Oracle SQL dynamic pivot











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Imagine there are two tables table like so:



Table 1



ID   |   NAME   |   ATTR_A   |   ATTR_B   |   ATTR_C   |    ATTR_D
------------------------------------------------------------------
0 Bob 1 2 3 0
1 Jane 2 0 4 3
2 Richard 1 3 5 4


Table 2



ID   |   TABLE_1_FK   |   ATTR_E
--------------------------------
0 0 A
1 1 B
2 1 C
3 1 D
4 2 E
5 2 F


Using a left join to merge table 1 and 2, the resulting table will be



Joined Table



NAME    |   ATTR_E    |   ATTR_A   |   ATTR_B   |   ATTR_C   |   ATTR_D
-----------------------------------------------------------------------
Bob A 1 2 3 0
Jane B 2 0 4 3
Jane C 2 0 4 3
Jane D 2 0 4 3
Richard E 1 3 5 4
Richard F 1 3 5 4


Question




3 out of the 4 attributes (ATTR_A,B,C,D) in table 1 will always have a value greater than 0, create a new table so that if the attribute in table 1 has a value greater than 0, display it under a "table 1 attr" column, with its value under a "table 1 attr_value" column. Along with that, create a new column to describe the attribute selected.




The resulting table would be like so:



Result



NAME    |    ATTR_E    |    "table 1 attr"    |    "table 1 value"   |   "attr description"
---------------------------------------------------------------------------------------------
Bob A ATTR_A 1 Apple
Bob A ATTR_B 2 Banana
Bob A ATTR_C 3 Carrot
Jane B ATTR_A 2 Apple
Jane B ATTR_C 4 Carrot
Jane B ATTR_D 3 Durian
Jane C ATTR_A 2 Apple
Jane C ATTR_C 4 Carrot
Jane C ATTR_D 3 Durian
Jane D ATTR_A 2 Apple
Jane D ATTR_C 4 Carrot
Jane D ATTR_D 3 Durian
Richard E ATTR_A 1 Apple
Richard E ATTR_B 3 Banana
Richard E ATTR_C 5 Carrot
Richard E ATTR_D 4 Durian
Richard F ATTR_A 1 Apple
Richard F ATTR_B 3 Banana
Richard F ATTR_C 5 Carrot
Richard F ATTR_D 4 Durian









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  • 1




    What have you tried so far?
    – Josh Eller
    Nov 19 at 16:15










  • I'm really confused. Where does the description come from?
    – Gordon Linoff
    Nov 19 at 16:24















up vote
0
down vote

favorite












Imagine there are two tables table like so:



Table 1



ID   |   NAME   |   ATTR_A   |   ATTR_B   |   ATTR_C   |    ATTR_D
------------------------------------------------------------------
0 Bob 1 2 3 0
1 Jane 2 0 4 3
2 Richard 1 3 5 4


Table 2



ID   |   TABLE_1_FK   |   ATTR_E
--------------------------------
0 0 A
1 1 B
2 1 C
3 1 D
4 2 E
5 2 F


Using a left join to merge table 1 and 2, the resulting table will be



Joined Table



NAME    |   ATTR_E    |   ATTR_A   |   ATTR_B   |   ATTR_C   |   ATTR_D
-----------------------------------------------------------------------
Bob A 1 2 3 0
Jane B 2 0 4 3
Jane C 2 0 4 3
Jane D 2 0 4 3
Richard E 1 3 5 4
Richard F 1 3 5 4


Question




3 out of the 4 attributes (ATTR_A,B,C,D) in table 1 will always have a value greater than 0, create a new table so that if the attribute in table 1 has a value greater than 0, display it under a "table 1 attr" column, with its value under a "table 1 attr_value" column. Along with that, create a new column to describe the attribute selected.




The resulting table would be like so:



Result



NAME    |    ATTR_E    |    "table 1 attr"    |    "table 1 value"   |   "attr description"
---------------------------------------------------------------------------------------------
Bob A ATTR_A 1 Apple
Bob A ATTR_B 2 Banana
Bob A ATTR_C 3 Carrot
Jane B ATTR_A 2 Apple
Jane B ATTR_C 4 Carrot
Jane B ATTR_D 3 Durian
Jane C ATTR_A 2 Apple
Jane C ATTR_C 4 Carrot
Jane C ATTR_D 3 Durian
Jane D ATTR_A 2 Apple
Jane D ATTR_C 4 Carrot
Jane D ATTR_D 3 Durian
Richard E ATTR_A 1 Apple
Richard E ATTR_B 3 Banana
Richard E ATTR_C 5 Carrot
Richard E ATTR_D 4 Durian
Richard F ATTR_A 1 Apple
Richard F ATTR_B 3 Banana
Richard F ATTR_C 5 Carrot
Richard F ATTR_D 4 Durian









share|improve this question




















  • 1




    What have you tried so far?
    – Josh Eller
    Nov 19 at 16:15










  • I'm really confused. Where does the description come from?
    – Gordon Linoff
    Nov 19 at 16:24













