# 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

## 16 Replies to “OR aggregate”

1. 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 . . . .

2. 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.

3. according to the name, it seems to be an internal function used for materialized views

4. 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

5. the same without subquery

``````
select n
from mytable
model
return updated rows
dimension by (row_number() over (order by null) rn)
measures (n)
rules iterate (10000)
until (presentv(n[ITERATION_NUMBER+3],1,2)=2)
( n[1] = n[1] + nvl(n[ITERATION_NUMBER+2],0) -
bitand(n[1],nvl(n[ITERATION_NUMBER+2],0)))
/
``````

6. 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 π

Regards,
Rob.

7. 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)

``````
with mytable as (
select 1 p, 63 n from dual union all
select 2,1 from dual union all
select 2,1 from dual union all
select 2,null from dual union all
select 2,1 from dual)
select p,n
from mytable
model
return updated rows
partition by (p)
dimension by (row_number() over (partition by p order by null) rn)
measures (n)
rules iterate (10000)
until (presentv(n[ITERATION_NUMBER+3],1,2)=2)
(
n[1] = n[1] + nvl(n[ITERATION_NUMBER+2],0) -
bitand(n[1],nvl(n[ITERATION_NUMBER+2],0))
)
/

P          N
---------- ----------
1         63
2          1
``````

8. Good points.

Although the preconditions were not mentioned, your solution handles them and is therefore more robust.

Regards,
Rob.

9. Aketi Jyuuzou says:

Oh
Good solution!

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;

10. Aketi Jyuuzou says:

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;

11. for AGGREGATE_BITXOR, I would then use :

``````select
mod(sum(sign(bitand(32,n))),2)*32+
mod(sum(sign(bitand(16,n))),2)*16+
mod(sum(sign(bitand( 8,n))),2)* 8+
mod(sum(sign(bitand( 4,n))),2)* 4+
mod(sum(sign(bitand( 2,n))),2)* 2+
mod(sum(sign(bitand( 1,n))),2) n
from t;
``````

12. wow! this is exactly the same, I like this π

arigato!