Subnet (or CIDR) IP control at T-SQL - tsql

Question

I don't know how to exactly explain, but there is a problem about selecting and comparing query IP subnet. For example, there is a list for IP address and I have another CIDR/subnet mask list (X.X.X.0/24 etc). How can I learn that each IP address in first list is in CIDR/subnet mask list via T-SQL?
For example:
IP: 172.28.112.23 -> false
IP: 172.28.111.33 -> true
IP List Output:
SubNet Output:

Answer 1

You want to do exactly what a computer would do to determine if an ip address is in a subnet- ie:
1) convert the network address, subnet mask and test address to binary.
2) Check if (Network Address & Subnet Mask) = (Test Address & Subnet mask)
(& represents bitwise AND)
If this comparison is true the test address is within the subnet
The key to understanding this is to realise that IP addresses (and subnet masks) are just 32 bit numbers.
A bitwise and between 2 32 bit numbers creates a new 32 bit number with a 1 in the position where there was a 1 in both of the 2 numbers being compared, and a 0 otherwise.
EG: 1010 & 1100 = 1000 because the first digit is 1 in both numbers (yielding a 1 in the result for the first digit), but the 2nd 3rd and 4th digits are not (so give 0 in the result for the 2nd 3rd and 4th digits).
SQL Server cannot do a bitwise and between 2 binary numbers unfortunately, but it works fine between decimal representations (ie when converted to BIGINT datatype).
Therefore I'd propose you create a function that converts your IP addresses to BIGINT datatype firstly
CREATE FUNCTION dbo.fnIPtoBigInt
(
#Ipaddress NVARCHAR(15) -- should be in the form '123.123.123.123'
)
RETURNS BIGINT
AS
BEGIN
DECLARE #part1 AS NVARCHAR(3)
DECLARE #part2 AS NVARCHAR(3)
DECLARE #part3 AS NVARCHAR(3)
DECLARE #part4 AS NVARCHAR(3)
SELECT #part1 = LEFT(#Ipaddress, CHARINDEX('.',#Ipaddress) - 1)
SELECT #Ipaddress = SUBSTRING(#Ipaddress, LEN(#part1) + 2, 15)
SELECT #part2 = LEFT(#Ipaddress, CHARINDEX('.',#Ipaddress) - 1)
SELECT #Ipaddress = SUBSTRING(#Ipaddress, LEN(#part2) + 2, 15)
SELECT #part3 = LEFT(#Ipaddress, CHARINDEX('.',#Ipaddress) - 1)
SELECT #part4 = SUBSTRING(#Ipaddress, LEN(#part3) + 2, 15)
DECLARE #ipAsBigInt AS BIGINT
SELECT #ipAsBigInt =
(16777216 * (CAST(#part1 AS BIGINT)))
+ (65536 * (CAST(#part2 AS BIGINT)))
+ (256 * (CAST(#part3 AS BIGINT)))
+ (CAST(#part4 AS BIGINT))
RETURN #ipAsBigInt
END
GO
Then you can easily implement a function to test if an address is in a subnet:
CREATE FUNCTION dbo.fnIsIpaddressInSubnet
(
#networkAddress NVARCHAR(15), -- 'eg: '192.168.0.0'
#subnetMask NVARCHAR(15), -- 'eg: '255.255.255.0' for '/24'
#testAddress NVARCHAR(15) -- 'eg: '192.168.0.1'
)
RETURNS BIT AS
BEGIN
RETURN CASE WHEN (dbo.fnIPtoBigInt(#networkAddress) & dbo.fnIPtoBigInt(#subnetMask))
= (dbo.fnIPtoBigInt(#testAddress) & dbo.fnIPtoBigInt(#subnetMask))
THEN 1 ELSE 0 END
END
To make this a bit easier for you you'll probably want a function that can convert '/24' to a BigInt too.
'/24' is a shorthand way of writing 255.255.255.0 - ie a 32bit number with the first 24bits set to 1 (and the remaining 8 bits set to 0)
CREATE FUNCTION dbo.fnSubnetBitstoBigInt
(
#SubnetBits TINYINT -- max = 32
)
RETURNS BIGINT
AS
BEGIN
DECLARE #multiplier AS BIGINT = 2147483648
DECLARE #ipAsBigInt AS BIGINT = 0
DECLARE #bitIndex TINYINT = 1
WHILE #bitIndex <= #SubnetBits
BEGIN
SELECT #ipAsBigInt = #ipAsBigInt + #multiplier
SELECT #multiplier = #multiplier / 2
SELECT #bitIndex = #bitIndex + 1
END
RETURN #ipAsBigInt
END
GO
If you create the following additional function the conversion becomes easy
CREATE FUNCTION dbo.fnIsIpaddressInSubnetShortHand
(
#network NVARCHAR(18), -- 'eg: '192.168.0.0/24'
#testAddress NVARCHAR(15) -- 'eg: '192.168.0.1'
)
RETURNS BIT AS
BEGIN
DECLARE #networkAddress NVARCHAR(15)
DECLARE #subnetBits TINYINT
SELECT #networkAddress = LEFT(#network, CHARINDEX('/', #network) - 1)
SELECT #subnetBits = CAST(SUBSTRING(#network, LEN(#networkAddress) + 2, 2) AS TINYINT)
RETURN CASE WHEN (dbo.fnIPtoBigInt(#networkAddress) & dbo.fnSubnetBitstoBigInt(#subnetBits))
= (dbo.fnIPtoBigInt(#testAddress) & dbo.fnSubnetBitstoBigInt(#subnetBits))
THEN 1 ELSE 0 END
END
i.e.
SELECT dbo.fnIsIpaddressInSubnetShorthand('192.168.2.0/24','192.168.3.91') -- returns 0
SELECT dbo.fnIsIpaddressInSubnetShorthand('192.168.2.0/24','192.168.2.91') -- returns 1

