Oracle pivot on a column with delimited data

oracleoracle-10gpivot

My data is like:

keycol,col1,col2,col3
1,a;b;c,some data,some other data
2,x,some data,some other data
3,y;z,some data,some other data

Where a column is delimited in the source system with semicolons.

And I want to pivot it to:

1,a,some data,some other data
1,b,some data,some other data
1,c,some data,some other data
2,x,some data,some other data
3,y,some data,some other data
3,z,some data,some other data

I found a technique here, but I can't quite get it to work:

CREATE TABLE yt
    (keycol int, col1 varchar2(5), col2 varchar2(9), col3 varchar2(15))
;

INSERT ALL 
    INTO yt (keycol, col1, col2, col3)
         VALUES (1, 'a;b;c', 'some data', 'some other data')
SELECT * FROM dual
;

INSERT ALL 
    INTO yt (keycol, col1, col2, col3)
         VALUES (2, 'x', 'some data', 'some other data')
SELECT * FROM dual
;

INSERT ALL 
    INTO yt (keycol, col1, col2, col3)
         VALUES (3, 'y;z', 'some data', 'some other data')
SELECT * FROM dual
;

I thought I could include the keycol in the CONNECT BY this way to get parallel recursive chains, but I guess it doesn't work like that. I'm pretty sure I've done this with recursive CTEs in SQL Server.

SELECT keycol
  ,trim(regexp_substr(col1, '[^;]+', 1, level)) col1
  ,col2 
  ,col3
FROM yt t
CONNECT BY keycol = PRIOR keycol AND instr(col1, ';', 1, level - 1) > 0

http://sqlfiddle.com/#!4/3d378

FWIW, I'm on Oracle 10g.

Best Answer

Using the code from this link as a starting point, this should give you the result without the need for a DISTINCT:

select t.keycol,
    regexp_substr ( t.col1, '[^;]+', 1, n) as col1,
    t.col2,
    t.col3
from yt t 
cross join
(
    select level n
    from 
    (
        select max(length(col1) - length(replace(col1, ';'))) as max_semi
        from yt
    )
    connect by level <= 1 + max_semi
) c
where c.n <= 1 + length(col1) - length(replace(col1, ';'))
order by t.keycol, col1;

See SQL Fiddle with Demo

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