Select record between two IP ranges - sql

Question

I have a table which stores a ID, Name, Code, IPLow, IPHigh such as:
1, Lucas, 804645, 192.130.1.1, 192.130.1.254
2, Maria, 222255, 192.168.2.1, 192.168.2.254
3, Julia, 123456, 192.150.3.1, 192.150.3.254
Now, if I have an IP address 192.168.2.50, how can I retrieve the matching record?
Edit
Based on Gordon's answer (which I'm getting compilation errors) this is what I have:
select PersonnelPC.*
from (select PersonnelPC.*,
(
cast(parsename(iplow, 4)*1000000000 as decimal(12, 0)) +
cast(parsename(iplow, 3)*1000000 as decimal(12, 0)) +
cast(parsename(iplow, 2)*1000 as decimal(12, 0)) +
(parsename(iplow, 1))
) as iplow_decimal,
(
cast(parsename(iphigh, 4)*1000000000 as decimal(12, 0)) +
cast(parsename(iphigh, 3)*1000000 as decimal(12, 0)) +
cast(parsename(iphigh, 2)*1000 as decimal(12, 0)) +
(parsename(iphigh, 1))
) as iphigh_decimal
from PersonnelPC
) PersonnelPC
where 192168002050 between iplow_decimal and iphigh_decimal;
but this gives me an error:
Msg 8115, Level 16, State 2, Line 1
Arithmetic overflow error converting expression to data type int.
Any ideas?

Answer 1

Painfully. SQL Server has lousy string manipulation functions. It does, however, offer parsename(). This approach converts the IP address to a large decimal value for the comparison:
select t.*
from (select t.*,
(cast(parsename(iplow, 4)*1000000000.0 as decimal(12, 0)) +
cast(parsename(iplow, 3)*1000000.0 as decimal(12, 0)) +
cast(parsename(iplow, 2)*1000.0 as decimal(12, 0)) +
cast(parsename(iplow, 1) as decimal(12, 0))
) as iplow_decimal,
(cast(parsename(iphigh, 4)*1000000000.0 as decimal(12, 0)) +
cast(parsename(iphigh, 3)*1000000.0 as decimal(12, 0)) +
cast(parsename(iphigh, 2)*1000.0 as decimal(12, 0)) +
cast(parsename(iphigh, 1) as decimal(12, 0))
) as iphigh_decimal
from t
) t
where 192168002050 between iplow_decimal and iphigh_decimal;
I should note that IP addresses are often stored in the database as the 4-byte unsigned integers. This makes comparisons much easier . . . although you need complicated logic (usually wrapped in a function) to convert the values to a readable format.

Answer 2

Try this simple way checking range
DECLARE #IP NVARCHAR(30)='192.168.500.1'
SELECT * FROM
Branches
WHERE
CAST (PARSENAME(#IP,4) AS INT)>=CAST(PARSENAME(IPLow,4) AS INT) AND CAST(PARSENAME(#IP,3) AS INT)>=CAST(PARSENAME(IPLow,3) AS INT) AND CAST(PARSENAME(#IP,2) AS INT)>=CAST(PARSENAME(IPLow,2) AS INT) AND CAST(PARSENAME(#IP,1) AS INT)>=CAST(PARSENAME(IPLow,1) AS INT)
AND
CAST(PARSENAME( #IP,4) AS INT) <= CAST(PARSENAME(IPHigh ,4) AS INT) AND CAST(PARSENAME(#IP ,3) AS INT) <=CAST(PARSENAME(IPHigh ,3) AS INT) AND CAST(PARSENAME(#IP ,2) AS INT) <=CAST(PARSENAME(IPHigh ,2) AS INT) AND CAST(PARSENAME(#IP ,1) AS INT)<=CAST(PARSENAME(IPHigh ,1) AS INT)
AS Per #Ed Haper Comment Cast is needed.

Answer 3

With this function you can transform any IP address to a form where each part has 3 digits. With this you could do a normal alphanumeric compare. if you want you could return BIGINT too...
CREATE FUNCTION dbo.IPWidth3(#IP VARCHAR(100))
RETURNS VARCHAR(15)
BEGIN
DECLARE #RetVal VARCHAR(15);
WITH Splitted AS
(
SELECT CAST('<x>' + REPLACE(#IP,'.','</x><x>') + '</x>' AS XML) AS IPSplitted
)
SELECT #RetVal = STUFF(
(
SELECT '.' + REPLACE(STR(Part.value('.','int'),3),' ','0')
FROM Splitted.IPSplitted.nodes('/x') AS One(Part)
FOR XML PATH('')
),1,1,'')
FROM Splitted;
RETURN #RetVal;
END
GO
DECLARE #IP VARCHAR(100)='192.43.2.50';
SELECT dbo.IPWidth3(#IP);
The result
192.043.002.050
To reflect Ed Harper's comment here the same function returning a DECIMAL(12,0):
CREATE FUNCTION dbo.IP_as_Number(#IP VARCHAR(100))
RETURNS DECIMAL(12,0)
BEGIN
DECLARE #RetVal DECIMAL(12,0);
WITH Splitted AS
(
SELECT CAST('<x>' + REPLACE(#IP,'.','</x><x>') + '</x>' AS XML) AS IPSplitted
)
SELECT #RetVal =
CAST((
SELECT REPLACE(STR(Part.value('.','int'),3),' ','0')
FROM Splitted.IPSplitted.nodes('/x') AS One(Part)
FOR XML PATH('')
) AS DECIMAL(12,0))
FROM Splitted;
RETURN #RetVal;
END
GO
DECLARE #IP VARCHAR(100)='192.43.2.50';
SELECT dbo.IP_as_Number(#IP);

