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How collpase all subgroups into one line and keep the same order
sql union with an aggregation componentHow do I limit the number of rows returned by an Oracle query after ordering?Why does this left join evaluate to a cross join?ORDER BY 2 values that might be NULLHow to import an SQL file using the command line in MySQL?How to order an already ordered subquerySQL - Update Record Based on Entity's Previous RecordOracle Find and Rewrite Consecutive RowsDay Number In a PatternHow to calculate number of Orders for given date range
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
This is a simplified version of my table
+----+----------+------------+------------+
| ID | Category | Start Date | End Date |
+----+----------+------------+------------+
| 1 | 'Alpha' | 2018/04/12 | 2018/04/15 |
| 2 | null | 2018/04/17 | 2018/04/21 |
| 3 | 'Gamma' | 2018/05/02 | 2018/05/07 |
| 4 | 'Gamma' | 2018/05/09 | 2018/05/11 |
| 5 | 'Gamma' | 2018/05/11 | 2018/05/17 |
| 6 | 'Alpha' | 2018/05/17 | 2018/05/23 |
| 7 | 'Alpha' | 2018/05/23 | 2018/05/24 |
| 8 | null | 2018/05/24 | 2018/06/02 |
| 9 | 'Beta' | 2018/06/12 | 2018/06/16 |
| 10 | 'Beta' | 2018/06/16 | 2018/06/20 |
+----+----------+------------+------------+
All Start Date are unique, not nullable and they have the same order as the IDs (if a and b are IDs and a < b then StartDate[a] < StartDate[b]). The Start Date is not always equal to the End Date of the previous row for the same Category (look at id 3 and 4).
I'm looking for a query that will give me the following result
+----------+------------+------------+
| Category | Start Date | End Date |
+----------+------------+------------+
| 'Alpha' | 2018/04/12 | 2018/04/15 |
| null | 2018/04/17 | 2018/04/21 |
| 'Gamma' | 2018/05/02 | 2018/05/17 |
| 'Alpha' | 2018/05/17 | 2018/05/24 |
| null | 2018/05/24 | 2018/06/02 |
| 'Beta' | 2018/06/12 | 2018/06/20 |
+----------+------------+------------+
Note: The End Date will be equal to End Date of the last row in the subgroup (same continuous Category).
sql oracle gaps-and-islands
add a comment |
This is a simplified version of my table
+----+----------+------------+------------+
| ID | Category | Start Date | End Date |
+----+----------+------------+------------+
| 1 | 'Alpha' | 2018/04/12 | 2018/04/15 |
| 2 | null | 2018/04/17 | 2018/04/21 |
| 3 | 'Gamma' | 2018/05/02 | 2018/05/07 |
| 4 | 'Gamma' | 2018/05/09 | 2018/05/11 |
| 5 | 'Gamma' | 2018/05/11 | 2018/05/17 |
| 6 | 'Alpha' | 2018/05/17 | 2018/05/23 |
| 7 | 'Alpha' | 2018/05/23 | 2018/05/24 |
| 8 | null | 2018/05/24 | 2018/06/02 |
| 9 | 'Beta' | 2018/06/12 | 2018/06/16 |
| 10 | 'Beta' | 2018/06/16 | 2018/06/20 |
+----+----------+------------+------------+
All Start Date are unique, not nullable and they have the same order as the IDs (if a and b are IDs and a < b then StartDate[a] < StartDate[b]). The Start Date is not always equal to the End Date of the previous row for the same Category (look at id 3 and 4).
