如何在 MySQL 中生成 UUIDv4？

Question

MySQL 的 UUID 函数 returns a UUIDv1 GUID. I'm looking for an easy way to generate random GUIDs (i.e. UUIDv4) in SQL.

Answer 1

我花了很多时间寻找解决方案并提出了以下建议 mysql 使用标准 MySQL 生成随机 UUID（即 UUIDv4）的函数职能。我正在回答我自己的问题来分享它，希望它会成为有用。

-- Change delimiter so that the function body doesn't end the function declaration
DELIMITER //

CREATE FUNCTION uuid_v4()
    RETURNS CHAR(36) NO SQL
BEGIN
    -- Generate 8 2-byte strings that we will combine into a UUIDv4
    SET @h1 = LPAD(HEX(FLOOR(RAND() * 0xffff)), 4, '0');
    SET @h2 = LPAD(HEX(FLOOR(RAND() * 0xffff)), 4, '0');
    SET @h3 = LPAD(HEX(FLOOR(RAND() * 0xffff)), 4, '0');
    SET @h6 = LPAD(HEX(FLOOR(RAND() * 0xffff)), 4, '0');
    SET @h7 = LPAD(HEX(FLOOR(RAND() * 0xffff)), 4, '0');
    SET @h8 = LPAD(HEX(FLOOR(RAND() * 0xffff)), 4, '0');

    -- 4th section will start with a 4 indicating the version
    SET @h4 = CONCAT('4', LPAD(HEX(FLOOR(RAND() * 0x0fff)), 3, '0'));

    -- 5th section first half-byte can only be 8, 9 A or B
    SET @h5 = CONCAT(HEX(FLOOR(RAND() * 4 + 8)),
                LPAD(HEX(FLOOR(RAND() * 0x0fff)), 3, '0'));

    -- Build the complete UUID
    RETURN LOWER(CONCAT(
        @h1, @h2, '-', @h3, '-', @h4, '-', @h5, '-', @h6, @h7, @h8
    ));
END
//
-- Switch back the delimiter
DELIMITER ;

注意：使用的伪随机数生成（MySQL的RAND）不是密码安全，因此有一些偏差会增加冲突风险。

Answer 2

如果您正在使用数据库并且没有创建函数的权限，这里是与上面相同的版本，它可以作为 SQL 表达式使用：

SELECT LOWER(CONCAT(
    LPAD(HEX(FLOOR(RAND() * 0xffff)), 4, '0'), 
    LPAD(HEX(FLOOR(RAND() * 0xffff)), 4, '0'), '-',
    LPAD(HEX(FLOOR(RAND() * 0xffff)), 4, '0'), '-', 
    '4',
    LPAD(HEX(FLOOR(RAND() * 0x0fff)), 3, '0'), '-', 
    HEX(FLOOR(RAND() * 4 + 8)), 
    LPAD(HEX(FLOOR(RAND() * 0x0fff)), 3, '0'), '-', 
    LPAD(HEX(FLOOR(RAND() * 0xffff)), 4, '0'),
    LPAD(HEX(FLOOR(RAND() * 0xffff)), 4, '0'),
    LPAD(HEX(FLOOR(RAND() * 0xffff)), 4, '0')));

Answer 3

两个现有答案都依赖于 MySQL RAND() 函数：

RAND() is not meant to be a perfect random generator. It is a fast way to generate random numbers on demand that is portable between platforms for the same MySQL version.

在实践中，这意味着使用此函数生成的 UUID 可能（并且将会）有偏差，并且碰撞发生的频率可能比预期的要高。

解决方案

可以使用 random_bytes() 函数在 MySQL 端生成安全的 UUID V4：

This function returns a binary string of len random bytes generated using the random number generator of the SSL library.

所以我们可以将函数更新为：

CREATE FUNCTION uuid_v4s()
    RETURNS CHAR(36)
BEGIN
    -- 1th and 2nd block are made of 6 random bytes
    SET @h1 = HEX(RANDOM_BYTES(4));
    SET @h2 = HEX(RANDOM_BYTES(2));

    -- 3th block will start with a 4 indicating the version, remaining is random
    SET @h3 = SUBSTR(HEX(RANDOM_BYTES(2)), 2, 3);

    -- 4th block first nibble can only be 8, 9 A or B, remaining is random
    SET @h4 = CONCAT(HEX(FLOOR(ASCII(RANDOM_BYTES(1)) / 64)+8),
                SUBSTR(HEX(RANDOM_BYTES(2)), 2, 3));

    -- 5th block is made of 6 random bytes
    SET @h5 = HEX(RANDOM_BYTES(6));

    -- Build the complete UUID
    RETURN LOWER(CONCAT(
        @h1, '-', @h2, '-4', @h3, '-', @h4, '-', @h5
    ));
END

这应该会生成足够随机的 UUID V4，而不用担心冲突。

注意： 遗憾的是 MariaDB 不支持 RANDOM_BYTES()（参见 https://mariadb.com/kb/en/function-differences-between-mariadb-105-and-mysql-80/#miscellaneous）

测试

我创建了以下测试场景：插入随机 UUID v4 作为 table 的主键，直到创建 40.000.000 行。发现冲突时，该行将更新并递增 collisions 列：

INSERT INTO test (uuid) VALUES (uuid_v4()) ON DUPLICATE KEY UPDATE collisions=collisions+1;

各函数4000万行后的碰撞总和为：

+----------+----------------+
| RAND()   | RANDOM_BYTES() |
+----------+----------------+
|       55 |              0 |
+----------+----------------+

随着行数的增加，这两种情况下的冲突次数都会增加。

Answer 4

在不创建数据库函数的情况下使用 RANDOM_BYTES 改编 Elias Soares 的答案：

SELECT LOWER(CONCAT(
    HEX(RANDOM_BYTES(4)), '-',
    HEX(RANDOM_BYTES(2)), '-4',
    SUBSTR(HEX(RANDOM_BYTES(2)), 2, 3), '-',
    CONCAT(HEX(FLOOR(ASCII(RANDOM_BYTES(1)) / 64)+8),SUBSTR(HEX(RANDOM_BYTES(2)), 2, 3)), '-',
    HEX(RANDOM_BYTES(6))
))

如何在 MySQL 中生成 UUIDv4？

How to generate a UUIDv4 in MySQL?

mysql

uuid

解决方案

测试