Similar to this question (Sql Server convert integer to binary string, but I would like to convert a varbinary(16) to its text version instead.
I am doing something tragically wrong as the results of my code will attest.
create function GetGuidBinaryString (@value varbinary(16))
returns varchar(128)
as
begin
declare @vsresult varchar(128)
declare @inti int
select @inti = 1开发者_JAVA百科28, @vsresult = ''
while @inti>0
begin
select @vsresult=convert(char(1), @value % 2)+@vsresult
select @value = convert(int, (@value / 2)), @inti=@inti-1
end
return @vsresult
end
create table #values (binvalue varchar(128))
delete from #values
declare @intcount int
select @intcount = 0
while @intcount < 100
begin
insert into #values select dbo.GetGuidBinaryString(convert(varbinary(16),convert(bigint,2147483640) + @intcount))
select @intcount = @intcount+1
end
select * from #values
Perhaps there is some implicit conversion I am doing in the function as the function only works correctly for positive integers.
@value % 2
and @value / 2
is doing an implicit conversion.
select @value = convert(int, (@value / 2))
is doing an explicit conversion to int so here you are getting a negative int for the values stored in varbinary(16) that after the division converted to bigint is larger than 2,147,483,647.
% for a negative int will give you a -1.
I do not think it is possible to convert varbinary(16) to binary using % and /. They only work on int/bigint etc.
Here is a conversion routine that works for positive bigint values. I do not know what representation you would expect for negative bigint values. Convert your varbinary(16) field to bigint in the call to the function and perhaps it does what you want. I am sure it does not work for all possible values that you can store in a varbinary(16) field.
create function BigIntToBin (@v bigint)
returns varchar(256)
as
begin
declare @res varchar(256)
declare @i int
set @i = 128
set @res = ''
while @i > 0
begin
if @v % 2 = 0
set @res = '0' + @res
else
set @res = '1' + @res
set @v = @v / 2
set @i = @i - 1
end
return @res
end
精彩评论