Question

在sql中它看起来像这样：

select to_char(d,'WW')    as Week
     , to_char(d,'YYYY')  as Year
     , to_char(d,'DD')    as "Day of Friday"
     , d
from (
      select (level-1)*7 + to_date('02/01/2015','DD/MM/YYYY') d
        from dual
      connect by level <= 53*5);

但我需要一个Postgres脚本......

Answer 1

你试过extract(year from <your date field>)吗？看date-time functions

Answer 2

Postgres日期时间掩码几乎相同。在你的情况下，我猜他们是平等的：

select to_char(d,'WW')    as Week
     , to_char(d,'YYYY')  as Year
     , to_char(d,'DD')    as "Day of Friday"
     , d
from to_date('02/01/2015','DD/MM/YYYY') d

给出：

01; 2015; 12; 2015-01-12

除非预计周不是第一个和'星期五'和＃34;不是一个月的一天......

Answer 3

感谢您的回复

这就是我想要的：

SELECT 
     extract('week' from d) as "Week"
    ,extract('year' from d) as "Year"
    ,extract('day' from d)  as "Friday Number"
FROM generate_series('2015-01-02'::date, '2015-12-31'::date, '1 week'::interval) d

如何返回周数+年+周五周数

3 个答案: