摘要: 1 -- 判断指定日期是星期几 2 select 3 etl_dt 4 ,case when dayofweek(etl_dt) = 1 then 7 else dayofweek(etl_dt) - 1 end as week1 -- 周一至周七 5 ,datediff(etl_dt,'1970-01-05') % 7 + 1 ... 阅读全文
posted @ 2018-04-16 23:31 chenzechao 阅读(779) 评论(0) 推荐(0)