oracle 周函数推算日期,使用oracle日期函数获取周数？

仔细阅读：

W - 周月(1-5)的第1点就开始的 月的第一天和第七结束。

WITH t AS (SELECT TO_DATE ('07-07-2016', 'dd-mm-rrrr') d FROM DUAL)

SELECT TO_CHAR (d, 'w') week,

TO_CHAR (d, 'd') day_no,

TO_CHAR (d, 'Day') Day

FROM t

WEEK DAY_NO DAY

---- ------ ----------

1 5 Donnerstag

1 row selected.

它可以使你怎么想的那样，独立NLS_TERRITORY的：

WITH t AS (SELECT TO_DATE ('07-08-2016', 'dd-mm-rrrr') d FROM DUAL)

SELECT trunc(TO_CHAR (d+1, 'j')/7)-trunc(TO_CHAR (trunc(trunc(d,'mm'),'IW')+1, 'j')/7)+1 week,

TO_CHAR (d, 'd') day_no,

TO_CHAR (d, 'Day') Day

FROM t

WEEK DAY_NO DAY

---------- ------ ----------

2 1 Sonntag

1 row selected.

但是这将是有趣的，当一个月的第一天是星期六。你将在本月的第六周。

WITH t AS (SELECT TO_DATE ('30-11-2014', 'dd-mm-rrrr') d FROM DUAL)

SELECT trunc(TO_CHAR (d+1, 'j')/7)-trunc(TO_CHAR (trunc(trunc(d,'mm'),'IW')+1, 'j')/7)+1 week,

TO_CHAR (d, 'd') day_no,

TO_CHAR (d, 'Day') Day

FROM t

WEEK DAY_NO DAY

---------- ------ ----------

6 1 Sonntag

1 row selected.