表MySQL的更新冲突如下:具有独特的键
CREATE TABLE `ToursCartsItems` (
`Id` int(10) unsigned NOT NULL auto_increment,
`UserId` char(40) default NULL,
`TourId` int(10) unsigned NOT NULL,
`CreatedAt` int(10) unsigned default NULL,
PRIMARY KEY (`Id`),
UNIQUE KEY `UniqueUserProduct` (`UserId`,`TourId`)
) ENGINE=MyISAM AUTO_INCREMENT=5 DEFAULT CHARSET=latin1 ROW_FORMAT=FIXED
// simple sample data
INSERT INTO
ToursCartsItems (UserId, TourId)
VALUES
("old", 1), ("old", 2), ("new", 1), ("new", 3);
所以一个用户可以拥有许多旅行团(别提是什么旅行团)。 UserId字段是char
而不是int
,因为用户可能未登录,在这种情况下会话ID被使用。
当用户登录时,其用户ID将发生变化。所以简单的更新将是
UPDATE ToursCartsItems SET UserId="new" WHERE UserId="old"
-- In reality, the new UserId would be an integer, but never mind that.
但是,这可能会给一个重复的输入密钥。旧用户和新用户都具有相同的导览,我们应该在更新之前删除一个。
所以我已经试过
UPDATE ToursCartsItems
SET UserId="in"
WHERE UserId="out"
AND (TourId NOT IN (SELECT TourId FROM ToursCartsItems WHERE UserId="in")
);
DELETE FROM ToursCartsItems WHERE UserId="old";
和
TRUNCATE ToursCartsItems;
INSERT INTO ToursCartsItems (UserId, TourId) VALUES ("old", 1), ("old", 2), ("new", 1), ("new", 3);
DELETE FROM ToursCartsItems WHERE UserId="old" AND TourId IN (SELECT TourId FROM ToursCartsItems WHERE UserId="new");
UPDATE ToursCartsItems SET UserId="new" WHERE UserId="old";
两个给我的错误。有没有办法在SQL查询本身要做到这一点,还是我只需要做
SELECT * FROM ToursCartsItems WHERE UserId IN ("old", "new")
,然后做必要的计算自己在PHP?
请问您可以分享您获得第一次更新的错误吗? – 2010-12-15 19:13:56