我花了很多时间寻找解决方案，并提出了以下mysql函数，该函数使用标准MySQL函数生成一个随机UUID（即UUIDv4）。我正在回答我自己的问题，以期分享它，希望对大家有用。

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

CREATE FUNCTION uuid_v4()
    RETURNS CHAR(36)
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：所使用的伪随机数生成（MySQL’s RAND）在密码上不是安全的，因此存在一些偏差，可能会增加冲突风险。