Answer 4

Use below to fetch the ipLow / IPHigh in 4 columns. You can use those columns to compare Ips.
DECLARE#ip VARCHAR(50)='192.168.0.81'
SELECT (SUBSTRING((#ip), 0,
patindex('%.%',
(#ip))))
,
substring((REPLACE(#ip, (SUBSTRING((#ip), 0,
patindex('%.%',
(#ip)) + 1)),
'')),
0,
patindex('%.%',
((REPLACE(#ip, (SUBSTRING((#ip), 0,
patindex('%.%',
(#ip)) + 1)),
''))))),
SUBSTRING((SUBSTRING(#ip, LEN((SUBSTRING((#ip), 0,
patindex('%.%',
(#ip))))) + 2 + LEN(substring((REPLACE(#ip, (SUBSTRING((#ip), 0,
patindex('%.%',
(#ip)) + 1)),
'')),
0,
patindex('%.%',
((REPLACE(#ip, (SUBSTRING((#ip), 0,
patindex('%.%',
(#ip)) + 1)),
'')))))) + 1,
LEN(#IP) - 1 - LEN(reverse(SUBSTRING(reverse(#ip), 0,
patindex('%.%',
reverse(#ip))))))), 0,
PATINDEX('%.%',
(SUBSTRING(#ip, LEN((SUBSTRING((#ip), 0,
patindex('%.%',
(#ip))))) + 2 + LEN(substring((REPLACE(#ip, (SUBSTRING((#ip), 0,
patindex('%.%',
(#ip)) + 1)),
'')),
0,
patindex('%.%',
((REPLACE(#ip, (SUBSTRING((#ip), 0,
patindex('%.%',
(#ip)) + 1)),
'')))))) + 1,
LEN(#IP) - 1 - LEN(reverse(SUBSTRING(reverse(#ip), 0,
patindex('%.%',
reverse(#ip))))))
))),
reverse(SUBSTRING(reverse(#ip), 0,
patindex('%.%',
reverse(#ip))))

Answer 5

Consider something like this example to convert the address into a number.
CREATE FUNCTION dbo.IPAddressAsNumber (#IPAddress AS varchar(15))
RETURNS bigint
BEGIN
RETURN
CONVERT (bigint,
CONVERT(varbinary(1), CONVERT(int, PARSENAME(#IPAddress, 4))) +
CONVERT(varbinary(1), CONVERT(int, PARSENAME(#IPAddress, 3))) +
CONVERT(varbinary(1), CONVERT(int, PARSENAME(#IPAddress, 2))) +
CONVERT(varbinary(1), CONVERT(int, PARSENAME(#IPAddress, 1))) )
END
and with that you could use standard operators like BETWEEN to find rows within the range you have in the table
DECLARE #t table (ID int, Name varchar(50), Code int, IPLow varchar(15), IPHigh varchar(15))
INSERT INTO #t VALUES
(1, 'Lucas', 804645, '192.130.1.1', '192.130.1.254'),
(2, 'Maria', 222255, '192.168.2.1', '192.168.2.254'),
(3, 'Julia', 123456, '192.150.3.1', '192.150.3.254')
SELECT * FROM #t
WHERE dbo.IPAddressAsNumber('192.168.2.50')
BETWEEN dbo.IPAddressAsNumber(IPLow) AND dbo.IPAddressAsNumber(IPHigh)
The scheme essentially uses PARSENAME to isolate each part of the address, converts each part into a SQL binary string, concatenating the strings together to get a single SQL binary string representing the address, and shows the result as a bigint.
In a textual representation of hexadecimal values think of this as smashing the 4 parts together 192(0xC0) + 168(0xA8) + 2(0x02) + 50(0x32) into 0xC0A80232. When you turn that combined string into its binary digits (0s and 1s) you would end up with something that could be thought of as the address in a binary form used by the network stack in address routing and subnet masking tables. When you turn that into a number in the form of an unsigned integer (or in this case a bigint) you get 3232236082.
Interestingly this scheme gives you a "number" that can be used in place of the original address in lots of ways. You can for example ping the number 2130706433 instead of the address 127.0.0.1 -- the name resolver in Windows will convert it similarly to how DNS is used to find the address of a hostname.
For the sake of completeness, here is another function that can be used to convert the number form back into the standard string form
CREATE FUNCTION dbo.IPAddressFromNumber (#IPNumber AS bigint)
RETURNS varchar(15)
BEGIN
RETURN
CONVERT (varchar(15),
CONVERT(varchar(3), CONVERT(int, SUBSTRING(CONVERT(varbinary(4), #IPNumber), 1,1))) + '.' +
CONVERT(varchar(3), CONVERT(int, SUBSTRING(CONVERT(varbinary(4), #IPNumber), 2,1))) + '.' +
CONVERT(varchar(3), CONVERT(int, SUBSTRING(CONVERT(varbinary(4), #IPNumber), 3,1))) + '.' +
CONVERT(varchar(3), CONVERT(int, SUBSTRING(CONVERT(varbinary(4), #IPNumber), 4,1))) )
END

