如何在Oracle中获得一周的第一天和最后一天？

Question

我需要从一些具有以下格式的字符串中获取第一天和一周中的最后一天:

'201118'

2011年是一年,18是一周的数字.知道一周的数量,我如何获得一周的第一天和最后一天？

我该怎么做呢？

Answer 1

周

select TRUNC(sysdate, 'iw') AS iso_week_start_date,
       TRUNC(sysdate, 'iw') + 7 - 1/86400 AS iso_week_end_date
from dual;

月

select 
TRUNC (sysdate, 'mm') AS month_start_date,
LAST_DAY (TRUNC (sysdate, 'mm')) + 1 - 1/86400 AS month_end_date
from dual;

Answer 2

如果您使用的是Oracle,此代码可以帮助您:

select 
TRUNC(sysdate, 'YEAR') Start_of_the_year,
TRUNC(sysdate, 'MONTH') Start_of_the_month,
TRUNC(sysdate, 'DAY') start_of_the_week,
TRUNC(sysdate+365, 'YEAR')-1 End_of_the_year,
TRUNC(sysdate+30, 'MONTH')-1 End_of_the_month,
TRUNC(sysdate+6, 'DAY')-1 end_of_the_week
from dual;


select 
TRUNC(sysdate, 'YEAR') Start_of_the_year,
TRUNC(sysdate+365, 'YEAR')-1 End_of_the_year,
TRUNC(sysdate, 'MONTH') Start_of_the_month,
TRUNC(sysdate+30, 'MONTH')-1 End_of_the_month,
TRUNC(sysdate, 'DAY')+1 start_of_the_week,  -- starting Monday
TRUNC(sysdate+6, 'DAY') end_of_the_week     -- finish Sunday
from dual;

Answer 3

假设您的意思是相对于一年的第一天的周数......

SELECT first_day_of_week, first_day_of_week+6 last_day_of_week
FROM (
  SELECT TO_DATE(YEAR||'0101','YYYYMMDD') + 7 * (week-1) first_day_of_week
  FROM (
    SELECT substr(yearweek,1,4) YEAR, to_number(substr(yearweek,5)) week
    FROM (
      SELECT '201118' yearweek FROM dual
    )
  )
)
;