'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 |