如何在Oracle中生成GUID？

Question

是否可以在Insert语句中自动生成GUID？

另外,我应该使用什么类型的字段来存储此GUID？

Answer 1

您可以使用SYS_GUID()函数在insert语句中生成GUID:

insert into mytable (guid_col, data) values (sys_guid(), 'xxx');

用于存储GUID的首选数据类型是RAW(16).

正如Gopinath回答:

 select sys_guid() from dual
 union all
 select sys_guid() from dual
 union all 
 select sys_guid() from dual

你得到

88FDC68C75DDF955E040449808B55601
88FDC68C75DEF955E040449808B55601
88FDC68C75DFF955E040449808B55601

正如托尼安德鲁斯所说,只有一个角色不同

88FDC68C75D D F955E040449808B55601
88FDC68C75D E F955E040449808B55601
88FDC68C75D F F955E040449808B55601

也许有用:http: //feuerthoughts.blogspot.com/2006/02/watch-out-for-sequential-oracle-guids.html

Answer 2

您还可以在表的create语句中包含guid作为默认值,例如:

create table t_sysguid
( id     raw(16) default sys_guid() primary key
, filler varchar2(1000)
)
/

见这里:http://rwijk.blogspot.com/2009/12/sysguid.html

Answer 3

示例位于：http : //www.orafaq.com/usenet/comp.databases.oracle.server/2006/12/20/0646.htm

SELECT REGEXP_REPLACE(SYS_GUID(), '(.{8})(.{4})(.{4})(.{4})(.{12})', '\1-\2-\3-\4-\5') MSSQL_GUID  FROM DUAL 

结果：

6C7C9A50-3514-4E77-E053-B30210AC1082 

Answer 4

如果您需要非顺序引导，您可以sys_guid()通过哈希函数发送结果（请参阅/sf/answers/1577439041/）。这个想法是保留原始创作中使用的任何独特性，并获得具有更多改组位的东西。

例如：

LOWER(SUBSTR(STANDARD_HASH(SYS_GUID(), 'SHA1'), 0, 32))  

显示默认顺序 guid 与通过哈希发送它的示例：

SELECT LOWER(SYS_GUID()) AS OGUID FROM DUAL
UNION ALL
SELECT LOWER(SYS_GUID()) AS OGUID FROM DUAL
UNION ALL
SELECT LOWER(SYS_GUID()) AS OGUID FROM DUAL
UNION ALL
SELECT LOWER(SYS_GUID()) AS OGUID FROM DUAL
UNION ALL
SELECT LOWER(SUBSTR(STANDARD_HASH(SYS_GUID(), 'SHA1'), 0, 32)) AS OGUID FROM DUAL
UNION ALL
SELECT LOWER(SUBSTR(STANDARD_HASH(SYS_GUID(), 'SHA1'), 0, 32)) AS OGUID FROM DUAL
UNION ALL
SELECT LOWER(SUBSTR(STANDARD_HASH(SYS_GUID(), 'SHA1'), 0, 32)) AS OGUID FROM DUAL
UNION ALL
SELECT LOWER(SUBSTR(STANDARD_HASH(SYS_GUID(), 'SHA1'), 0, 32)) AS OGUID FROM DUAL  

输出

SELECT LOWER(SYS_GUID()) AS OGUID FROM DUAL
UNION ALL
SELECT LOWER(SYS_GUID()) AS OGUID FROM DUAL
UNION ALL
SELECT LOWER(SYS_GUID()) AS OGUID FROM DUAL
UNION ALL
SELECT LOWER(SYS_GUID()) AS OGUID FROM DUAL
UNION ALL
SELECT LOWER(SUBSTR(STANDARD_HASH(SYS_GUID(), 'SHA1'), 0, 32)) AS OGUID FROM DUAL
UNION ALL
SELECT LOWER(SUBSTR(STANDARD_HASH(SYS_GUID(), 'SHA1'), 0, 32)) AS OGUID FROM DUAL
UNION ALL
SELECT LOWER(SUBSTR(STANDARD_HASH(SYS_GUID(), 'SHA1'), 0, 32)) AS OGUID FROM DUAL
UNION ALL
SELECT LOWER(SUBSTR(STANDARD_HASH(SYS_GUID(), 'SHA1'), 0, 32)) AS OGUID FROM DUAL  

Answer 5

通过在insert语句中自动生成一个guid来表达你的意思并不清楚,但我猜你正在尝试做类似以下的事情:

INSERT INTO MY_TAB (ID, NAME) VALUES (SYS_GUID(), 'Adams');
INSERT INTO MY_TAB (ID, NAME) VALUES (SYS_GUID(), 'Baker');

在这种情况下,我认为ID列应该声明为RAW(16);

我正在做这件事.我没有方便测试的Oracle实例,但我认为这就是你想要的.

Answer 6

sys_guid() 是一个糟糕的选择，正如其他答案所提到的。生成 UUID 并避免序列值的一种方法是自己生成随机的十六进制字符串：

select regexp_replace(
    to_char(
        DBMS_RANDOM.value(0, power(2, 128)-1),
        'FM0xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx'),
    '([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') from DUAL;