Eddie Awad’s Blog - Latest Comments in SQL problem

Re: SQL problem

evan — Wed, 11 Jun 2008 01:03:59 -0000

select * from t a where a.id in (&1) and not exists (
select * from table(sys.KU$_OBJNUMSET(&1)) b where not exists (
select * from t c where c.id=b.column_value))

Re: SQL problem

Jeffrey Kemp — Thu, 13 Oct 2005 01:16:21 -0000

After re-reading the original post it appears it was not a requirement to limit the rows returned unless all the query rows existed in it; so my last comment is probably superfluous if it doesn't matter if all the rows get returned.

Another case of needing to get the exact user requirements...

Re: SQL problem

Jeffrey Kemp — Thu, 13 Oct 2005 01:10:30 -0000

If you remove the "id in (select id from u)" (or the equivalent "exists (...)", won't the query return values not in the query set? I.E.:

SQL> select * from t;
ID
----------
1
2
3

SQL> select * from u;
ID
----------
1
2
3

SQL> insert into t values (4);
1 row created.

SQL> select * from t
where not exists (select id from u minus select id from t);
ID
----------
1
2
4
3

SQL> select * from t where id in (select id from u)
and not exists (select id from u minus select id from t);
ID
----------
1
2
3

-- in the first query, 4 is incorrectly returned, yes?

Re: SQL problem

Robert Vollman — Wed, 12 Oct 2005 13:32:43 -0000

Great post! That is an interesting SQL puzzler!

Re: SQL problem

Partha — Wed, 12 Oct 2005 07:12:04 -0000

There is a problem with Gary's solution as it would give duplicate values if the collection contained any duplications. As per the requirement, we cannot have duplicate rows returned from t even if the values appear duplicate in the input list or collection(whatever you might call it).

Jeff's solution was what I wanted to acheive, but really could not get it working. I got around to it using Set Union, but not a very neat solution. Please note the trick for removing duplicate values was using a Set operation. Jeff's solution could be further improved by removing the where id in clause completely. The Set minus operation does the job neatly.

So using the improvement on Jeff's solution and using any method to manipulate the input list, the solution could be as below :

select * from t
where not exists (
select to_number(column_name)
from Input_Table
minus
select id from t)

The *`Input_Table`* can be computed from input list of values either using PL/SQL method (Partha), `sys.dbms_debug_vc2coll` (Gary) or temporary table method (Jeff). This would be driven by the assumptions you make of the input data set.

The *`column_name`* would be `column_name` in case of PL/SQL method, `column_value` in case of `dbms_debug_vc2coll` method or the column name of the temporary table in case of temporary table method.

Re: SQL problem

Gary — Wed, 12 Oct 2005 02:35:18 -0000

It uses a collection type (but in my case I use the predefined one).
Depends on whether they want to pass the list as a single variable (which will need parsing into its components using PL/SQL as per Partha's solution).

create table t(id number);

insert into t values(1);

insert into t values(2);

select * from
(select t.id, x.column_value x,
count(distinct x.column_value) over () count_x,
count(distinct t.id) over () count_t
from t, table(sys.dbms_debug_vc2coll(1,2,3)) x
where t.id(+) = to_number(x.column_value))
where count_x = count_t
/

select * from
(select t.id, x.column_value x,
count(distinct x.column_value) over () count_x,
count(distinct t.id) over () count_t
from t, table(sys.dbms_debug_vc2coll(1,2)) x
where t.id(+) = to_number(x.column_value))
where count_x = count_t
/

select * from
(select t.id, x.column_value x,
count(distinct x.column_value) over () count_x,
count(distinct t.id) over () count_t
from t, table(sys.dbms_debug_vc2coll(1)) x
where t.id(+) = to_number(x.column_value))
where count_x = count_t
/

Re: SQL problem

Jeffrey Kemp — Wed, 12 Oct 2005 01:33:59 -0000

Perhaps simplistic, but here's my take on this problem:

jeffkemponoracle.blogspot.com/2005/10/sql-problem.html

Re: SQL problem

Mr. Ed — Wed, 12 Oct 2005 00:52:25 -0000

OK, another (shorter) try at posting this (id_domain contains 1..1000):

select id
from (
select t.id, count(distinct t.id) over () cnt1, cnt2
from (
select id, count(id) over () cnt2
from id_domain
where id in (1,2,17,17)
) p, id_tab t
where t.id = p.id
)
where cnt1 = cnt2;

select id
from (
select id
from id_tab
where id in (1,2,17,17)
)
where not exists (
select null
from id_domain
where id in (1,2,17,17)
and id not in (
select id
from id_tab
)
);

Re: SQL problem

Mr. Ed — Wed, 12 Oct 2005 00:46:14 -0000

How about these two methods:

create table id_tab (id number);
insert into id_tab select rn
from (select rownum rn from dba_objects) where mod(rn, 20) != 0;
insert into id_tab select rn
from (select rownum rn from dba_objects) where mod(rn, 10) != 0;
insert into id_tab select rn
from (select rownum rn from dba_objects) where mod(rn, 5) != 0;
commit;

create table id_domain (id number);
insert into id_domain select rownum from dba_objects where rownum

Re: SQL problem

Eddie Awad — Tue, 11 Oct 2005 13:23:04 -0000

Partha, thanks, your solution works great (even though it needed some PL/SQL help). However, a minor modification to the query is needed to make it work for both numeric and alphanumeric input. In addition to removing the `to_number`, a `to_char` needs to be added as well. that way, it'll work for both numeric and alphanumeric all the time:

scott: select * from t
2 where 0=(select cnt_a-cnt_b
3 from (select count(*) cnt_a
4 from (select column_value
5 from table(cast (my_comma_to_table('1,a,2') as tablist))
6 union select id from t)) a,
7 (select count(*) cnt_b from t) b
8 )
9 /
from (select column_value
*
ERROR at line 4:
ORA-01790: expression must have same datatype as
corresponding expression

no rows selected

Re: SQL problem

Partha — Tue, 11 Oct 2005 11:59:29 -0000

There is a more generic way, but it would involve creating a SQL type and a function to convert comma-seperated list to a collection. Then using set logic you can satisfy the requirement and it will handle duplicate values as well in the input. The solution given below has been tested using the table structure as you have shown above, but it can handle any type of table:

create or replace type tablist as table of varchar2(255);

create or replace function my_comma_to_table(csv varchar2)
return tablist is
l_string long default csv || ',';
l_data tablist := tablist();
n number;
begin
loop
exit when l_string is null;
n := instr( l_string, ',' );
l_data.extend;
l_data(l_data.count) :=
ltrim( rtrim( substr( l_string, 1, n-1 ) ) );
l_string := substr( l_string, n+1 );
end loop;
return l_data;
end my_comma_to_table;

select * from t
where 0=(select cnt_a-cnt_b
from (select count(*) cnt_a
from (select to_number(column_value)
from table(cast (my_comma_to_table('1,2,2') as tablist))
union select id from t)) a,
(select count(*) cnt_b from t) b
)
/

The above would handle even when the list contains alphanumeric values. In that case just remove the to\_number function in the select query.

Regards
Partha