Oracle中生成uuid的方法

Oracle中生成uuid的方法
 
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在Oracle SQL 提供了一个生成uuid的函数sys_guid:
http://download.oracle.com/docs/cd/E11882_01/server.112/e17118/functions187.htm#i79194
http://en.wikipedia.org/wiki/Universally_unique_identifier
 
SQL> select sys_guid() from dual ;
SYS_GUID()
--------------------------------
A8F662B86E7413FEE040970A437C6BD7
 
但该函数如下问题：
     1.返回类型为RAW
     2.没有- (dash)分隔符
     3.返回的字母大写
 
为了使产生的uuid符合rfc 4122的标准(http://tools.ietf.org/html/rfc4122)
创建函数如下，测试通过。
 
CREATE OR REPLACE
FUNCTION get_uuid
RETURN VARCHAR
IS
guid VARCHAR (50);
BEGIN
guid := lower(RAWTOHEX(sys_guid()));
RETURN
substr(guid,1,8)||'-'||substr(guid,9,4)||'-'||substr(guid,13,4)||'-'||substr(guid,17,4)||'-'||substr(guid,21,12);
END get_uuid;
 
SQL> select get_uuid from dual ;
 
GET_UUID
--------------------------------------------------------------------------------
a8f662b8-6e7a-13fe-e040-970a437c6bd7
 
SQL> /
 
GET_UUID
--------------------------------------------------------------------------------
a8f662b8-6e7b-13fe-e040-970a437c6bd7
 
SQL> /
 
GET_UUID
--------------------------------------------------------------------------------
a8f662b8-6e7c-13fe-e040-970a437c6bd7