Query using Analytical Function

Question

/*Hi there, i have the following table:*/
create table test (charcode varchar2(10),numberv number,father varchar2(10));
insert into test values('1',null,null); // Level 1
insert into test values('1.1',null,'1'); // Level 2
insert into test values('1.1.1',10,'1.1'); // Level 3
insert into test values('1.1.2',50,'1.1'); // Level 3
insert into test values('1.2',100,'1'); // Level 2
I want the following result
charcode , numberv
1 160
1.1 60
1.1.1 10
1.1.2 50
1.2 100
The real table has more than 1,000,000 and i have more then 8 levels, without analytical sum, it spend more than 3 hours to process, and i know that with analytical way it should spend less than 15 minutes.
I have a ideia on how to start: /*
select charcode,sum(numberv) over(partition father order by charcode desc)
order by charcode desc;
/* but this result it´s far way from what i expect ( the charcode 1 wasn´t populate and other values appears different values than i want
Thanks
Hélio*/

Answer 1

here is your query. it will work for charcode between 1 and 9999.9999.9999.9999
of course you could extend it
this use regexp, but you could also use substr and instr.
select charcode,
numberv,
decode(length(charcode)-length(replace(charcode,'.')),
    0,
    sum(NUMBERV) over (partition by
        substr(
            replace(regexp_replace(
            regexp_replace(
            regexp_replace('.'||replace(charcode,'.','..')||'.',
            '\.([0-9]{1})\.' , '.000\1.'),
            '\.([0-9]{2})\.' , '.00\1.'),
            '\.([0-9]{3})\.' , '.0\1.'),'.')
        ,1, 4)
    1,
    sum(NUMBERV) over (partition by
        substr(
            replace(regexp_replace(
            regexp_replace(
            regexp_replace('.'||replace(charcode,'.','..')||'.',
            '\.([0-9]{1})\.' , '.000\1.'),
            '\.([0-9]{2})\.' , '.00\1.'),
            '\.([0-9]{3})\.' , '.0\1.'),'.')
        ,5, 4)
    2,
    sum(NUMBERV) over (partition by
        substr(
            replace(regexp_replace(
            regexp_replace(
            regexp_replace('.'||replace(charcode,'.','..')||'.',
            '\.([0-9]{1})\.' , '.000\1.'),
            '\.([0-9]{2})\.' , '.00\1.'),
            '\.([0-9]{3})\.' , '.0\1.'),'.')
        ,9, 4)
    3,
    sum(NUMBERV) over (partition by
        substr(
            replace(regexp_replace(
            regexp_replace(
            regexp_replace('.'||replace(charcode,'.','..')||'.',
            '\.([0-9]{1})\.' , '.000\1.'),
            '\.([0-9]{2})\.' , '.00\1.'),
            '\.([0-9]{3})\.' , '.0\1.'),'.')
        ,13, 4)
    4,
    sum(NUMBERV) over (partition by
        substr(
            replace(regexp_replace(
            regexp_replace(
            regexp_replace('.'||replace(charcode,'.','..')||'.',
            '\.([0-9]{1})\.' , '.000\1.'),
            '\.([0-9]{2})\.' , '.00\1.'),
            '\.([0-9]{3})\.' , '.0\1.'),'.')
        ,17, 4)
) "sum"
from test
order by
replace(regexp_replace(
    regexp_replace(
    regexp_replace('.'||replace(charcode,'.','..')||'.',
    '\.([0-9]{1})\.' , '.000\1.'),
    '\.([0-9]{2})\.' , '.00\1.'),
    '\.([0-9]{3})\.' , '.0\1.'),'.')
CHARCODE      NUMBERV        sum
1                            160
1.1                           60
1.1.1              10         10
1.1.2              50         50
1.2               100        100I hope you appreciate !
Regards
Laurent