Split IPv4 address into 4 numbers in Oracle sql

user_8795235 0
Viewed 1515

I'm trying to split a given IPv4 address into four numbers.

In SQL Server this query works well for me:

select CAST (PARSENAME('10.20.30.40',4) AS INT)

result: 10

select CAST (PARSENAME('10.20.30.40',3) AS INT)

result: 20

and so on.

I need the equivalent syntax in Oracle SQL, but can't find it. Any idea?

oracle sql-server sql
3 Answers

You could use regexp_substr:

select ip,
       regexp_substr(ip, '\d+',1,1) as first_octet,
       regexp_substr(ip, '\d+',1,2) as second_octet,
       regexp_substr(ip, '\d+',1,3) as third_octet,
       regexp_substr(ip, '\d+',1,4) as fourth_octet
from  (select '10.20.30.40' AS ip from dual )ips;

Rextester Demo

user_50708790

You can use simple string functions (INSTR and SUBSTR) that are much faster than regular expressions:

SQL Fiddle

Oracle 11g R2 Schema Setup:

CREATE TABLE sample_data ( ip_address ) AS
SELECT '10.20.30.40' FROM DUAL

Query 1:

SELECT TO_NUMBER(
         SUBSTR( ip_address, 1, first_sep - 1 )
       ) AS ClassA,
       TO_NUMBER(
         SUBSTR( ip_address, first_sep + 1, second_sep - first_sep )
       ) AS ClassB,
       TO_NUMBER(
         SUBSTR( ip_address, second_sep + 1, third_sep - second_sep )
       ) AS ClassC,
       TO_NUMBER(
         SUBSTR( ip_address, third_sep + 1 )
       ) AS ClassD
FROM   (
  SELECT ip_address,
         INSTR( ip_address, '.', 1, 1 ) AS first_sep,
         INSTR( ip_address, '.', 1, 2 ) AS second_sep,
         INSTR( ip_address, '.', 1, 3 ) AS third_sep
  FROM   sample_data
)

Results:

| CLASSA | CLASSB | CLASSC | CLASSD |
|--------|--------|--------|--------|
|     10 |     20 |     30 |     40 |
user_15092640

In case you need all in one function, this is another solution:

SELECT REGEXP_SUBSTR('10.20.30.40', '\d+', 1, LEVEL) as octet, level
FROM dual 
CONNECT BY LEVEL <= 4;


OCTET   LEVEL
10      1
20      2
30      3
40      4
user_30272660
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