如何在 Oracle 中生成 GUID？

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最佳答案

You can use the SYS_GUID() function to generate a GUID in your insert statement:

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

The preferred datatype for storing GUIDs is RAW(16).

As Gopinath answer:

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

You get

88FDC68C75DDF955E040449808B55601
88FDC68C75DEF955E040449808B55601
88FDC68C75DFF955E040449808B55601

As Tony Andrews says, differs only at one character

88FDC68C75DDF955E040449808B55601
88FDC68C75DEF955E040449808B55601
88FDC68C75DFF955E040449808B55601

Maybe useful: http://feuerthoughts.blogspot.com/2006/02/watch-out-for-sequential-oracle-guids.html

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It is not clear what you mean by auto-generate a guid into an insert statement but at a guess, I think you are trying to do something like the following:

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

In that case I believe the ID column should be declared as RAW(16);

I am doing this off the top of my head. I don't have an Oracle instance handy to test against, but I think that is what you want.

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You can also include the guid in the create statement of the table as default, for example:

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

See here: http://rwijk.blogspot.com/2009/12/sysguid.html

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You can run the following query

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

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sys_guid() is a poor option, as other answers have mentioned. One way to generate UUIDs and avoid sequential values is to generate random hex strings yourself:

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;

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you can use function bellow in order to generate your UUID

create or replace FUNCTION RANDOM_GUID
RETURN VARCHAR2 IS


RNG    NUMBER;
N      BINARY_INTEGER;
CCS    VARCHAR2 (128);
XSTR   VARCHAR2 (4000) := NULL;
BEGIN
CCS := '0123456789' || 'ABCDEF';
RNG := 15;


FOR I IN 1 .. 32 LOOP
N := TRUNC (RNG * DBMS_RANDOM.VALUE) + 1;
XSTR := XSTR || SUBSTR (CCS, N, 1);
END LOOP;


RETURN SUBSTR(XSTR, 1, 4) || '-' ||
SUBSTR(XSTR, 5, 4)        || '-' ||
SUBSTR(XSTR, 9, 4)        || '-' ||
SUBSTR(XSTR, 13,4)        || '-' ||
SUBSTR(XSTR, 17,4)        || '-' ||
SUBSTR(XSTR, 21,4)        || '-' ||
SUBSTR(XSTR, 24,4)        || '-' ||
SUBSTR(XSTR, 28,4);
END RANDOM_GUID;

Example of GUID genedrated by the function above:
8EA4-196D-BC48-9793-8AE8-5500-03DC-9D04

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Example found on: 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

Result:

6C7C9A50-3514-4E77-E053-B30210AC1082

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If you need non-sequential guids you can send the sys_guid() results through a hashing function (see https://stackoverflow.com/a/22534843/1462295 ). The idea is to keep whatever uniqueness is used from the original creation, and get something with more shuffled bits.

For instance:

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

Example showing default sequential guid vs sending it through a hash:

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

output

80c32a4fbe405707e0531e18980a1bbb
80c32a4fbe415707e0531e18980a1bbb
80c32a4fbe425707e0531e18980a1bbb
80c32a4fbe435707e0531e18980a1bbb
c0f2ff2d3ef7b422c302bd87a4588490
d1886a8f3b4c547c28b0805d70b384f3
a0c565f3008622dde3148cfce9353ba7
1c375f3311faab15dc6a7503ce08182c

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I would recommend using Oracle's "dbms_crypto.randombytes" function.

Why?
This function returns a RAW value containing a cryptographically secure pseudo-random sequence of bytes, which can be used to generate random material for encryption keys.

select REGEXP_REPLACE(dbms_crypto.randombytes(16), '(.{8})(.{4})(.{4})(.{4})(.{12})', '\1-\2-\3-\4-\5') from dual;

You should not use the function "sys_guid" because only one character changes.

ALTER TABLE locations ADD (uid_col RAW(16));


UPDATE locations SET uid_col = SYS_GUID();


SELECT location_id, uid_col FROM locations
ORDER BY location_id, uid_col;


LOCATION_ID UID_COL
----------- ----------------------------------------------------------------
1000 09F686761827CF8AE040578CB20B7491
1100 09F686761828CF8AE040578CB20B7491
1200 09F686761829CF8AE040578CB20B7491
1300 09F68676182ACF8AE040578CB20B7491
1400 09F68676182BCF8AE040578CB20B7491
1500 09F68676182CCF8AE040578CB20B7491

https://docs.oracle.com/database/121/SQLRF/functions202.htm#SQLRF06120

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Creating a 350 characters GUID:

dbms_random.STRING ('a', 350) - returning string in mixed case alpha characters

dbms_random.STRING ('x', 350) - returning string in uppercase alpha-numeric characters