Reputation: 5478
My app internally uses RAW(16) guids generated via sys_guid().
Recently I started associating Active Directory users where I get guid's in the format of: 00000000-0000-0000-0000-00000000000
Are the following two functions doing this conversion correctly?
Found the following site:
create or replace
FUNCTION RAWTOGUID
( RawData IN RAW
) RETURN VARCHAR AS
BEGIN
declare HexData varchar(32) := rawtohex(RawData);
begin
return
substr(HexData, 7, 2)
|| substr(HexData, 5, 2)
|| substr(HexData, 3, 2)
|| substr(HexData, 1, 2)
|| '-'
|| substr(HexData, 11, 2)
|| substr(HexData, 9, 2)
|| '-'
|| substr(HexData, 15, 2)
|| substr(HexData, 13, 2)
|| '-'
|| substr(HexData, 17, 4)
|| '-'
|| substr(HexData, 21, 12);
end;
END RAWTOGUID;
Adding in the following site:
http://dbaspot.com/oracle-server/69226-guid-char-conversion-function.html
Came up with this function to do the inverse:
create or replace
FUNCTION GUIDTORAW
( HexData IN VARCHAR
) RETURN RAW AS
BEGIN
declare StringData varchar(32) := TRANSLATE(HexData,'0{-}','0');
begin
return
hextoraw(substr(StringData, 7, 2)
|| substr(StringData, 5, 2)
|| substr(StringData, 3, 2)
|| substr(StringData, 1, 2)
|| substr(StringData, 11, 2)
|| substr(StringData, 9, 2)
|| substr(StringData, 15, 2)
|| substr(StringData, 13, 2)
|| substr(StringData, 17, 4)
|| substr(StringData, 21, 12));
end;
END GUIDTORAW;
They convert back and forth, but am I actually respecting the endianness or have the right order at all?
Upvotes: 5
Views: 2894
Reputation: 18068
Quoting from the UUID standard:
The structure of UUIDs is:
Field Data Type Octet Note # time_low unsigned 32 0-3 The low field of the bit integer timestamp time_mid unsigned 16 4-5 The middle field of the bit integer timestamp time_hi_and_version unsigned 16 6-7 The high field of the bit integer timestamp multiplexed with the version number clock_seq_hi_and_rese unsigned 8 8 The high field of the rved bit integer clock sequence multiplexed with the variant clock_seq_low unsigned 8 9 The low field of the bit integer clock sequence node unsigned 48 10-15 The spatially unique bit integer node identifier
In the absence of explicit application or presentation protocol specification to the contrary, a UUID is encoded as a 128-bit object, as follows:
The fields are encoded as 16 octets, with the sizes and order of the fields defined above, and with each field encoded with the Most Significant Byte first (known as network byte order). Note that the field names, particularly for multiplexed fields, follow historical practice.
0 1 2 3 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+ | time_low | +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+ | time_mid | time_hi_and_version | +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+ |clk_seq_hi_res | clk_seq_low | node (0-1) | +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+ | node (2-5) | +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
Quoting from the Guid documentation:
public Guid(int a, short b, short c, byte[] d)
Guid(1,2,3,new byte[]{0,1,2,3,4,5,6,7})
creates a Guid that corresponds to "00000001-0002-0003-0001-020304050607".
According to Wikipedia's Endianness article Windows stores numbers in little endian --> least significant byte first.
The mapping 1,2,3,new byte[]{0,1,2,3,4,5,6,7} to "00000001-0002-0003-0001-020304050607", shows us that the numbers are displayed big-endian, as in the UUID standard, however, the byte array is provided in the same order as display - no need to swap the bytes.
So Guids are displayed as:
{time_low (4B) - time_mid (2B) - time_hi_and_version (2B) - clock_sq_hi_and_reserved(1B), clock_seq_low (1B) - node (6B)}
In little endian this results in byte order (byte[] does not count as a number as is therefore :
{3,2,1,0 - 5,4 - 7,6 - 8,9 - 10,11,12,13,14,15}
which results in hexidecimal character order (each byte is 2 hexidecimal digits):
{6,7,4,5,2,3,0,1 - 10,11,8,9 - 14,15,12,13 - 16,17,18,19 - 20,21,22,23,24,25,26,27,28,29,30,31}
In Oracle, the substr string function is 1-based, therefore the Oracle string indexes are:
{7,8,5,6,3,4,1,2 - 11,12,9,10 - 15,16,13,14 - 17,18,19,20 - 21,22,23,24,25,26,27,28,29,30,31,32}
Which results in the command
substr(HexData, 7, 2)
|| substr(HexData, 5, 2)
|| substr(HexData, 3, 2)
|| substr(HexData, 1, 2)
|| '-'
|| substr(HexData, 11, 2)
|| substr(HexData, 9, 2)
|| '-'
|| substr(HexData, 15, 2)
|| substr(HexData, 13, 2)
|| '-'
|| substr(HexData, 17, 4)
|| '-'
|| substr(HexData, 21, 12);
And transposed (after removing '-'s):
{7,8,5,6,3,4,1,2, 11,12,9,10, 15,16,13,14, 17,18,19,20, 21,22,23,24,25,26,27,28,29,30,31,32}
is reversed back into
{1,2,3,4,5,6,7,8, 9,10,11,12, 13,14,15,16, 17,18,19,20, 21,22,23,24,25,26,27,28,29,30,31,32}
using the same function (without adding '-'s) - converting BE to LE and LE to BE have the same swaps since the bytes are simply reversed and reversing reversed bytes results in non-reversed bytes.
Upvotes: 6