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Imagine there are two tables table like so:



Table 1



ID   |   NAME   |   ATTR_A   |   ATTR_B   |   ATTR_C   |    ATTR_D
------------------------------------------------------------------
0 Bob 1 2 3 0
1 Jane 2 0 4 3
2 Richard 1 3 5 4


Table 2



ID   |   TABLE_1_FK   |   ATTR_E
--------------------------------
0 0 A
1 1 B
2 1 C
3 1 D
4 2 E
5 2 F


Using a left join to merge table 1 and 2, the resulting table will be



Joined Table



NAME    |   ATTR_E    |   ATTR_A   |   ATTR_B   |   ATTR_C   |   ATTR_D
-----------------------------------------------------------------------
Bob A 1 2 3 0
Jane B 2 0 4 3
Jane C 2 0 4 3
Jane D 2 0 4 3
Richard E 1 3 5 4
Richard F 1 3 5 4


Question




3 out of the 4 attributes (ATTR_A,B,C,D) in table 1 will always have a value greater than 0, create a new table so that if the attribute in table 1 has a value greater than 0, display it under a "table 1 attr" column, with its value under a "table 1 attr_value" column. Along with that, create a new column to describe the attribute selected.




The resulting table would be like so:



Result



NAME    |    ATTR_E    |    "table 1 attr"    |    "table 1 value"   |   "attr description"
---------------------------------------------------------------------------------------------
Bob A ATTR_A 1 Apple
Bob A ATTR_B 2 Banana
Bob A ATTR_C 3 Carrot
Jane B ATTR_A 2 Apple
Jane B ATTR_C 4 Carrot
Jane B ATTR_D 3 Durian
Jane C ATTR_A 2 Apple
Jane C ATTR_C 4 Carrot
Jane C ATTR_D 3 Durian
Jane D ATTR_A 2 Apple
Jane D ATTR_C 4 Carrot
Jane D ATTR_D 3 Durian
Richard E ATTR_A 1 Apple
Richard E ATTR_B 3 Banana
Richard E ATTR_C 5 Carrot
Richard E ATTR_D 4 Durian
Richard F ATTR_A 1 Apple
Richard F ATTR_B 3 Banana
Richard F ATTR_C 5 Carrot
Richard F ATTR_D 4 Durian









share|improve this question















Imagine there are two tables table like so:



Table 1



ID   |   NAME   |   ATTR_A   |   ATTR_B   |   ATTR_C   |    ATTR_D
------------------------------------------------------------------
0 Bob 1 2 3 0
1 Jane 2 0 4 3
2 Richard 1 3 5 4


Table 2



ID   |   TABLE_1_FK   |   ATTR_E
--------------------------------
0 0 A
1 1 B
2 1 C
3 1 D
4 2 E
5 2 F


Using a left join to merge table 1 and 2, the resulting table will be



Joined Table



NAME    |   ATTR_E    |   ATTR_A   |   ATTR_B   |   ATTR_C   |   ATTR_D
-----------------------------------------------------------------------
Bob A 1 2 3 0
Jane B 2 0 4 3
Jane C 2 0 4 3
Jane D 2 0 4 3
Richard E 1 3 5 4
Richard F 1 3 5 4


Question




3 out of the 4 attributes (ATTR_A,B,C,D) in table 1 will always have a value greater than 0, create a new table so that if the attribute in table 1 has a value greater than 0, display it under a "table 1 attr" column, with its value under a "table 1 attr_value" column. Along with that, create a new column to describe the attribute selected.




The resulting table would be like so:



Result



NAME    |    ATTR_E    |    "table 1 attr"    |    "table 1 value"   |   "attr description"
---------------------------------------------------------------------------------------------
Bob A ATTR_A 1 Apple
Bob A ATTR_B 2 Banana
Bob A ATTR_C 3 Carrot
Jane B ATTR_A 2 Apple
Jane B ATTR_C 4 Carrot
Jane B ATTR_D 3 Durian
Jane C ATTR_A 2 Apple
Jane C ATTR_C 4 Carrot
Jane C ATTR_D 3 Durian
Jane D ATTR_A 2 Apple
Jane D ATTR_C 4 Carrot
Jane D ATTR_D 3 Durian
Richard E ATTR_A 1 Apple
Richard E ATTR_B 3 Banana
Richard E ATTR_C 5 Carrot
Richard E ATTR_D 4 Durian
Richard F ATTR_A 1 Apple
Richard F ATTR_B 3 Banana
Richard F ATTR_C 5 Carrot
Richard F ATTR_D 4 Durian






sql database oracle






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edited Nov 19 at 16:13

























asked Nov 19 at 16:05









Dowopq

32




32








  • 1




    What have you tried so far?
    – Josh Eller
    Nov 19 at 16:15










  • I'm really confused. Where does the description come from?
    – Gordon Linoff
    Nov 19 at 16:24














