How to generate a version 4 (random) UUID on Oracle?

Question

This blog explains, that the output of sys_guid() is not random for every system:

http://feuerthoughts.blogspot.de/2006/02/watch-out-for-sequential-orac

Answer 1

there are some pure plsql functions written by me and one of my friend that generates uuid version 4 and formats any type of GUIDs. also formatters written in two way. one concating string and one use regex for formatting uuid

CREATE OR REPLACE FUNCTION RANDOM_UUD_RAW
  RETURN RAW IS V_UUID RAW(16);
  BEGIN V_UUID := SYS.DBMS_CRYPTO.Randombytes(16);
    V_UUID := UTL_RAW.Overlay(UTL_RAW.Bit_or(UTL_RAW.Bit_and(UTL_RAW.Substr(V_UUID, 7, 1), '0F'), '40'), V_UUID, 7, 1);
    V_UUID := UTL_RAW.Overlay(UTL_RAW.Bit_or(UTL_RAW.Bit_and(UTL_RAW.Substr(V_UUID, 9, 1), '3F'), '80'), V_UUID, 9, 1);
    RETURN V_UUID;
  END RANDOM_UUD_RAW; --
CREATE OR REPLACE FUNCTION UUID_FORMATTER_CONCAT(V_UUID RAW)
  RETURN VARCHAR2 IS V_STR VARCHAR2(36);
  BEGIN V_STR := lower(SUBSTR(V_UUID, 1, 8) || '-' || SUBSTR(V_UUID, 9, 4) || '-' || SUBSTR(V_UUID, 13, 4) || '-' || SUBSTR(V_UUID, 17, 4) || '-' || SUBSTR(V_UUID, 21));
    RETURN V_STR;
  END UUID_FORMATTER_CONCAT; --
CREATE OR REPLACE FUNCTION UUID_FORMATTER_REGEX(V_UUID RAW)
  RETURN VARCHAR2 IS V_STR VARCHAR2(36);
  BEGIN V_STR := lower(regexp_replace(V_UUID, '(.{8})(.{4})(.{4})(.{4})(.{12})', '\1-\2-\3-\4-\5'));
    RETURN V_STR;
  END UUID_FORMATTER_REGEX; --
CREATE OR REPLACE FUNCTION RANDOM_UUID_STR
  RETURN VARCHAR2 AS BEGIN RETURN UUID_FORMATTER_CONCAT(RANDOM_UUD_RAW());
  END RANDOM_UUID_STR; --
CREATE OR REPLACE FUNCTION RANDOM_UUID_STR_REGEX
  RETURN VARCHAR2 AS BEGIN RETURN UUID_FORMATTER_REGEX(RANDOM_UUD_RAW());
  END RANDOM_UUID_STR_REGEX;