I'm looking for a query that will give me the following result
+----------+------------+------------+
| Category | Start Date | End Date |
+----------+------------+------------+
| 'Alpha' | 2018/04/12 | 2018/04/15 |
| null | 2018/04/17 | 2018/04/21 |
| 'Gamma' | 2018/05/02 | 2018/05/17 |
| 'Alpha' | 2018/05/17 | 2018/05/24 |
| null | 2018/05/24 | 2018/06/02 |
| 'Beta' | 2018/06/12 | 2018/06/20 |
+----------+------------+------------+
Note: The End Date will be equal to End Date of the last row in the subgroup (same continuous Category).
sql oracle gaps-and-islands
add a comment |
This is a simplified version of my table
+----+----------+------------+------------+
| ID | Category | Start Date | End Date |
+----+----------+------------+------------+
| 1 | 'Alpha' | 2018/04/12 | 2018/04/15 |
| 2 | null | 2018/04/17 | 2018/04/21 |
| 3 | 'Gamma' | 2018/05/02 | 2018/05/07 |
| 4 | 'Gamma' | 2018/05/09 | 2018/05/11 |
| 5 | 'Gamma' | 2018/05/11 | 2018/05/17 |
| 6 | 'Alpha' | 2018/05/17 | 2018/05/23 |
| 7 | 'Alpha' | 2018/05/23 | 2018/05/24 |
| 8 | null | 2018/05/24 | 2018/06/02 |
| 9 | 'Beta' | 2018/06/12 | 2018/06/16 |
| 10 | 'Beta' | 2018/06/16 | 2018/06/20 |
+----+----------+------------+------------+
All Start Date are unique, not nullable and they have the same order as the IDs (if a and b are IDs and a < b then StartDate[a] < StartDate[b]). The Start Date is not always equal to the End Date of the previous row for the same Category (look at id 3 and 4).
I'm looking for a query that will give me the following result
+----------+------------+------------+
| Category | Start Date | End Date |
+----------+------------+------------+
| 'Alpha' | 2018/04/12 | 2018/04/15 |
| null | 2018/04/17 | 2018/04/21 |
| 'Gamma' | 2018/05/02 | 2018/05/17 |
| 'Alpha' | 2018/05/17 | 2018/05/24 |
| null | 2018/05/24 | 2018/06/02 |
| 'Beta' | 2018/06/12 | 2018/06/20 |
+----------+------------+------------+
Note: The End Date will be equal to End Date of the last row in the subgroup (same continuous Category).
sql oracle gaps-and-islands
This is a simplified version of my table
+----+----------+------------+------------+
| ID | Category | Start Date | End Date |
+----+----------+------------+------------+
| 1 | 'Alpha' | 2018/04/12 | 2018/04/15 |
| 2 | null | 2018/04/17 | 2018/04/21 |
| 3 | 'Gamma' | 2018/05/02 | 2018/05/07 |
| 4 | 'Gamma' | 2018/05/09 | 2018/05/11 |
| 5 | 'Gamma' | 2018/05/11 | 2018/05/17 |
| 6 | 'Alpha' | 2018/05/17 | 2018/05/23 |
| 7 | 'Alpha' | 2018/05/23 | 2018/05/24 |
| 8 | null | 2018/05/24 | 2018/06/02 |
| 9 | 'Beta' | 2018/06/12 | 2018/06/16 |
| 10 | 'Beta' | 2018/06/16 | 2018/06/20 |
+----+----------+------------+------------+
All Start Date are unique, not nullable and they have the same order as the IDs (if a and b are IDs and a < b then StartDate[a] < StartDate[b]). The Start Date is not always equal to the End Date of the previous row for the same Category (look at id 3 and 4).
