In one of the recent projects, I realize the bottleneck of the query was an inline function which was converting Hex to Decimal. I optimized the inline function and reduced the query running time to one-tenth of the original running time. Later, I was eager to find out the script my blog readers might be using for hex to decimal conversion. Please leave your comments here and I will consider all the valid answers and publish with due credit to the author in one of the future posts. If the script you have posted here is not your original script, I suggest that you include the source as well.
Reference: Pinal Dave (https://blog.sqlauthority.com)
25 Comments. Leave new
Here is one way to do it:
create function fn_HexToIntnt(@str varchar(16))
returns bigint as begin
select @str=upper(@str)
declare @i int, @len int, @char char(1), @output bigint
select @len=len(@str)
,@i=@len
,@output=case
when @len>0
then 0
end
while (@i>0)
begin
select @char=substring(@str,@i,1), @output=@output
+(ASCII(@char)
-(case
when @char between ‘A’ and ‘F’
then 55
else
case
when @char between ‘0’ and ‘9’
then 48 end
end))
*power(16.,@len-@i)
,@i=@i-1
end
return @output
end
thx, this is the only one that worked for me, for an RFID reader returning decimals from 8 hex characters others ways of doing it returned 10% negative and inconsistent values
hmm..
What about using built-in function Convert?
SELECT CONVERT(INT, 0x00000100)
SELECT CONVERT(VARBINARY(8), 256)
I do not pretend to be the author, but i am using this for a long time
I’ve lately interested in Sql Server’s CLR support so here goes.
First I enable CLR support on Sql Server.
EXEC sp_CONFIGURE ‘clr enabled’ , ‘1’
GO
RECONFIGURE;
GO
Then I create a simple DLL assembly called DBClasses with a static class named StringFormatFunctions and a static method HexToInt.
public partial class StringFormatFunctions
{
[SqlFunction(IsDeterministic = true)]
public static int HexToInt(string input)
{
return int.Parse(input, System.Globalization.NumberStyles.HexNumber);
}
}
Next I load it to Sql Server.
CREATE ASSEMBLY DBClasses
FROM ‘C:\path\DBClasses.dll’
WITH PERMISSION_SET = SAFE
Finally I create a little function for it.
CREATE FUNCTION HexToInt(@input nvarchar)
RETURNS int
AS
EXTERNAL NAME DBClasses.StringFormatFunctions.HexToInt
And now I can start to use it.
SELECT dbo.HexToInt(‘ffff’)
If it is integer make use of implicit convertion
SELECT CONVERT(VARBINARY(8), 659604),0x000A1094*1
If it is decimal, use cast function
SELECT CONVERT(VARBINARY(8), 659604.1),cast(0x07010001C9A56400 as decimal(12,2))
If you are using SQL Server 2008, you can simply use the built-in function CONVERT, with the style 1:
SELECT CONVERT(varchar(100),0x123456789abc123,1)+’?’
SELECT CONVERT(varbinary,’0x0123456789ABC123′,1)+0x0
Notice that on SQL Server 2005, the above queries will not yield an error, but you will get different results (the same as if we used style 0).
Source: https://docs.microsoft.com/en-us/sql/t-sql/functions/cast-and-convert-transact-sql?view=sql-server-2017
Razvan
OK, I’m going to assume that we are starting with a string holding hex digits, and we wnat to convert that into a int.
declare @hex varchar(10);
set @hex = ‘BD12’;
declare @retval int;
set @reval = — 48402 via some magic.