Answer 2

This is not an answer in itself, but a way to make one of the functions in James S answer easier to read and possibly more efficient.
SQL Server has a function to handle getting parts from database object names. Those names are 4 parts [Server].[Database].[Schema].[Object]. So the following allows you to get the schema name. The index works from the right
SELECT PARSENAME('[myServer].[master].[sys].[objects]', 2)
There's nothing to say you can't use that for an IP address. And as it's so fundamental to how SQL operates I assume it has been hella optimised.
CREATE FUNCTION dbo.fnIPtoBigInt
(
#Ipaddress NVARCHAR(15) -- should be in the form '123.123.123.123'
)
RETURNS BIGINT
AS
BEGIN
DECLARE #ipAsBigInt AS BIGINT
SELECT #ipAsBigInt =
(16777216 * (CAST(PARSENAME(#Ipaddress, 4) AS BIGINT)))
+ (65536 * (CAST(PARSENAME(#Ipaddress, 3) AS BIGINT)))
+ (256 * (CAST(PARSENAME(#Ipaddress, 2) AS BIGINT)))
+ (CAST(PARSENAME(#Ipaddress, 1) AS BIGINT))
RETURN #ipAsBigInt
END
GO

Answer 3

Complete solution
CREATE
OR ALTER FUNCTION dbo.IPv4SubnetContainsIPAddress (
#net AS VARCHAR(15),
#mask AS VARCHAR(15),
#ip AS VARCHAR(15)
) RETURNS tinyint AS BEGIN DECLARE #result AS tinyint IF LEN(#mask) <= 2
SELECT
#mask = m
FROM
(
VALUES
(0, '0.0.0.0'),
(1, '128.0.0.0'),
(2, '192.0.0.0'),
(3, '224.0.0.0'),
(4, '240.0.0.0'),
(5, '248.0.0.0'),
(6, '252.0.0.0'),
(7, '254.0.0.0'),
(8, '255.0.0.0'),
(9, '255.128.0.0'),
(10, '255.192.0.0'),
(11, '255.224.0.0'),
(12, '255.240.0.0'),
(13, '255.248.0.0'),
(14, '255.252.0.0'),
(15, '255.254.0.0'),
(16, '255.255.0.0'),
(17, '255.255.128.0'),
(18, '255.255.192.0'),
(19, '255.255.224.0'),
(20, '255.255.240.0'),
(21, '255.255.248.0'),
(22, '255.255.252.0'),
(23, '255.255.254.0'),
(24, '255.255.255.0'),
(25, '255.255.255.128'),
(26, '255.255.255.192'),
(27, '255.255.255.224'),
(28, '255.255.255.240'),
(29, '255.255.255.248'),
(30, '255.255.255.252'),
(31, '255.255.255.254'),
(32, '255.255.255.255')
) AS o (i, m)
WHERE
i = #mask
SELECT
#result = IIF(Count(*) = 4, 1, 0)
FROM
(
SELECT
*,
IIF(
o_ip BETWEEN o_subnet
AND o_broadcast,
1,
0
) AS eq
FROM
(
SELECT
*,
o_net & o_mask AS o_subnet,
o_net | (255 - o_mask) AS o_broadcast
FROM
(
SELECT
o_net,
o_mask,
o_ip
FROM
(
VALUES
(1, CAST(PARSENAME(#net, 4) AS INTEGER)),
(2, CAST(PARSENAME(#net, 3) AS INTEGER)),
(3, CAST(PARSENAME(#net, 2) AS INTEGER)),
(4, CAST(PARSENAME(#net, 1) AS INTEGER))
) AS c1 (i, o_net)
LEFT JOIN (
SELECT
i,
o_mask
FROM
(
VALUES
(1, CAST(PARSENAME(#mask, 4) AS INTEGER)),
(2, CAST(PARSENAME(#mask, 3) AS INTEGER)),
(3, CAST(PARSENAME(#mask, 2) AS INTEGER)),
(4, CAST(PARSENAME(#mask, 1) AS INTEGER))
) AS c2 (i, o_mask)
) AS c2 ON c1.i = c2.i
LEFT JOIN (
SELECT
i,
o_ip
FROM
(
VALUES
(1, CAST(PARSENAME(#ip, 4) AS INTEGER)),
(2, CAST(PARSENAME(#ip, 3) AS INTEGER)),
(3, CAST(PARSENAME(#ip, 2) AS INTEGER)),
(4, CAST(PARSENAME(#ip, 1) AS INTEGER))
) AS c3 (i, o_ip)
) AS c3 ON c1.i = c3.i
) AS t
) AS t
) AS t
WHERE
eq = 1 RETURN #result END
GO
SELECT
dbo.IPv4SubnetContainsIPAddress('192.168.64.0', '255.255.224.0', '192.168.40.1') -- returns 0
SELECT
dbo.IPv4SubnetContainsIPAddress('192.168.64.0', '19', '192.168.40.1') -- returns 0
SELECT
dbo.IPv4SubnetContainsIPAddress('192.168.64.0', '255.255.192.0', '192.168.80.1') -- returns 1
SELECT
dbo.IPv4SubnetContainsIPAddress('192.168.64.0', '18', '192.168.80.1') -- returns 1