I'm looking for a query that will give me the following result
+----------+------------+------------+
| Category | Start Date | End Date |
+----------+------------+------------+
| 'Alpha' | 2018/04/12 | 2018/04/15 |
| null | 2018/04/17 | 2018/04/21 |
| 'Gamma' | 2018/05/02 | 2018/05/17 |
| 'Alpha' | 2018/05/17 | 2018/05/24 |
| null | 2018/05/24 | 2018/06/02 |
| 'Beta' | 2018/06/12 | 2018/06/20 |
+----------+------------+------------+
Note: The End Date will be equal to End Date of the last row in the subgroup (same continuous Category).
sql oracle gaps-and-islands
sql oracle gaps-and-islands
edited Mar 25 at 21:45
Barbaros Özhan
18.6k7 gold badges16 silver badges35 bronze badges
18.6k7 gold badges16 silver badges35 bronze badges
asked Mar 25 at 21:37
Dominique FortinDominique Fortin
1,6658 silver badges17 bronze badges
1,6658 silver badges17 bronze badges
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
This is a gaps-and-islands problem. I think you can use the difference of row numbers:
select category, min(startdate), max(enddate)
from (select t.*,
row_number() over (order by id) as seqnum,
row_number() over (partition by category order by id) as seqnum_c
from t
) t
group by category, (seqnum - seqnum_c)
order by min(startdate);
I'm surprised how simple it is.
– Dominique Fortin
Mar 25 at 22:28
add a comment |
This is a gaps and islands
question, you can use such a logic below
select category, min(start_date) as start_date, max(end_date) as end_date
from
(
select tt.*, sum(grp) over (order by id, start_date) sm
from
(
with t( ID, Category, Start_Date, End_Date) as
(
select 1 , 'Alpha' , date'2018-04-12',date'2018-04-15' from dual union all
select 2 , null , date'2018-04-17',date'2018-04-21' from dual union all
select 3 , 'Gamma' , date'2018-05-02',date'2018-05-07' from dual union all
select 4 , 'Gamma' , date'2018-05-09',date'2018-05-11' from dual union all
select 5 , 'Gamma' , date'2018-05-11',date'2018-05-17' from dual union all
select 6 , 'Alpha' , date'2018-05-17',date'2018-05-23' from dual union all
select 7 , 'Alpha' , date'2018-05-23',date'2018-05-24' from dual union all
select 8 , null , date'2018-05-24',date'2018-06-02' from dual union all
select 9 , 'Beta' , date'2018-06-12',date'2018-06-16' from dual union all
select 10 , 'Beta' , date'2018-06-16',date'2018-06-20' from dual
)
select id, Category,
decode(nvl(lag(end_date) over
(order by end_date),start_date),start_date,0,1)
as grp, --> means prev. value equals or not
row_number() over (order by id, end_date) as rn, start_date, end_date
from t
) tt
order by rn
)
group by Category, sm
order by end_date;
CATEGORY START_DATE END_DATE
Alpha 12.04.2018 15.04.2018
NULL 17.04.2018 21.04.2018
Gamma 02.05.2018 07.05.2018
Gamma 09.05.2018 17.05.2018
Alpha 17.05.2018 24.05.2018
NULL 24.05.2018 02.06.2018
Beta 12.06.2018 20.06.2018
How will you get the second Alpha row?
– Dominique Fortin
Mar 25 at 21:43
@DominiqueFortin oh ok, I didn't notice that this is a gaps-and-islands question
– Barbaros Özhan
Mar 25 at 21:44
There shouldn't be 2 Gamma rows in the result.
– Dominique Fortin
Mar 25 at 22:14
@DominiqueFortin I know, but couldn't figure out yet :)
– Barbaros Özhan
Mar 25 at 22:15
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
This is a gaps-and-islands problem. I think you can use the difference of row numbers:
select category, min(startdate), max(enddate)
from (select t.*,
row_number() over (order by id) as seqnum,
row_number() over (partition by category order by id) as seqnum_c
from t
) t
group by category, (seqnum - seqnum_c)
order by min(startdate);
I'm surprised how simple it is.
– Dominique Fortin
Mar 25 at 22:28
add a comment |
This is a gaps-and-islands problem. I think you can use the difference of row numbers:
select category, min(startdate), max(enddate)
from (select t.*,
row_number() over (order by id) as seqnum,
row_number() over (partition by category order by id) as seqnum_c
from t
) t
group by category, (seqnum - seqnum_c)
order by min(startdate);
I'm surprised how simple it is.