Answer 6

select *
from ip a
join ip_details b
on a.ip_address >= b.ip_start
and a.ip_address <= b.ip_end;
In this, table "a" contains list of IP address and table "b" contains the IP ranges.
Instead of converting the ip address to numeric we can directly compare the string, it will do a byte by byte comparison.
This is working for me(PostgreSQL).

Answer 7

I was thinking along the lines of Gordon's answer, then realized you don't actually need to mess with numbers. If you zero-pad each part of the address, a string comparison works:
DECLARE #search varchar(50) = '192.168.2.50';
WITH DATA AS (
SELECT * FROM ( values
(1, 'Lucas', '192.130.1.1', '192.130.1.254'),
(2, 'Maria', '192.168.2.1', '192.168.2.254'),
(3, 'Julia', '192.150.3.1', '192.150.3.254')
) AS tbl (ID,Name,IPLow,IPHigh)
)
SELECT *
FROM DATA
WHERE REPLACE(STR(PARSENAME( #search, 4 ), 3, 0), ' ', '0')
+ REPLACE(STR(PARSENAME( #search, 3 ), 3, 0), ' ', '0')
+ REPLACE(STR(PARSENAME( #search, 2 ), 3, 0), ' ', '0')
+ REPLACE(STR(PARSENAME( #search, 1 ), 3, 0), ' ', '0')
BETWEEN
REPLACE(STR(PARSENAME( IPLow, 4 ), 3, 0), ' ', '0')
+ REPLACE(STR(PARSENAME( IPLow, 3 ), 3, 0), ' ', '0')
+ REPLACE(STR(PARSENAME( IPLow, 2 ), 3, 0), ' ', '0')
+ REPLACE(STR(PARSENAME( IPLow, 1 ), 3, 0), ' ', '0')
AND
REPLACE(STR(PARSENAME( IPHigh, 4 ), 3, 0), ' ', '0')
+ REPLACE(STR(PARSENAME( IPHigh, 3 ), 3, 0), ' ', '0')
+ REPLACE(STR(PARSENAME( IPHigh, 2 ), 3, 0), ' ', '0')
+ REPLACE(STR(PARSENAME( IPHigh, 1 ), 3, 0), ' ', '0')
You can, of course, put this inside a UDF for simplicity, though watch out for the performance hit on large queries.
CREATE FUNCTION dbo.IP_Comparable(#IP varchar(50))
RETURNS varchar(50)
WITH SCHEMABINDING
BEGIN
RETURN REPLACE(STR(PARSENAME( #IP, 4 ), 3, 0), ' ', '0')
+ REPLACE(STR(PARSENAME( #IP, 3 ), 3, 0), ' ', '0')
+ REPLACE(STR(PARSENAME( #IP, 2 ), 3, 0), ' ', '0')
+ REPLACE(STR(PARSENAME( #IP, 1 ), 3, 0), ' ', '0')
END
GO
DECLARE #search varchar(50) = '192.168.2.50';
WITH DATA AS (
SELECT * FROM ( values
(1, 'Lucas', '192.130.1.1', '192.130.1.254'),
(2, 'Maria', '192.168.2.1', '192.168.2.254'),
(3, 'Julia', '192.150.3.1', '192.150.3.254')
) AS tbl (ID,Name,IPLow,IPHigh)
)
SELECT *
FROM DATA
WHERE dbo.IP_Comparable(#search) BETWEEN dbo.IP_Comparable(IPLow) AND dbo.IP_Comparable(IPHigh)
This will avoid the issue you're having with integer overflows.

Answer 8

Depends on which record you are looking for the high or the low.
select * from table where IPlow like '192.168.2.50' or IPHigh like '192.168.2.50'