Now, this being the wacky world of SQL, the best solution is gonna be something involving a set operation on tables, so let’s start with a permanaent table mapping hex digits to their values:
CREATE TABLE [HexDigits](
[Digit] [char](1) NOT NULL,
[Value] [int] NOT NULL,
CONSTRAINT [PK_HexDigits] PRIMARY KEY CLUSTERED ([Digit] ASC) )
INSERT INTO [HexDigits] VALUES (‘0’, 0);
INSERT INTO [HexDigits] VALUES (‘1’, 1);
— etc
INSERT INTO [HexDigits] VALUES (‘E’, 14);
INSERT INTO [HexDigits] VALUES (‘F’, 15);
Then, given a similar (presumably temporary) table mapping digits in the string we want translated to their digit position (count from the right), such as this,
CREATE TABLE [HexString](
[Digit] [char](1) NOT NULL,
[Pos] [int] NOT NULL
CONSTRAINT [PK_HexString] PRIMARY KEY CLUSTERED ([Digit] ASC) )
INSERT INTO [HexString] VALUES (‘B’, 4);
INSERT INTO [HexString] VALUES (‘D’, 3);
INSERT INTO [HexString] VALUES (‘1’, 2);
INSERT INTO [HexString] VALUES (‘2’, 1);
Then we have the simple query
select SUM(POWER(16,(pos-1))*Value)
From HexString s, HexDigits d
where s.Digit = d.Digit
The trick here will be the convert the original string (“BD12”, in this example), into the rows if the HexString table. I don’t know of a good way to do that, but I’m sure someone knows a simple way to do that.
There is no ‘hex’ data type in sql server that we would convert from or to. We have hex literal (eg. 0xFFFF) which is int data type. We can have hex digits stored as ascii characters in varchar (e.g. ‘FFFF’, bytes 70,70,70,70 ), or as binary digits in varbinary (bytes 255, 255). You did not specify which sql type did you really want to convert from. If you want to convert from hex string to integer number, here it is:
DECLARE @hexstr VARCHAR(10); SET @hexstr = ‘ffff’
DECLARE @rez BIGINT; SET @rez = 0
WHILE @hexstr ”
BEGIN
SET @rez = @rez * 16 + CHARINDEX(LEFT(@hexstr,1),’0123456789ABCDEF’) – 1
SET @hexstr = SUBSTRING(@hexstr,2,100)
END
PRINT @rez
Without loop, in one SELECT statement:
DECLARE @hexstr VARCHAR(10); SET @hexstr = ‘ffff’
DECLARE @rez BIGINT;
SELECT @rez = ISNULL(@rez,0) * 16 + CHARINDEX(substring(@hexstr,n.number+1,1),’0123456789ABCDEF’) – 1
FROM MASTER..spt_values n WHERE n.TYPE=’P’ AND n.number<len(@hexstr)
PRINT @rez
Vedran Kesegic
beautiful code on the one without the loop! I couldn’t get the first code with the loop to work, but the second one is awesome! Thanks for depositing this here!
Hi guys,
Here’s an online tool I use to convert hex to decimal: hex to decimal converter
Pretty cool!
David
Hi All,
Please answer this,
select convert(varbinary(2),unicode(N’B’))
go
declare @tmp_var varchar(10)
SET @tmp_var = convert(varbinary(2),unicode(N’B’))
select @tmp_var ‘tmp_var’
Execute this query….
When a varbinary value is assigned to a varchar type, then it shows NULL. Why???
By
Biju.K.S
You can’t do that. Why do you want to do that?
No practical use..!
I just want to know that…
Thanks for your reply..:)
Hi,
please run this,
select convert(varbinary(2),unicode(N’B’))
go
declare @tmp_var varchar(10)
SET @tmp_var = sys.fn_varbintohexstr(convert(varbinary(2),unicode(N’B’)))
select @tmp_var ‘tmp_var’
;)
Thanks,
Biju
Hi,
please run this,
select convert(varbinary(2),unicode(N’B’))
go
declare @tmp_var varchar(10)
SET @tmp_var = sys.fn_varbintohexstr(convert(varbinary(2),unicode(N’B’)))
select @tmp_var ‘tmp_var’
;)
Thanks,
Biju
A Little late to the party, however when reading books online for convert I noticed you could apply styles when converting to binary.
SELECT CONVERT(VARBINARY(8), 65535),CAST(0x0000FFFF AS INT),
— If the value is a string without the 0x
CAST(CONVERT(VARBINARY, ‘ffff’, 2) AS INT),
CAST(CONVERT(VARBINARY, ‘0000ffff’, 2) AS INT),
— or if your string has the 0x
CAST(CONVERT(VARBINARY, ‘0x0000FFFF’, 1) AS INT),
CAST(CONVERT(VARBINARY, ‘0xFFFF’, 1) AS INT)
0x0000FFFF,65535,65535,65535,65535,65535
building on Ray’s contribution above, here is the generic case:
declare @hex varchar(64) = ‘0x00FF’
select cast( CONVERT(VARBINARY,’0x’+right(‘00000000’+replace(@hex,’x’,”),8),1) as int)
This could easily be made into a function.
Moderator, please remove the “commented” values after the declare statement above. The two comment dashes got replaced by another symbol. Thanks
hi.
i want to convert hex to readable string how to do pls help me its very urgent.
i have hex like “0605040B8423F0660601AE02056A00”
want to convert in human readable string.
pls help …..
Here is another solution
SELECT CONVERT(int, CONVERT(varbinary, ‘0xFF’, 1)) returns 255
so replace ‘0xFF’ with any other hex value and try. Remember to always include ‘0x’ in the beginning.
today your website posts have helped me lots of times to resolve problem from my manager.
you’re the best!!
thank you
CREATE TABLE #t
(
ip varchar(200)
)
insert into #t
select ‘C0A8019A’
union all
select ‘0A0B0028’
union all
select ‘0A0B2531’
union all
select ‘0A0B62CF’
union all
select ‘0A0B415F’
union all
select ‘0A0B62CF’
union all
select ‘0A0B2531’
union all
select ‘0A0B62CF’
union all
select ‘0A0B2531’
union all
select ‘0A0B62CF’
union all
select ‘0A0B415F’
union all
select ‘0A0B81B0′
select ip,substring(ip,1,2) A1
,substring(ip,len(substring(ip,1,2))+1,2) B1
,substring(ip,len(substring(ip,len(substring(ip,1,2))+1,2))+3,2) C1
,substring(ip,len(substring(ip,len(substring(ip,len(substring(ip,1,2))+1,2))+3,2))+5,2) D1
,case when left(substring(ip,1,2),1) =’A’ Then 10 * 16
when left(substring(ip,1,2),1) =’B’ Then 11 * 16
when left(substring(ip,1,2),1) =’C’ Then 12 * 16
when left(substring(ip,1,2),1) =’D’ Then 13 * 16
when left(substring(ip,1,2),1) =’E’ Then 14 * 16
when left(substring(ip,1,2),1) =’F’ Then 15 * 16
ELSE left(substring(ip,1,2),1) * 16
END
+
case when Right(substring(ip,1,2),1) =’A’ Then 10
when Right(substring(ip,1,2),1) =’B’ Then 11
when Right(substring(ip,1,2),1) =’C’ Then 12
when Right(substring(ip,1,2),1) =’D’ Then 13
when Right(substring(ip,1,2),1) =’E’ Then 14
when Right(substring(ip,1,2),1) =’F’ Then 15
ELSE Right(substring(ip,1,2),1)
END AS A
,case when left(substring(ip,len(substring(ip,1,2))+1,2),1) =’A’ Then 10 * 16
when left(substring(ip,len(substring(ip,1,2))+1,2),1) =’B’ Then 11 * 16
when left(substring(ip,len(substring(ip,1,2))+1,2),1) =’C’ Then 12 * 16
when left(substring(ip,len(substring(ip,1,2))+1,2),1) =’D’ Then 13 * 16
when left(substring(ip,len(substring(ip,1,2))+1,2),1) =’E’ Then 14 * 16
when left(substring(ip,len(substring(ip,1,2))+1,2),1) =’F’ Then 15 * 16
ELSE left(substring(ip,len(substring(ip,1,2))+1,2),1) * 16
END
+
case when Right(substring(ip,len(substring(ip,1,2))+1,2),1) =’A’ Then 10
when Right(substring(ip,len(substring(ip,1,2))+1,2),1) =’B’ Then 11
when Right(substring(ip,len(substring(ip,1,2))+1,2),1) =’C’ Then 12
when Right(substring(ip,len(substring(ip,1,2))+1,2),1) =’D’ Then 13
when Right(substring(ip,len(substring(ip,1,2))+1,2),1) =’E’ Then 14
when Right(substring(ip,len(substring(ip,1,2))+1,2),1) =’F’ Then 15
ELSE Right(substring(ip,len(substring(ip,1,2))+1,2),1)
END AS B
,case when left(substring(ip,len(substring(ip,len(substring(ip,1,2))+1,2))+3,2),1) =’A’ Then 10 * 16
when left(substring(ip,len(substring(ip,len(substring(ip,1,2))+1,2))+3,2),1) =’B’ Then 11 * 16
when left(substring(ip,len(substring(ip,len(substring(ip,1,2))+1,2))+3,2),1) =’C’ Then 12 * 16
when left(substring(ip,len(substring(ip,len(substring(ip,1,2))+1,2))+3,2),1) =’D’ Then 13 * 16
when left(substring(ip,len(substring(ip,len(substring(ip,1,2))+1,2))+3,2),1) =’E’ Then 14 * 16
when left(substring(ip,len(substring(ip,len(substring(ip,1,2))+1,2))+3,2),1) =’F’ Then 15 * 16
ELSE left(substring(ip,len(substring(ip,len(substring(ip,1,2))+1,2))+3,2),1) * 16
END
+
case when Right(substring(ip,len(substring(ip,len(substring(ip,1,2))+1,2))+3,2),1) =’A’ Then 10
when Right(substring(ip,len(substring(ip,len(substring(ip,1,2))+1,2))+3,2),1) =’B’ Then 11
when Right(substring(ip,len(substring(ip,len(substring(ip,1,2))+1,2))+3,2),1) =’C’ Then 12
when Right(substring(ip,len(substring(ip,len(substring(ip,1,2))+1,2))+3,2),1) =’D’ Then 13
when Right(substring(ip,len(substring(ip,len(substring(ip,1,2))+1,2))+3,2),1) =’E’ Then 14
when Right(substring(ip,len(substring(ip,len(substring(ip,1,2))+1,2))+3,2),1) =’F’ Then 15
ELSE Right(substring(ip,len(substring(ip,len(substring(ip,1,2))+1,2))+3,2),1)
END AS C
,case when left(substring(ip,len(substring(ip,len(substring(ip,len(substring(ip,1,2))+1,2))+3,2))+5,2),1) =’A’ Then 10 * 16
when left(substring(ip,len(substring(ip,len(substring(ip,len(substring(ip,1,2))+1,2))+3,2))+5,2),1) =’B’ Then 11 * 16
when left(substring(ip,len(substring(ip,len(substring(ip,len(substring(ip,1,2))+1,2))+3,2))+5,2),1) =’C’ Then 12 * 16
when left(substring(ip,len(substring(ip,len(substring(ip,len(substring(ip,1,2))+1,2))+3,2))+5,2),1) =’D’ Then 13 * 16
when left(substring(ip,len(substring(ip,len(substring(ip,len(substring(ip,1,2))+1,2))+3,2))+5,2),1) =’E’ Then 14 * 16
when left(substring(ip,len(substring(ip,len(substring(ip,len(substring(ip,1,2))+1,2))+3,2))+5,2),1) =’F’ Then 15 * 16
ELSE left(substring(ip,len(substring(ip,len(substring(ip,len(substring(ip,1,2))+1,2))+3,2))+5,2),1) * 16
END
+
case when Right(substring(ip,len(substring(ip,len(substring(ip,len(substring(ip,1,2))+1,2))+3,2))+5,2),1) =’A’ Then 10
when Right(substring(ip,len(substring(ip,len(substring(ip,len(substring(ip,1,2))+1,2))+3,2))+5,2),1) =’B’ Then 11
when Right(substring(ip,len(substring(ip,len(substring(ip,len(substring(ip,1,2))+1,2))+3,2))+5,2),1) =’C’ Then 12
