Question

是否可以在Insert语句中自动生成GUID？

此外，我应该使用哪种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');

存储GUID的首选数据类型是RAW（16）。

正如Gopinath的回答：

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

你得到了

88FDC68C75DDF955E040449808B55601
  88FDC68C75DEF955E040449808B55601
  88FDC68C75DFF955E040449808B55601

正如Tony Andrews所说， 仅在一个角色上有所不同

88FDC68C75D的 d F955E040449808B55601
  88FDC68C75D <强>电子 F955E040449808B55601
  88FDC68C75D的˚F F955E040449808B55601

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

Answer 2

您还可以将guid包含在表的create语句中作为默认值，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

通过自动生成一个插入语句的guid并不清楚你的意思但猜测，我认为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 4

正如其他答案所提到的那样，

sys_guid（）是一个糟糕的选择。生成UUID并避免顺序值的一种方法是自己生成随机十六进制字符串：

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;

Answer 5

您可以运行以下查询

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

Answer 6

您可以使用下面的函数来生成您的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;

通过上述功能生成的GUID示例：
的 8EA4-196D-BC48-9793-8AE8-5500-03DC-9D04

Answer 7

示例： 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 8

如果需要非顺序引导，则可以通过哈希函数发送sys_guid()结果（请参见https://stackoverflow.com/a/22534843/1462295）。这个想法是保留原始创作中使用的任何唯一性，并获得一些经过改组的位。

例如：

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

示例显示默认顺序guid与通过哈希将其发送：

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

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

Answer 9

我建议使用Oracle的“ dbms_crypto.randombytes”功能。

为什么？ 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;

如果仅更改一个字符，则不要使用“ sys_guid”功能。

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

Answer 10

创建一个350个字符的GUID：

dbms_random.STRING（'a'，350）-返回混合大小写字母字符的字符串

dbms_random.STRING（'x'，350）-返回大写字母数字字符的字符串

如何在Oracle中生成GUID？

10 个答案: