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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;








1















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).










share|improve this question






























    1















    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).










    share|improve this question


























      1












      1








      1








      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).










      share|improve this question
















      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






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      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






















          2 Answers
          2






          active

          oldest

          votes


















          1














          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);





          share|improve this answer























          • I'm surprised how simple it is.

            – Dominique Fortin
            Mar 25 at 22:28


















          0














          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





          share|improve this answer

























          • 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













          Your Answer






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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1














          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);





          share|improve this answer























          • I'm surprised how simple it is.

            – Dominique Fortin
            Mar 25 at 22:28















          1














          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);





          share|improve this answer























          • I'm surprised how simple it is.

            – Dominique Fortin
            Mar 25 at 22:28













          1












          1








          1







          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);





          share|improve this answer













          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);






          share|improve this answer












          share|improve this answer



          share|improve this answer










          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

















          • 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













          0














          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





          share|improve this answer

























          • 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















          0














          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





          share|improve this answer

























          • 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













          0












          0








          0







          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





          share|improve this answer















          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






          share|improve this answer














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          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

















          • 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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