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

https://stackoverflow.com/a/10899320/1194307

The following function use sys_guid() and transform it into uuid format:

create or replace function random_uuid return VARCHAR2 is
  v_uuid VARCHAR2(40);
begin
  select regexp_replace(rawtohex(sys_guid()), '([A-F0-9]{8})([A-F0-9]{4})([A-F0-9]{4})([A-F0-9]{4})([A-F0-9]{12})', '\1-\2-\3-\4-\5') into v_uuid from dual;
  return v_uuid;
end random_uuid;

It do not need create dbms_crypto package and grant it.