– Dominique Fortin
Mar 25 at 22:28
add a comment |
This is a gaps-and-islands problem. I think you can use the difference of row numbers:
select category, min(startdate), max(enddate)
from (select t.*,
row_number() over (order by id) as seqnum,
row_number() over (partition by category order by id) as seqnum_c
from t
) t
group by category, (seqnum - seqnum_c)
order by min(startdate);
This is a gaps-and-islands problem. I think you can use the difference of row numbers:
select category, min(startdate), max(enddate)
from (select t.*,
row_number() over (order by id) as seqnum,
row_number() over (partition by category order by id) as seqnum_c
from t
) t
group by category, (seqnum - seqnum_c)
order by min(startdate);
answered Mar 25 at 22:06
Gordon LinoffGordon Linoff
829k38 gold badges339 silver badges445 bronze badges
829k38 gold badges339 silver badges445 bronze badges
I'm surprised how simple it is.
– Dominique Fortin
Mar 25 at 22:28
add a comment |
I'm surprised how simple it is.
– Dominique Fortin
Mar 25 at 22:28
I'm surprised how simple it is.
– Dominique Fortin
Mar 25 at 22:28
I'm surprised how simple it is.
– Dominique Fortin
Mar 25 at 22:28
add a comment |
This is a gaps and islands
question, you can use such a logic below
select category, min(start_date) as start_date, max(end_date) as end_date
from
(
select tt.*, sum(grp) over (order by id, start_date) sm
from
(
with t( ID, Category, Start_Date, End_Date) as
(
select 1 , 'Alpha' , date'2018-04-12',date'2018-04-15' from dual union all
select 2 , null , date'2018-04-17',date'2018-04-21' from dual union all
select 3 , 'Gamma' , date'2018-05-02',date'2018-05-07' from dual union all
select 4 , 'Gamma' , date'2018-05-09',date'2018-05-11' from dual union all
select 5 , 'Gamma' , date'2018-05-11',date'2018-05-17' from dual union all
select 6 , 'Alpha' , date'2018-05-17',date'2018-05-23' from dual union all
select 7 , 'Alpha' , date'2018-05-23',date'2018-05-24' from dual union all
select 8 , null , date'2018-05-24',date'2018-06-02' from dual union all
select 9 , 'Beta' , date'2018-06-12',date'2018-06-16' from dual union all
select 10 , 'Beta' , date'2018-06-16',date'2018-06-20' from dual
)
select id, Category,
decode(nvl(lag(end_date) over
(order by end_date),start_date),start_date,0,1)
as grp, --> means prev. value equals or not
row_number() over (order by id, end_date) as rn, start_date, end_date
from t
) tt
order by rn
)
group by Category, sm
order by end_date;
CATEGORY START_DATE END_DATE
Alpha 12.04.2018 15.04.2018
NULL 17.04.2018 21.04.2018
Gamma 02.05.2018 07.05.2018
Gamma 09.05.2018 17.05.2018
Alpha 17.05.2018 24.05.2018
NULL 24.05.2018 02.06.2018
Beta 12.06.2018 20.06.2018
How will you get the second Alpha row?
– Dominique Fortin
Mar 25 at 21:43
@DominiqueFortin oh ok, I didn't notice that this is a gaps-and-islands question
– Barbaros Özhan
Mar 25 at 21:44
There shouldn't be 2 Gamma rows in the result.
