Oracle sql 按当前日期对工作日进行排序 [英] Oracle sql sort week days by current day

问题描述

我正在尝试根据以下顺序对日期进行排序:星期六、星期日、星期一、星期二、星期三、星期四、星期五.我正在尝试使用案例:

I am trying to sort the days based on the order: Saturday, Sunday, Monday, Tuesday, Wednesday, Thursday, Friday. I am trying using case:

select day,
 CASE day
  WHEN  1 THEN 1
  WHEN  2 THEN 2
  WHEN  3 THEN 3
  WHEN  4 THEN 4
  WHEN  5 THEN 5
  WHEN  6 THEN 6
  WHEN  7 THEN 7
  else 0
  END as day_nr
from week where day in (1,2,3,4,5,6,7)
order by day_nr asc

当我选择一周中的所有日子时，这没问题.但是，如果我只想要 1、5、6 天，则排序不正确.获取第一天 - 星期一.如何进行?

This is ok when I select all the days of the week. But if I want only for the day 1,5,6 the ordering is not correct. Gets the first day -Monday. How to proceed?

推荐答案

在 Oracle 中，第 1 天默认为星期日.

In Oracle day 1 is Sunday by default.

SELECT * FROM
(
 SELECT trunc(sysdate) + LEVEL-1 my_dt
      , TO_CHAR(trunc(sysdate) + LEVEL-1, 'DY') Wk_Day
      , TO_CHAR(trunc(sysdate) + LEVEL-2, 'D' ) Day#
   FROM dual 
 CONNECT BY LEVEL <= 10
 )
WHERE Day# IN (1,5,6)
ORDER BY my_dt, Day#
/

MY_DT    WK_DAY    DAY#
------------------------
5/10/2013    FRI    5
5/11/2013    SAT    6
5/13/2013    MON    1
5/17/2013    FRI    5
5/18/2013    SAT    6

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