mySQL触发器在控制台插入后工作,但在脚本插入之后不起作用

2024-10-06 12:34:12 发布

您现在位置:Python中文网/ 问答频道 /正文

我的扳机有问题。在

我设置了一个触发器,用于在表中插入后更新其他表。在

如果我从MySQL控制台进行插入,一切都很好,但是如果我从外部python脚本插入相同的数据,触发器什么也不做,如下所示。在

我尝试将定义器更改为“user”@“%”和“root”@“%”,但它仍然没有任何作用。在

mysql> select vid_visit,vid_money from videos where video_id=487;
+-----------+-----------+
| vid_visit | vid_money |
+-----------+-----------+
|        21 |     0.297 |
+-----------+-----------+
1 row in set (0,01 sec)

mysql> INSERT INTO `table`.`validEvents` ( `id` , `campaigns_id` , `video_id` , `date` , `producer_id` , `distributor_id` , `money_producer` , `money_distributor` , `type` ) VALUES ( NULL , '30', '487', '2010-05-20 01:20:00', '1', '0', '0.009', '0.000', 'PRE' );
Query OK, 1 row affected (0,00 sec)

mysql> select vid_visit,vid_money from videos where video_id=487;                                                                  

+-----------+-----------+
| vid_visit | vid_money |
+-----------+-----------+
|        22 |     0.306 |
+-----------+-----------+

DROP TRIGGER IF EXISTS `updateVisitAndMoney`//
CREATE TRIGGER `updateVisitAndMoney` BEFORE INSERT ON `validEvents`
 FOR EACH ROW BEGIN
    if (NEW.type = 'PRE') THEN
                SET @eventcash=NEW.money_producer + NEW.money_distributor;
        UPDATE campaigns SET cmp_visit_distributed = cmp_visit_distributed + 1 , cmp_money_distributed = cmp_money_distributed + NEW.money_producer + NEW.money_distributor WHERE idcampaigns = NEW.campaigns_id;
        UPDATE offer_producer SET ofp_visit_procesed = ofp_visit_procesed + 1 , ofp_money_procesed = ofp_money_procesed + NEW. money_producer WHERE ofp_video_id = NEW.video_id AND ofp_money_procesed = NEW. campaigns_id;
        UPDATE videos SET vid_visit = vid_visit + 1 , vid_money = vid_money + @eventcash WHERE video_id = NEW.video_id;

        if (NEW.distributor_id != '') then
            UPDATE agreements SET visit_procesed = visit_procesed + 1, money_producer = money_producer + NEW.money_producer, money_distributor = money_distributor + NEW.money_distributor WHERE id_campaigns = NEW. campaigns_id AND id_video = NEW.video_id AND ag_distributor_id = NEW.distributor_id;
            UPDATE eventForDay SET visit = visit + 1, money = money + NEW. money_distributor WHERE date = SYSDATE()  AND campaign_id = NEW. campaigns_id AND user_id = NEW.distributor_id;
            UPDATE eventForDay SET visit = visit + 1, money = money + NEW.money_producer WHERE date = SYSDATE() AND campaign_id = NEW. campaigns_id AND user_id= NEW.producer_id;
        ELSE
            UPDATE eventForDay SET visit = visit + 1, money = money + NEW. money_producer WHERE date = SYSDATE() AND campaign_id = NEW. campaigns_id AND user_id = NEW.producer_id;
        END IF;
    END IF;
END
//

Tags: producerandidnewvideoupdatevisitwhere
3条回答

我认为更可能的是遇到了未捕获的错误,而不是触发器没有执行,特别是因为它从控制台成功执行。在

您需要隔离错误发生的位置-触发器本身或调用脚本中。在

在python脚本中,打印出python发送给MySQL执行的SQL语句,以确保它的构造如您所期望的那样——例如,如果新建.type不等于'PRE',触发器将已执行,但不会导致任何更新。在

还要确保在插入时检查错误。我不是一个python程序员,所以我不能告诉你这是怎么做到的,但是{a1}似乎是你想要的。在

如果这两种方法都没有导致问题,请注释掉整个if (NEW.type = 'PRE') THEN块并进行简单的修改,例如设置新建.type“调试”。在确保触发器确实执行之后,继续重新测试,添加更多的实际代码,直到隔离问题为止。在

另外,与马科斯评论,如果脚本没有自动提交成功完成后,我会感到惊讶。事实上,我会对任何脚本/语言做出这样的声明。在

对于将来发现这个问题的人来说,我猜解决方案是将mysql autocommit设置为true。 打开连接查询后,请执行以下操作:

设置自动提交=1

Shell命令只在控制台中运行,而不在实际服务器上运行。您可以使用UDF或polling来完成所需的任务。在

相关问题 更多 >