Question

我的MySQL数据库包含两个表：user和coupon（一对一的关系）。

我想选择所有没有优惠券并创建新优惠券的用户（随机且唯一）。

user TABLE:
___________________________________
|  id   |   name   |   coupon_id  |
-----------------------------------
   1        John         5
   2        Mary         (null)  // Need to create one coupon.
   3        Doe          2
   4        Max          (null)  // Need to create one coupon.
   5        Rex          1
   7        Bill         (null)  // Need to create one coupon.


coupon TABLE:
______________________________________________
|  id   |   code (random 6-chars - unique)   |
----------------------------------------------
   1        80k2ni
   2        0akdne
   5        nk03jd

快捷方式：

选择没有优惠券的所有用户：SELECT * from user WHERE coupon_id IS NULL;

生成一个随机的6个字符串（MySQL）：LEFT(sha1(rand()), 6)。

Answer 1

假设您不介意在下一个coupon_id从6开始继续向上，可以按照以下方式完成此操作（请参阅SQL Fiddle Demo）：

-- Declare and set variables
SET @id_for_insert = (SELECT MAX(`id`) FROM `coupon`);
SET @id_for_update = @id_for_insert;

-- Insert new coupons
INSERT INTO `coupon` (id, code)
SELECT @id_for_insert := @id_for_insert + 1, LEFT(SHA1(RAND()), 6)
FROM `user`
WHERE coupon_id IS NULL;

-- Update users that don't already have a coupon with the newly created coupons
UPDATE `user`
SET coupon_id = @id_for_update := @id_for_update + 1
WHERE coupon_id IS NULL;

Answer 2

这样的事情可能是：

Insert into coupon 
select distinct id,LEFT(sha1(rand()), 6) 
from user WHERE coupon_id IS NULL

在此之后，您可以使用简单的连接更新来更新用户表中的优惠券。

Answer 3

//check if works inform me i will be glad
CREATE PROCEDURE `syncCouponUser` ()
BEGIN
  INSERT INTO coupon(id,code)//create coupons var no-used user ids
    select
    id,
    LEFT(sha1(rand())
    from user
    where coupon_id IS NULL ;

     UPDATE user //add coupons from coupon table with matches
    SET coupon_id=(
    select c.coupon_id
    from coupon c join user u 
    on c.id=u.id);

END

插入一对一关系

3 个答案: