Obtaining the aggregate sum of one value from multiple columns in sqlite












0















I have a table the equivalent of;



box, fruit1, fruit2, fruit3, no1, no2, no3



1, apple, banana, orange, 12, 23, 7



2, pear, apple, banana, 5, 15, 11



...



I want to run a query which will aggregate the total number of apples, oranges and bananas. It should result in (for boxes 1 and 2):



fruit, no



apples, 27



bananas, 34



oranges, 7



pears, 5



I obviously can't do a simple sum() and group by because I want to group by the fruit as it appears in any column, not just a single one.



Also, in the real life version, there are more than 100 different types of 'fruit' so doing a case...when selection for each one isn't really viable.










share|improve this question



























    0















    I have a table the equivalent of;



    box, fruit1, fruit2, fruit3, no1, no2, no3



    1, apple, banana, orange, 12, 23, 7



    2, pear, apple, banana, 5, 15, 11



    ...



    I want to run a query which will aggregate the total number of apples, oranges and bananas. It should result in (for boxes 1 and 2):



    fruit, no



    apples, 27



    bananas, 34



    oranges, 7



    pears, 5



    I obviously can't do a simple sum() and group by because I want to group by the fruit as it appears in any column, not just a single one.



    Also, in the real life version, there are more than 100 different types of 'fruit' so doing a case...when selection for each one isn't really viable.










    share|improve this question

























      0












      0








      0








      I have a table the equivalent of;



      box, fruit1, fruit2, fruit3, no1, no2, no3



      1, apple, banana, orange, 12, 23, 7



      2, pear, apple, banana, 5, 15, 11



      ...



      I want to run a query which will aggregate the total number of apples, oranges and bananas. It should result in (for boxes 1 and 2):



      fruit, no



      apples, 27



      bananas, 34



      oranges, 7



      pears, 5



      I obviously can't do a simple sum() and group by because I want to group by the fruit as it appears in any column, not just a single one.



      Also, in the real life version, there are more than 100 different types of 'fruit' so doing a case...when selection for each one isn't really viable.










      share|improve this question














      I have a table the equivalent of;



      box, fruit1, fruit2, fruit3, no1, no2, no3



      1, apple, banana, orange, 12, 23, 7



      2, pear, apple, banana, 5, 15, 11



      ...



      I want to run a query which will aggregate the total number of apples, oranges and bananas. It should result in (for boxes 1 and 2):



      fruit, no



      apples, 27



      bananas, 34



      oranges, 7



      pears, 5



      I obviously can't do a simple sum() and group by because I want to group by the fruit as it appears in any column, not just a single one.



      Also, in the real life version, there are more than 100 different types of 'fruit' so doing a case...when selection for each one isn't really viable.







      sqlite aggregate-functions






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      share|improve this question










      asked Nov 22 '18 at 10:19









      IsaacsonIsaacson

      966




      966
























          1 Answer
          1






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          0














          Call off the dogs, I've got it;



          select
          fruit,
          sum(number),
          from
          (
          select
          fruit1 as fruit,
          sum(no1) as number,
          from fruit_table
          group by fruit1
          union
          select
          fruit2 as fruit,
          sum(no2) as number,
          from fruit_table
          group by fruit2
          union
          select
          fruit3 as fruit,
          sum(no3) as number,
          from fruit_table
          group by fruit3
          union
          select
          )
          Group by fruit



          Create a union of all the different tables then aggregate from that unionC






          share|improve this answer























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            1 Answer
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            active

            oldest

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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

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            active

            oldest

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            0














            Call off the dogs, I've got it;



            select
            fruit,
            sum(number),
            from
            (
            select
            fruit1 as fruit,
            sum(no1) as number,
            from fruit_table
            group by fruit1
            union
            select
            fruit2 as fruit,
            sum(no2) as number,
            from fruit_table
            group by fruit2
            union
            select
            fruit3 as fruit,
            sum(no3) as number,
            from fruit_table
            group by fruit3
            union
            select
            )
            Group by fruit



            Create a union of all the different tables then aggregate from that unionC






            share|improve this answer




























              0














              Call off the dogs, I've got it;



              select
              fruit,
              sum(number),
              from
              (
              select
              fruit1 as fruit,
              sum(no1) as number,
              from fruit_table
              group by fruit1
              union
              select
              fruit2 as fruit,
              sum(no2) as number,
              from fruit_table
              group by fruit2
              union
              select
              fruit3 as fruit,
              sum(no3) as number,
              from fruit_table
              group by fruit3
              union
              select
              )
              Group by fruit



              Create a union of all the different tables then aggregate from that unionC






              share|improve this answer


























                0












                0








                0







                Call off the dogs, I've got it;



                select
                fruit,
                sum(number),
                from
                (
                select
                fruit1 as fruit,
                sum(no1) as number,
                from fruit_table
                group by fruit1
                union
                select
                fruit2 as fruit,
                sum(no2) as number,
                from fruit_table
                group by fruit2
                union
                select
                fruit3 as fruit,
                sum(no3) as number,
                from fruit_table
                group by fruit3
                union
                select
                )
                Group by fruit



                Create a union of all the different tables then aggregate from that unionC






                share|improve this answer













                Call off the dogs, I've got it;



                select
                fruit,
                sum(number),
                from
                (
                select
                fruit1 as fruit,
                sum(no1) as number,
                from fruit_table
                group by fruit1
                union
                select
                fruit2 as fruit,
                sum(no2) as number,
                from fruit_table
                group by fruit2
                union
                select
                fruit3 as fruit,
                sum(no3) as number,
                from fruit_table
                group by fruit3
                union
                select
                )
                Group by fruit



                Create a union of all the different tables then aggregate from that unionC







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Nov 22 '18 at 10:46









                IsaacsonIsaacson

                966




                966
































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