一周的第一天（从星期日开始）

I found that there is a function last_day for last day of month, and date_part(dow, date) for numeric day of week starting with Sunday, but I am trying to take a date, and get the first day of that week.

我发现每个月的最后一天有一个函数last_day，而date_part（dow，date）表示从星期日开始的数字星期几，但是我想要一个约会，并获得该周的第一天。

Meaning: if date='2018-02-14' then result should be '2018-02-11'.

含义：如果date ='2018-02-14'，则结果应为'2018-02-11'。

Any ideas?

有任何想法吗？

1 个解决方案

#1

You simply want to subtract the dow value from the current date.

您只想从当前日期中减去道具值。

select dateadd(d, -datepart(dow, my_date), my_date)  
from (select date('2018-02-14') as my_date)

> 2018-02-11 00:00:00.0

For example, if dow is 3 for 2018-02-14 - a Wednesday - you can subtract 3 days to get back to "day 0".

例如，如果2018-02-14（周三）的道绩为3，则可以减去3天以回到“第0天”。

There's also the date_trunc function which will truncate everything after a given datepart. This is a little clunky, and will only set you back to the previous Monday, not Sunday.

还有date_trunc函数，它会在给定的datepart之后截断所有内容。这有点笨重，只会让你回到上周一，而不是周日。

select date_trunc('week', my_date)
from (select date('2018-02-14') as my_date)

#1