Question

我有两张桌子表一：default_sellers_commissions有：

id
value - INT

表二：marketplace_saleperpartner有：

id
userID
commission - INT default '0.00'
commission_id

我需要的是在表marketplace_saleperpartner中插入新行后，使用来自marketplace_saleperpartner.commission default_sellers_commissions.value的{{1}}更新marketplace_saleperpartner.commission_id = default_sellers_commissions.id值。CREATE TRIGGER `marketplacedefaultcommission` AFTER INSERT ON `marketplace_saleperpartner` FOR EACH ROW UPDATE marketplace_saleperpartner SET `marketplace_saleperpartner`.`commision` = `default_sellers_commissions`.`value` WHERE `marketplace_saleperpartner`.`comm_id` = `default_sellers_commissions`.`id` AND `marketplace_saleperpartner`.`commision`='0.00';

所以我创建了一个触发器：

marketplace_saleperpartner.commision

P.S：
出于某种原因，我需要MySql Error: Can't update table in stored function/trigger because it is already used by statement which invoked this stored function/trigger第一次插入值=＆＃39; 0.00＆＃39;

现在，我在尝试将新行插入表marketplace_saleperpartner

时收到此错误

assertion

请帮忙吗？

Answer 1

您可以使用BEFORE INSERT触发器将marketplace_saleperpartner.commision设置为value marketplace_saleperpartner。

CREATE TRIGGER marketplacedefaultcommission
BEFORE INSERT ON marketplace_saleperpartner
FOR EACH ROW
BEGIN
    SET new.commission = (
        SELECT value
        FROM   default_sellers_commissions
        WHERE  id = new.commission_id
    );
END;

SQL Fiddle示例

创建After插入触发器后无法插入表中

1 个答案: