Increment a uniqueidentifier in TSQL - sql-server

Question

I am looking for a way to increment a uniqueidentifier by 1 in TSQL. For example, if the id is A6BC60AD-A4D9-46F4-A7D3-98B2A7237A9E, I'd like to be able to select A6BC60AD-A4D9-46F4-A7D3-98B2A7237A9F.
#rein It's for a data import. We have an intermediate table with IDs that we're generating records from, and we join on those IDs later in the import. Unfortunately, now some of those records generate a couple of records in the next table, so we need a new id that is reproducible.

Answer 1

The way you want to increment Guid is not correct for SQL Server as Guid is a structure with different byte order in the byte groups, please have a look at:
http://sqlblog.com/blogs/alberto_ferrari/archive/2007/08/31/how-are-guids-sorted-by-sql-server.aspx
and notice the following:
Now, when I run modified Alberto's query, I'm getting the following sequence:
3, 2, 1, 0, 5, 4, 7, 6, 9, 8, 15, 14, 13, 12, 11, 10
That means, that GUID's byte #3 is the least significant and GUID's byte #10 is the most significant [from SQL Server ORDER BY clause perspective].
Here is simple function to increment a uniqueidentifier accounting for this:
create function [dbo].[IncrementGuid](#guid uniqueidentifier)
returns uniqueidentifier
as
begin
declare #guid_binary binary(16), #b03 binary(4), #b45 binary(2), #b67 binary(2), #b89 binary(2), #bAF binary(6)
select #guid_binary = #guid
select #b03 = convert(binary(4), reverse(substring(#guid_binary,1,4)))
select #b45 = convert(binary(2), reverse(substring(#guid_binary,5,2)))
select #b67 = convert(binary(2), reverse(substring(#guid_binary,7,2)))
select #b89 = convert(binary(2), substring(#guid_binary,9,2))
select #bAF = convert(binary(6), substring(#guid_binary,11,6))
if (#b03 < 'FFFFFFFF')
begin
select #b03 = convert(binary(4), cast(#b03 as int) + 1)
end
else if (#b45 < 'FFFF')
begin
select #b45 = convert(binary(2), cast(#b45 as int) + 1)
end
else if (#b89 < 'FFFF')
begin
select #b89 = convert(binary(2), cast(#b89 as int) + 1)
end
else
begin
select #bAF = convert(binary(6), cast(#bAF as bigint) + 1)
end
return convert(binary(16), reverse(convert(char(4),#b03)) + reverse(convert(char(2),#b45)) + reverse(convert(char(2),#b67)) + convert(char(2),#b89) + convert(char(6),#bAF))
end
Note that bytes 6 and 7 are not incremented as they contain the Guid version bits.
But as others has pointed you really should not be doing this. In your case it might be better if you create a temp table for these Guids (with two columns: one integer as index and second one with generated Guids).

Answer 2

Here is one way I've come up with, but I'm hoping there is a better way.
LEFT([ID], 19) + RIGHT(CONVERT(uniqueidentifier, CONVERT(binary(16), CONVERT(binary(16), [ID]) + CONVERT(bigint, 1))), 17) AS 'MyNewID'

Answer 3

You can do this approach, but I'm not accounting for the case of overflowing lower 8 bytes.
declare #guid uniqueidentifier, #binaryUpper8 binary(8), #binaryLower8 binary(8), #binary16 binary(16), #bigint bigint
set #guid = 'A6BC60AD-A4D9-46F4-A7D3-98B2A7237A9E'
set #binary16 = cast(#guid as binary(16))
--harvest lower 8 bytes
select #binaryUpper8= substring(#binary16, 1, 8)
,#binaryLower8 = substring(#binary16, 9, 8)
set #bigint = cast(#binaryLower8 as bigint)
--increment
set #bigint = #bigint + 1
--convert back
set #binaryLower8 = cast(#bigint as binary(8))
set #binary16 = #binaryUpper8 + #binaryLower8
set #guid = cast(#binary16 as uniqueidentifier)
select #guid