Oracle REGEXP_REPLACE function to find decimal and special characters - sql

Question

I am working with table data that contains strings with decimal and back-slash like below:
info
1/2.2.2
2/1.1.1
3/1.1.11
I need to use a regular expression to replace the data like below:
info
1/2.2
2/1.1
3/1.1

Answer 1

Don't use a (slow) regular expression, use simple (faster) string functions instead:
SELECT info,
CASE
WHEN INSTR(info, '.', 1, 2) > 0
THEN SUBSTR(info, 1, INSTR(info, '.', 1, 2) - 1)
ELSE info
END AS part
FROM table_name;
Which, for the sample data:
CREATE TABLE table_name (info) AS
SELECT '1/2.2.2' FROM DUAL UNION ALL
SELECT '2/1.1.1' FROM DUAL UNION ALL
SELECT '3/1.1.11' FROM DUAL UNION ALL
SELECT '3/1.1' FROM DUAL;
Outputs:
INFO
PART
1/2.2.2
1/2.2
2/1.1.1
2/1.1
3/1.1.11
3/1.1
3/1.1
3/1.1
If you want to update the table then:
UPDATE table_name
SET info = SUBSTR(info, 1, INSTR(info, '.', 1, 2) - 1)
WHERE INSTR(info, '.', 1, 2) > 0
fiddle

Answer 2

For the sake of argument, here's a solution using REGEXP_SUBSTR(). REGEXP_SUBSTR() returns NULL if the pattern is not found. Thanks to MT0 for the CTE so I didn't have to type it up :-)
WITH table_name(ID, info) AS (
SELECT 1, '1/2.2.2' FROM DUAL UNION ALL
SELECT 2, '2/1.1.1' FROM DUAL UNION ALL
SELECT 3, '3/1.1.11' FROM DUAL UNION ALL
SELECT 4, '3/1.1' FROM DUAL UNION ALL
SELECT 5, '4/4' FROM DUAL)
SELECT ID, REGEXP_SUBSTR(info, '\d/\d\.\d') DATA
from table_name;
ID DATA
---------- --------
1 1/2.2
2 2/1.1
3 3/1.1
4 3/1.1
5
5 rows selected.