2015-10-06 25 views

回答

10

我花了相當一段時間尋找解決方案,並提出了以下 mysql函數,該函數使用標準MySQL 函數生成隨機UUID(即UUIDv4)。我正在回答我自己的問題,希望分享 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:使用的僞隨機數生成(MySQL的RAND)不 加密安全,因而具有一定的偏差,可以提高碰撞 風險。