– Dominique Fortin
Mar 25 at 22:14
@DominiqueFortin I know, but couldn't figure out yet :)
– Barbaros Özhan
Mar 25 at 22:15
add a comment |
This is a gaps and islands
question, you can use such a logic below
select category, min(start_date) as start_date, max(end_date) as end_date
from
(
select tt.*, sum(grp) over (order by id, start_date) sm
from
(
with t( ID, Category, Start_Date, End_Date) as
(
select 1 , 'Alpha' , date'2018-04-12',date'2018-04-15' from dual union all
select 2 , null , date'2018-04-17',date'2018-04-21' from dual union all
select 3 , 'Gamma' , date'2018-05-02',date'2018-05-07' from dual union all
select 4 , 'Gamma' , date'2018-05-09',date'2018-05-11' from dual union all
select 5 , 'Gamma' , date'2018-05-11',date'2018-05-17' from dual union all
select 6 , 'Alpha' , date'2018-05-17',date'2018-05-23' from dual union all
select 7 , 'Alpha' , date'2018-05-23',date'2018-05-24' from dual union all
select 8 , null , date'2018-05-24',date'2018-06-02' from dual union all
select 9 , 'Beta' , date'2018-06-12',date'2018-06-16' from dual union all
select 10 , 'Beta' , date'2018-06-16',date'2018-06-20' from dual
)
select id, Category,
decode(nvl(lag(end_date) over
(order by end_date),start_date),start_date,0,1)
as grp, --> means prev. value equals or not
row_number() over (order by id, end_date) as rn, start_date, end_date
from t
) tt
order by rn
)
group by Category, sm
order by end_date;
CATEGORY START_DATE END_DATE
Alpha 12.04.2018 15.04.2018
NULL 17.04.2018 21.04.2018
Gamma 02.05.2018 07.05.2018
Gamma 09.05.2018 17.05.2018
Alpha 17.05.2018 24.05.2018
NULL 24.05.2018 02.06.2018
Beta 12.06.2018 20.06.2018
How will you get the second Alpha row?
– Dominique Fortin
Mar 25 at 21:43
@DominiqueFortin oh ok, I didn't notice that this is a gaps-and-islands question
– Barbaros Özhan
Mar 25 at 21:44
There shouldn't be 2 Gamma rows in the result.
– Dominique Fortin
Mar 25 at 22:14
@DominiqueFortin I know, but couldn't figure out yet :)
– Barbaros Özhan
Mar 25 at 22:15
add a comment |
This is a gaps and islands
question, you can use such a logic below
select category, min(start_date) as start_date, max(end_date) as end_date
from
(
select tt.*, sum(grp) over (order by id, start_date) sm
from
(
with t( ID, Category, Start_Date, End_Date) as
(
select 1 , 'Alpha' , date'2018-04-12',date'2018-04-15' from dual union all
select 2 , null , date'2018-04-17',date'2018-04-21' from dual union all
select 3 , 'Gamma' , date'2018-05-02',date'2018-05-07' from dual union all
select 4 , 'Gamma' , date'2018-05-09',date'2018-05-11' from dual union all
select 5 , 'Gamma' , date'2018-05-11',date'2018-05-17' from dual union all
select 6 , 'Alpha' , date'2018-05-17',date'2018-05-23' from dual union all
select 7 , 'Alpha' , date'2018-05-23',date'2018-05-24' from dual union all
select 8 , null , date'2018-05-24',date'2018-06-02' from dual union all
select 9 , 'Beta' , date'2018-06-12',date'2018-06-16' from dual union all
select 10 , 'Beta' , date'2018-06-16',date'2018-06-20' from dual
)
select id, Category,
decode(nvl(lag(end_date) over
(order by end_date),start_date),start_date,0,1)
as grp, --> means prev. value equals or not
row_number() over (order by id, end_date) as rn, start_date, end_date
from t
) tt
order by rn
)
group by Category, sm
order by end_date;
CATEGORY START_DATE END_DATE
Alpha 12.04.2018 15.04.2018