when Right(substring(ip,len(substring(ip,len(substring(ip,len(substring(ip,1,2))+1,2))+3,2))+5,2),1) =’D’ Then 13
when Right(substring(ip,len(substring(ip,len(substring(ip,len(substring(ip,1,2))+1,2))+3,2))+5,2),1) =’E’ Then 14
when Right(substring(ip,len(substring(ip,len(substring(ip,len(substring(ip,1,2))+1,2))+3,2))+5,2),1) =’F’ Then 15
ELSE Right(substring(ip,len(substring(ip,len(substring(ip,len(substring(ip,1,2))+1,2))+3,2))+5,2),1)
END AS D
INTO #temp
from #t
select ip, convert(varchar,A)+’.’+ convert(varchar,B)+’.’+convert(varchar,C) +’.’+convert(varchar,D) IPAddress from #temp
Hi pinal,
please give me solution of following error
The TCP/IP connection to the host 11.01.0.45, port 1433 has failed. Error: “Address already in use: connect. Verify the connection properties, check that an instance of SQL Server is running on the host and accepting TCP/IP connections at the port, and that no firewall is blocking TCP connections to the port.”.
1000
com.websym.common.exception.DBException: org.hibernate.exception.JDBCConnectionException: Cannot open connection
Jayesh G
I have a solution using a tally table and a hex table (hex table idea from James Curran’s suggestion above) that handles strings both with and without a leading ‘0x’, and that is very fast.
For those who don’t know, a tally table is simply a table with one column filled with sequential numbers. Here is a simple way to create and load a tally table with powers of 2 numbers:
—————-
CREATE TABLE tally (n int)
DECLARE @i int,
@maxN int
SET @i = 16
INSERT INTO tally(n) VALUES(1)
WHILE @i > 1
BEGIN
SELECT @maxN = MAX(n)
FROM tally
INSERT INTO tally(n)
SELECT n + @maxN
FROM tally
SET @i = @i – 1
END
—————-
This builds a 32,768 entry tally table starting at one. Now we can use the tally table to simplify building the hex table, as well as in speeding the decoding of a hex string.
——————
CREATE TABLE hexDigits (
h char(1) NOT NULL PRIMARY KEY CLUSTERED,
v int NOT NULL
)
GO
DECLARE @s varchar(15)
SET @s = ‘123456789ABCDEF’
INSERT INTO hexDigits VALUES(‘0’, 0)
INSERT INTO hexDigits(h,v)
SELECT SUBSTRING(@s, t.n, 1), t.n
FROM tally AS t
WHERE t.n BETWEEN 1 and LEN(@s)
—————-
Now we can decode the hex string quickly with this function:
—————-
CREATE FUNCTION fnHex2Int (@s varchar(18))
RETURNS bigint
AS
BEGIN
DECLARE @ret bigint
SELECT
@s = REPLACE(SUBSTRING(@s + ‘.’, PATINDEX(‘%[^0xX]%’, @s + ‘.’), 18), ‘.’, 0)
IF @s NOT LIKE ‘%[^0-9A-Fa-f]%’
BEGIN
SELECT @ret = SUM(POWER(16,(num.n – 1)) * h.v)
FROM hexdigits AS h
RIGHT JOIN (
SELECT SUBSTRING(REVERSE(@s), t.n, 1) AS x,
t.n
FROM tally AS t
WHERE t.n BETWEEN 1 and LEN(@s)
) num
ON h.h = num.x
END
RETURN @ret
END
—————-
Now, you can test the function. This query:
SELECT dbo.fnHex2Int(‘0x00001000’) as h1,
dbo.fnHex2Int(‘00001000’) as h2,
dbo.fnHex2Int(‘0x00000000’) as h3,
dbo.fnHex2Int(‘0xray100’) as h4
Returned:
h1 h2 h3 h4
65536 65536 0 NULL