2 1
Anda kemudian dapat menggunakan CTE rekursif (dari 11gR2 dan seterusnya) berdasarkan itu:
with u as (
select t.*,
row_number() over (partition by "Description" order by "Start") as rn,
case when lag("SequentialOrder")
over (partition by "Description" order by "Start") < "SequentialOrder"
then 1 else 0 end as newblock
from test t
),
r ("Start", "Description", "MaximunRow", "SequentialOrder", rn, blocknum,
pos, lastmaxrow) as (
select u."Start", u."Description", u."MaximunRow", u."SequentialOrder", u.rn,
1, 1, u."MaximunRow"
from u
where rn = 1
union all
select u."Start", u."Description", u."MaximunRow", u."SequentialOrder", u.rn,
case when r.pos = r.lastmaxrow or u.newblock = 0
then r.blocknum + 1 else r.blocknum end,
case when r.pos = r.lastmaxrow or u.newblock = 0
then 1 else r.pos + 1 end,
case when r.pos = r.lastmaxrow or u.newblock = 0
then r.lastmaxrow else u."MaximunRow" end
from r
join u on u."Description" = r."Description" and u.rn = r.rn + 1
)
select * from r
order by "Start";
Start Description MaximunRow SequentialOrder RN BLOCKNUM POS LASTMAXROW
--------- ----------- ---------- --------------- --- ---------- ---- ----------
12-JUN-15 A 3 3 1 1 1 3
13-JUN-15 A 3 4 2 1 2 3
14-JUN-15 A 3 5 3 1 3 3
01-JUL-15 A 3 4 4 2 1 3
02-JUL-15 A 3 3 5 3 1 3
04-JUL-15 A 3 4 6 3 2 3
01-AUG-15 B 2 5 1 1 1 2
16-AUG-15 B 2 7 2 1 2 2
Ini menetapkan blocknum ke setiap baris, dengan yang dimulai dari satu untuk setiap deskripsi di anggota jangkar, dan bertambah di anggota rekursif baik jika newblock adalah nol (menunjukkan jeda urutan) atau jumlah anggota di blok adalah maksimum sebelumnya. (Saya mungkin tidak memiliki logika untuk 'maksimum sebelumnya' karena tidak jelas dalam pertanyaan.)
Anda kemudian dapat mengelompokkan berdasarkan deskripsi dan nomor blok yang dihasilkan:
with u as (
select t.*,
row_number() over (partition by "Description" order by "Start") as rn,
case when lag("SequentialOrder")
over (partition by "Description" order by "Start") < "SequentialOrder"
then 1 else 0 end as newblock
from test t
),
r ("Start", "Description", "MaximunRow", "SequentialOrder", rn, blocknum,
pos, lastmaxrow) as (
select u."Start", u."Description", u."MaximunRow", u."SequentialOrder", u.rn,
1, 1, u."MaximunRow"
from u
where rn = 1
union all
select u."Start", u."Description", u."MaximunRow", u."SequentialOrder", u.rn,
case when r.pos = r.lastmaxrow or u.newblock = 0
then r.blocknum + 1 else r.blocknum end,
case when r.pos = r.lastmaxrow or u.newblock = 0
then 1 else r.pos + 1 end,
case when r.pos = r.lastmaxrow or u.newblock = 0
then r.lastmaxrow else u."MaximunRow" end
from r
join u on u."Description" = r."Description" and u.rn = r.rn + 1
)
select min(r."Start") as "Start", max(r."Start") as "End", r."Description"
from r
group by r."Description", r.blocknum
order by r."Description", r.blocknum;
Start End Description
--------- --------- -----------
12-JUN-15 14-JUN-15 A
01-JUL-15 01-JUL-15 A
02-JUL-15 04-JUL-15 A
01-AUG-15 16-AUG-15 B
Data sampel Anda tidak memicu jeda baris maksimum karena Anda tidak memiliki urutan yang lebih panjang dari 3. Dengan beberapa data tambahan:
Insert into TEST ("Start","Description","MaximunRow","SequentialOrder") values (to_date('15-JUN-15','DD-MON-RR'),'A',3,7);
Insert into TEST ("Start","Description","MaximunRow","SequentialOrder") values (to_date('16-JUN-15','DD-MON-RR'),'A',3,8);
Insert into TEST ("Start","Description","MaximunRow","SequentialOrder") values (to_date('17-JUN-15','DD-MON-RR'),'A',3,10);
Insert into TEST ("Start","Description","MaximunRow","SequentialOrder") values (to_date('18-JUN-15','DD-MON-RR'),'A',3,12);
Insert into TEST ("Start","Description","MaximunRow","SequentialOrder") values (to_date('19-JUN-15','DD-MON-RR'),'A',3,13);
kueri yang sama mendapatkan:
Start End Description
--------- --------- -----------
12-JUN-15 14-JUN-15 A
15-JUN-15 17-JUN-15 A
18-JUN-15 19-JUN-15 A
01-JUL-15 01-JUL-15 A
02-JUL-15 04-JUL-15 A
01-AUG-15 16-AUG-15 B
sehingga Anda dapat melihatnya membelah pada perubahan urutan dan tentang memukul tiga baris di blok.
Anda bisa lolos hanya dengan CTE rekursif, dan bukan CTE perantara sebelumnya, dengan membandingkan urutan berurutan secara langsung dalam pernyataan kasus alih-alih menggunakan newblock; tetapi memiliki rn menemukan baris berikutnya lebih mudah daripada mencoba menemukan tanggal berikutnya karena tidak berdekatan.
