'I'm trying to join 2 tables that don't have same columns and using case select. Is this possible?

select 
    case 
        when code = number_1 
        then number_1=code
        when count(code)>=count(number_1)
        then number_1 = sum(code)
        else 'Null'
    end
from table_1, table_2 
ORDER BY code, number_1 ;

table 1

code value
0 None
1 R
2 W
4 C
8 D
16 U
32 Uown

Table 2

number
0
1
2
3
4
5
8
12
13
16
20
25
26
27
32
43
44
45
60
61
62
63
64
68
70

expected output is

number output
0 0
1 1
2 2
3 2,1
4 4
7 4,2,1
8 8
16 16
32 32
43 32,8,2,1
63 32,16,8,4,2,1
64 null
70 null


Solution 1:[1]

You can use the BITAND function in the JOIN condition:

SELECT t2."NUMBER",
       CASE SUM(t1.code)
       WHEN t2."NUMBER"
       THEN LISTAGG(t1.code, ',') WITHIN GROUP (ORDER BY t1.code DESC)
       END AS output,
       CASE SUM(t1.code)
       WHEN t2."NUMBER"
       THEN LISTAGG(t1.value, ',') WITHIN GROUP (ORDER BY t1.code DESC)
       END AS value_output
FROM   table_2 t2
       INNER JOIN table_1 t1
       ON (   t2."NUMBER" = t1.code
           OR (t1.code > 0 AND BITAND(t2."NUMBER", t1.code) = t1.code))
GROUP BY t2."NUMBER"

Which, for the sample data:

CREATE TABLE table_1 (code, value) AS
SELECT 0,  'None' FROM DUAL UNION ALL
SELECT 1,  'R' FROM DUAL UNION ALL
SELECT 2,  'W' FROM DUAL UNION ALL
SELECT 4,  'C' FROM DUAL UNION ALL
SELECT 8,  'D' FROM DUAL UNION ALL
SELECT 16, 'U' FROM DUAL UNION ALL
SELECT 32, 'Uown' FROM DUAL;

CREATE TABLE Table_2 ("NUMBER") AS
SELECT COLUMN_VALUE
FROM   SYS.ODCINUMBERLIST(
         0,1,2,3,4,5,8,12,13,16,20,25,26,27,32,43,44,45,60,61,62,63,64,68,70
       );

Outputs:

NUMBER OUTPUT VALUE_OUTPUT
0 0 None
1 1 R
2 2 W
3 2,1 W,R
4 4 C
5 4,1 C,R
8 8 D
12 8,4 D,C
13 8,4,1 D,C,R
16 16 U
20 16,4 U,C
25 16,8,1 U,D,R
26 16,8,2 U,D,W
27 16,8,2,1 U,D,W,R
32 32 Uown
43 32,8,2,1 Uown,D,W,R
44 32,8,4 Uown,D,C
45 32,8,4,1 Uown,D,C,R
60 32,16,8,4 Uown,U,D,C
61 32,16,8,4,1 Uown,U,D,C,R
62 32,16,8,4,2 Uown,U,D,C,W
63 32,16,8,4,2,1 Uown,U,D,C,W,R
68 null null
70 null null

db<>fiddle here

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1 MT0