Before we start with code let us take a sample IP address, does 127.0.0.1 look familiar? Yes that is your local IP address.

Here it is in decimal and binary
127 0 0 1
01111111 00000000 00000000 00000001

Now to convert, you would take the first value,
add the second value + 256
add the third value + (256 * 256) = 65536
add the fourth value + (256 * 256 * 256) =16777216

So in our case the select would be

T-SQL
1
2
3
4
5
select
1 +
0 * 256 +
0 * 65536 +
127 * 16777216
select
1 +
0 * 256 +
0 * 65536 +
127 * 16777216

which is
2130706433

So to convert from IP Adress to integer is very simple, you use PARSENAME to split it up and do the math. Here is the function.

T-SQL
1
2
3
4
5
6
7
8
9
10
11
CREATE FUNCTION dbo.IPAddressToInteger (@IP AS varchar(15))
RETURNS bigint
AS
BEGIN
 RETURN (CONVERT(bigint, PARSENAME(@IP,1)) +
         CONVERT(bigint, PARSENAME(@IP,2)) * 256 +
         CONVERT(bigint, PARSENAME(@IP,3)) * 65536 +
         CONVERT(bigint, PARSENAME(@IP,4)) * 16777216)
 
END
GO
CREATE FUNCTION dbo.IPAddressToInteger (@IP AS varchar(15))
RETURNS bigint
AS
BEGIN
 RETURN (CONVERT(bigint, PARSENAME(@IP,1)) +
         CONVERT(bigint, PARSENAME(@IP,2)) * 256 +
         CONVERT(bigint, PARSENAME(@IP,3)) * 65536 +
         CONVERT(bigint, PARSENAME(@IP,4)) * 16777216)

END
GO

But how do you get 127.0.0.1 out of 2130706433?
It is the reversed of what we did before (surprise) so instead of multiplying we will be dividing
Here is the funcion

T-SQL
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
CREATE FUNCTION dbo.IntegerToIPAddress (@IP AS bigint)
RETURNS varchar(15)
AS
BEGIN
 DECLARE @Octet1 bigint
 DECLARE @Octet2 tinyint
 DECLARE @Octet3 tinyint
 DECLARE @Octet4 tinyint
 DECLARE @RestOfIP bigint
 
 SET @Octet1 = @IP / 16777216
 SET @RestOfIP = @IP - (@Octet1 * 16777216)
 SET @Octet2 = @RestOfIP / 65536
 SET @RestOfIP = @RestOfIP - (@Octet2 * 65536)
 SET @Octet3 = @RestOfIP / 256
 SET @Octet4 = @RestOfIP - (@Octet3 * 256)
 
 RETURN(CONVERT(varchar, @Octet1) + '.' +
        CONVERT(varchar, @Octet2) + '.' +
        CONVERT(varchar, @Octet3) + '.' +
        CONVERT(varchar, @Octet4))
END
CREATE FUNCTION dbo.IntegerToIPAddress (@IP AS bigint)
RETURNS varchar(15)
AS
BEGIN
 DECLARE @Octet1 bigint
 DECLARE @Octet2 tinyint
 DECLARE @Octet3 tinyint
 DECLARE @Octet4 tinyint
 DECLARE @RestOfIP bigint

 SET @Octet1 = @IP / 16777216
 SET @RestOfIP = @IP - (@Octet1 * 16777216)
 SET @Octet2 = @RestOfIP / 65536
 SET @RestOfIP = @RestOfIP - (@Octet2 * 65536)
 SET @Octet3 = @RestOfIP / 256
 SET @Octet4 = @RestOfIP - (@Octet3 * 256)

 RETURN(CONVERT(varchar, @Octet1) + '.' +
        CONVERT(varchar, @Octet2) + '.' +
        CONVERT(varchar, @Octet3) + '.' +
        CONVERT(varchar, @Octet4))
END

Now let’s try this out, first run this

T-SQL
1
select dbo.IPAddressToInteger('127.0.0.1')
select dbo.IPAddressToInteger('127.0.0.1')

That returns 2130706433
Now run this

T-SQL
1
select dbo.IntegerToIPAddress(2130706433)
select dbo.IntegerToIPAddress(2130706433)

That returns 127.0.0.1

Thanks to K. Brian Kelley for the inspiration for this post, you can also check http://www.truthsolutions.com/ to see some of his books

And also check out the related Order IP Addresses wiki article which I wrote a while ago