Oracle IN condition without sort - database

Question

May I know is there any solution to get the result without ordering in Oracle? It is because when I execute the query as follows, it seems to automatically helps me to sort it by ID field.
SELECT ID FROM USER WHERE ID IN (5004, 5003, 5005, 5002, 5008);
Actual results Expected results
---5002 ---5004
---5003 ---5003
---5004 ---5005
---5005 ---5002
---5008 ---5008
Million thanks if you guys have solutions on this.

Answer 1

SELECT statements return the rows of their result sets in an unpredictable order unless you give an ORDER BY clause.
Certain DBMS products give the illusion that their result sets are in a predictable order. But if you rely on that you're bound to be disappointed.

Answer 2

This is one way I've seen in the past using INSTR:
SELECT *
FROM YourTable
WHERE ID IN (5004, 5003, 5005, 5002, 5008)
ORDER BY INSTR ('5004,5003,5005,5002,5008', id)
SQL Fiddle Demo
I've also seen use of CASE like this:
ORDER BY
CASE ID
WHEN 5004 THEN 1
WHEN 5003 THEN 2
WHEN 5005 THEN 3
WHEN 5002 THEN 4
WHEN 5008 THEN 5
END

Answer 3

if you want to keep the order as your in list, you can do something like this:
SQL> create type user_va as varray(1000) of number;
2 /
Type created.
SQL> with users as (select /*+ cardinality(a, 10) */ rownum r, a.column_value user_id
2 from table(user_va(11, 0, 19, 5)) a)
3 select d.user_id, d.username
4 from dba_users d
5 inner join users u
6 on u.user_id = d.user_id
7 order by u.r
8 /
USER_ID USERNAME
---------- ------------------------------
11 OUTLN
0 SYS
19 DIP
5 SYSTEM
i.e we put the elements into a varray and assign a rownum prior to merging the set. we can then order by that r to maintain the order of our in list. The cardinality hint just tells the optimizer how many rows are in the array (doesn't have to be dead on, just in the ballpark..as without this, it will assume 8k rows and may prefer a full scan over an index approach)
if you don't have privs to create a type and this is just some adhoc thing, there's a few public ones:
select owner, type_name, upper_bound max_elements, length max_size, elem_type_name
from all_Coll_types
where coll_type = 'VARYING ARRAY'
and elem_type_name in ('INTEGER', 'NUMBER');

Answer 4

There is no guarantee of sort order without an ORDER BY clause.

Answer 5

If your question is about why the ordering occurs then the answer is: Do you have an index or primary key defined on the column ID? If yes the database responds to your query with an index scan. That is: it looks up the IDs in the IN clause not in the table itself but in the index defined on your ID-column. Within the index the values are ordered.
To get more information about the execution of your query try Oracle's explain plan feature.
To get the values in a certain order you have to add an ORDER BY clause. One way of doing this would be
select ID
from USER
where ID in (5004, 5003, 5005, 5002, 5008)
order by
case ID
when 5004 then 1
when 5003 then 2
...
end;
A more general way would be to add an ORDERING column to your table:
select ID
from USER
where ID in (5004, 5003, 5005, 5002, 5008)
order by
ORDERING;

Answer 6

Another solution that I found here.
select ID
from USER
where ID in (5004, 5003, 5005, 5002, 5008)
order by decode(ID, 5002, 1, 5003, 2, 5004, 3, 5005, 4, 5008, 5);
order by decode(COLUMN NAME, VALUE, POSITION)
*Note: Only need to repeat the VALUE and POSITION
And yah, thanks for all the responds! I am really appreciate it.