  • 1




    What have you tried so far?
    – Josh Eller
    Nov 19 at 16:15










  • I'm really confused. Where does the description come from?
    – Gordon Linoff
    Nov 19 at 16:24








1




1




What have you tried so far?
– Josh Eller
Nov 19 at 16:15




What have you tried so far?
– Josh Eller
Nov 19 at 16:15












I'm really confused. Where does the description come from?
– Gordon Linoff
Nov 19 at 16:24




I'm really confused. Where does the description come from?
– Gordon Linoff
Nov 19 at 16:24












1 Answer
1






active

oldest

votes

















up vote
0
down vote



accepted










You need to UNPIVOT your joined result. I'm posting a working example exactly like your problem, so you replace with your tables and such as stated in the sql comments.



SELECT *
FROM
(
-- your join result here...
SELECT 1 col_id, 2 col_attrib_a, 0 col_attrib_b FROM dual UNION
SELECT 9 col_id, 0 col_attrib_a, 7 col_attrib_b FROM dual
)
UNPIVOT
(
value
FOR value_type IN (col_attrib_a, col_attrib_b) -- the name of the attrib columns here...
)
WHERE 1=1
AND value > 0
;





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

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    active

    oldest

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    active

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    up vote
    0
    down vote



    accepted










    You need to UNPIVOT your joined result. I'm posting a working example exactly like your problem, so you replace with your tables and such as stated in the sql comments.



    SELECT *
    FROM
    (
    -- your join result here...
    SELECT 1 col_id, 2 col_attrib_a, 0 col_attrib_b FROM dual UNION
    SELECT 9 col_id, 0 col_attrib_a, 7 col_attrib_b FROM dual
    )
    UNPIVOT
    (
    value
    FOR value_type IN (col_attrib_a, col_attrib_b) -- the name of the attrib columns here...
    )
    WHERE 1=1
    AND value > 0
    ;





    share|improve this answer

























      up vote
      0
      down vote



      accepted










      You need to UNPIVOT your joined result. I'm posting a working example exactly like your problem, so you replace with your tables and such as stated in the sql comments.



      SELECT *
      FROM
      (
      -- your join result here...
      SELECT 1 col_id, 2 col_attrib_a, 0 col_attrib_b FROM dual UNION
      SELECT 9 col_id, 0 col_attrib_a, 7 col_attrib_b FROM dual
      )
      UNPIVOT
      (
      value
      FOR value_type IN (col_attrib_a, col_attrib_b) -- the name of the attrib columns here...
      )
      WHERE 1=1
      AND value > 0
      ;





      share|improve this answer























        up vote
        0
        down vote



        accepted







        up vote
        0
        down vote



        accepted






        You need to UNPIVOT your joined result. I'm posting a working example exactly like your problem, so you replace with your tables and such as stated in the sql comments.



        SELECT *
        FROM
        (
        -- your join result here...
        SELECT 1 col_id, 2 col_attrib_a, 0 col_attrib_b FROM dual UNION
        SELECT 9 col_id, 0 col_attrib_a, 7 col_attrib_b FROM dual
        )
        UNPIVOT
        (
        value
        FOR value_type IN (col_attrib_a, col_attrib_b) -- the name of the attrib columns here...
        )
        WHERE 1=1
        AND value > 0
        ;





        share|improve this answer












        You need to UNPIVOT your joined result. I'm posting a working example exactly like your problem, so you replace with your tables and such as stated in the sql comments.



        SELECT *
        FROM
        (
        -- your join result here...
        SELECT 1 col_id, 2 col_attrib_a, 0 col_attrib_b FROM dual UNION
        SELECT 9 col_id, 0 col_attrib_a, 7 col_attrib_b FROM dual
        )
        UNPIVOT
        (
        value
        FOR value_type IN (col_attrib_a, col_attrib_b) -- the name of the attrib columns here...
        )
        WHERE 1=1
        AND value > 0
        ;






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 19 at 16:24









        Felypp Oliveira

        1,676917




        1,676917






























             

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