我有一个名为 sales_data 的表,其中有 3 列(id int、udf varchar(20)、date_of_sale datetime)。 我试图通过将时间调整为 6 小时来查找 date_of_sale 列的工作日,现在我必须将 udf 列更新为与 date_of_sale 对应的工作日。我有一个选择查询的想法,但如何更新 udf 列?
select weekday(subtime(s.date_of_sale ,'6:0:0')) as putdata, CASE WHEN weekday(subtime(s.date_of_sale ,'6:0:0'))=0 THEN 'Sunday' WHEN weekday(subtime(s.date_of_sale ,'6:0:0'))=1 THEN 'Monday' WHEN weekday(subtime(s.date_of_sale ,'6:0:0'))=2 THEN 'Tuesday' WHEN weekday(subtime(s.date_of_sale ,'6:0:0'))=3 THEN 'Wednesday' WHEN weekday(subtime(s.date_of_sale ,'6:0:0'))=4 THEN 'Thursday' WHEN weekday(subtime(s.date_of_sale ,'6:0:0'))=5 THEN 'Friday' WHEN weekday(subtime(s.date_of_sale ,'6:0:0'))=6 THEN 'Saturday' END as udf from sales_data s;
添加生成列(https://dev.mysql.com/doc/refman/8.0/en/create-table- generated-columns.html)以简化处理,并避免数据不一致:
查看演示:https://dbfiddle.uk/2d5iIvBv
(我没有得到相同的工作日,也许是区域设置?)
您的上述查询已经差不多了。您只需要添加更新语句即可。
下面的查询应该适合您。