OR aggregate
you want to BIT_OR multiple rows. For example you have a table with 3 rows that you want to aggregate with BIT_OR
1010 (10)
1100 (12)
0110 (6)
=========
1110 (14)
Let’s try
with t as (
select 10 n from dual union all
select 12 from dual union all
select 6 from dual)
select
utl_raw.cast_to_binary_integer(
sys.mvaggrawbitor(
utl_raw.cast_from_binary_integer(
n
)
)
) N
from t;
N
---
14
It is that easy !
disclaimer: mvaggrawbitor is not documented
Hi Laurent,
>> It is that easy !
Ha! Maybe easy for you!
This is very impressive and elegent solution to a complex problem! You are a worthy scholar indeed . . . .
Hi Laurent,
Any idea why Oracle has mvaggrawbitor but no aggregated “and” and “not” operators? (At least on 10.2.0.2)
If they only have enough space for just one aggregating binary operator, XOR would be a better option.
according to the name, it seems to be an internal function used for materialized views
or:
SQL> with t as
2 ( select *
3 from mytable
4 model
5 dimension by (row_number() over (order by null) rn)
6 measures (n)
7 rules
8 ( n[any] order by rn desc = nvl(n[cv()+1],0) + n[cv()] – bitand(nvl(n[cv()+1],0),n[cv()])
9 )
10 )
11 select n
12 from t
13 where rn = 1
14 /
N
———-
14
the same without subquery
That’s even better.
And doing some final (?) little optimizations:
SQL> select n
2 from mytable
3 model
4 return updated rows
5 dimension by (rownum r)
6 measures (n)
7 rules iterate (10000) until n[iteration_number+3] is null
8 ( n[1] = n[1] + n[iteration_number+2] – bitand(n[1],n[iteration_number+2])
9 )
10 /
N
———-
14
Now, it really starts looking that easy
Regards,
Rob.
well, if the table contains only one row, it will return null and if the third row is a NULL, it will exit too early (therefore UNTIL PRESENTV)
Good points.
Although the preconditions were not mentioned, your solution handles them and is therefore more robust.
Regards,
Rob.
Oh
Good solution!
I post homeMadeBitOr
with t as (
select 10 n from dual union all
select 12 from dual union all
select 6 from dual)
select
sum(distinct(bitand(32,n)))+
sum(distinct(bitand(16,n)))+
sum(distinct(bitand( 8,n)))+
sum(distinct(bitand( 4,n)))+
sum(distinct(bitand( 2,n)))+
sum(distinct(bitand( 1,n))) as N
from t;
Nice approach, I wished I found it myself
oops
We can use solution which is more simple.
with t as (
select 10 n from dual union all
select 12 from dual union all
select 6 from dual)
select
max(bitand(32,n))+
max(bitand(16,n))+
max(bitand( 8,n))+
max(bitand( 4,n))+
max(bitand( 2,n))+
max(bitand( 1,n)) as N
from t;
this makes sense!
and for AGGREGATE_BITAND, we could use MIN then!
for AGGREGATE_BITXOR, I would then use :
http://oraclesqlpuzzle.hp.infoseek.co.jp/8-20.html
(written by Japanese language)
My site mentions these solutions
Thank you
wow! this is exactly the same, I like this
arigato!