在 TSQL 中增加唯一标识符

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 

请注意，字节 6 和 7 不会递增，因为它们包含 Guid 版本位.但正如其他人指出的那样，您确实不应该这样做.在您的情况下，如果您为这些 Guid 创建一个临时表可能会更好(有两列:一个整数作为索引，第二个带有生成的 Guid).