NULL 17.04.2018 21.04.2018
Gamma 02.05.2018 07.05.2018
Gamma 09.05.2018 17.05.2018
Alpha 17.05.2018 24.05.2018
NULL 24.05.2018 02.06.2018
Beta 12.06.2018 20.06.2018
This is a gaps and islands
question, you can use such a logic below
select category, min(start_date) as start_date, max(end_date) as end_date
from
(
select tt.*, sum(grp) over (order by id, start_date) sm
from
(
with t( ID, Category, Start_Date, End_Date) as
(
select 1 , 'Alpha' , date'2018-04-12',date'2018-04-15' from dual union all
select 2 , null , date'2018-04-17',date'2018-04-21' from dual union all
select 3 , 'Gamma' , date'2018-05-02',date'2018-05-07' from dual union all
select 4 , 'Gamma' , date'2018-05-09',date'2018-05-11' from dual union all
select 5 , 'Gamma' , date'2018-05-11',date'2018-05-17' from dual union all
select 6 , 'Alpha' , date'2018-05-17',date'2018-05-23' from dual union all
select 7 , 'Alpha' , date'2018-05-23',date'2018-05-24' from dual union all
select 8 , null , date'2018-05-24',date'2018-06-02' from dual union all
select 9 , 'Beta' , date'2018-06-12',date'2018-06-16' from dual union all
select 10 , 'Beta' , date'2018-06-16',date'2018-06-20' from dual
)
select id, Category,
decode(nvl(lag(end_date) over
(order by end_date),start_date),start_date,0,1)
as grp, --> means prev. value equals or not
row_number() over (order by id, end_date) as rn, start_date, end_date
from t
) tt
order by rn
)
group by Category, sm
order by end_date;
CATEGORY START_DATE END_DATE
Alpha 12.04.2018 15.04.2018
NULL 17.04.2018 21.04.2018
Gamma 02.05.2018 07.05.2018
Gamma 09.05.2018 17.05.2018
Alpha 17.05.2018 24.05.2018
NULL 24.05.2018 02.06.2018
Beta 12.06.2018 20.06.2018
edited Mar 25 at 22:17
answered Mar 25 at 21:39
Barbaros ÖzhanBarbaros Özhan
18.6k7 gold badges16 silver badges35 bronze badges
18.6k7 gold badges16 silver badges35 bronze badges
How will you get the second Alpha row?
– Dominique Fortin
Mar 25 at 21:43
@DominiqueFortin oh ok, I didn't notice that this is a gaps-and-islands question
– Barbaros Özhan
Mar 25 at 21:44
There shouldn't be 2 Gamma rows in the result.
– Dominique Fortin
Mar 25 at 22:14
@DominiqueFortin I know, but couldn't figure out yet :)
– Barbaros Özhan
Mar 25 at 22:15
add a comment |
How will you get the second Alpha row?
– Dominique Fortin
Mar 25 at 21:43
@DominiqueFortin oh ok, I didn't notice that this is a gaps-and-islands question
– Barbaros Özhan
Mar 25 at 21:44
There shouldn't be 2 Gamma rows in the result.
– Dominique Fortin
Mar 25 at 22:14
@DominiqueFortin I know, but couldn't figure out yet :)
– Barbaros Özhan
Mar 25 at 22:15
How will you get the second Alpha row?
– Dominique Fortin
Mar 25 at 21:43
How will you get the second Alpha row?
– Dominique Fortin
Mar 25 at 21:43
@DominiqueFortin oh ok, I didn't notice that this is a gaps-and-islands question
– Barbaros Özhan
Mar 25 at 21:44
@DominiqueFortin oh ok, I didn't notice that this is a gaps-and-islands question
– Barbaros Özhan
Mar 25 at 21:44
There shouldn't be 2 Gamma rows in the result.
– Dominique Fortin
Mar 25 at 22:14
There shouldn't be 2 Gamma rows in the result.
– Dominique Fortin
Mar 25 at 22:14
@DominiqueFortin I know, but couldn't figure out yet :)
– Barbaros Özhan
Mar 25 at 22:15
@DominiqueFortin I know, but couldn't figure out yet :)
– Barbaros Özhan
Mar 25 at 22:15
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