How to convert column with delimited string into rows

I have a string value in a single column, delimited by colon. The number of items in the string is variable. I need to select the data from this column into separate rows based on the delimiter. I can write a function with a loop but if there is a way to do this in SQL it would be better.
Table contains a column with data value:
12:130:1400
And I want to select data and return as:
12
130
1400
This in in Oracle 9i.
Please don't post "look for pivot or transpose in the forum" as that is not a helpful answer (I have already done that).
Thanks!
Message was edited by:
splinternet

SQL> create table mytable (id,value)
2 as
3 select 1, '12:130:1400' from dual union all
4 select 2, '483' from dual union all
5 select 3, '1:2:3:4:5:6:77:888' from dual union all
6 select 4, null from dual
7 /
Tabel is aangemaakt.
SQL> select id
2       , trim(':' from v) value
3       , substr
4         ( v
5         , instr(v,':',1,t.column_value) + 1
6         , instr(v,':',1,1 + t.column_value)
7           - instr(v,':',1,t.column_value) - 1
8         ) part
9    from ( select id, ':' || value || ':' v from mytable ) m
10       , table
11         ( cast
12           ( multiset
13             ( select level l
14                 from dual
15              connect by rownum <= length(m.v) - length(replace(m.v,':')) - 1
16             )
17           as sys.dbms_debug_vc2coll
18           )
19         ) t
20   order by m.id
21       , t.column_value
22 /
        ID VALUE                PART
         1 12:130:1400          12
         1 12:130:1400          130
         1 12:130:1400          1400
         2 483                  483
         3 1:2:3:4:5:6:77:888   1
         3 1:2:3:4:5:6:77:888   2
         3 1:2:3:4:5:6:77:888   3
         3 1:2:3:4:5:6:77:888   4
         3 1:2:3:4:5:6:77:888   5
         3 1:2:3:4:5:6:77:888   6
         3 1:2:3:4:5:6:77:888   77
         3 1:2:3:4:5:6:77:888   888
         4
13 rijen zijn geselecteerd.Regards,
Rob.

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