当前状态基于周数 ORACLE答案

【问题标题】：Current status based on week number ORACLE当前状态基于周数 ORACLE
【发布时间】：2019-04-07 04:07:17
【问题描述】：

我收到此查询是为了获取从年初 (01/01/2018) 到明年年底 (31/12/2019) 的所有天数。

SELECT MYDATE,
TO_CHAR(NR_OF_SUNDAYS + 1,'FM09') WEEK_NUM,
FROM
(
SELECT MYDATE,
( (TRUNC(MYDATE,'DAY') - TRUNC(TRUNC(MYDATE,'YYYY'),'DAY')) / 7 ) +
CASE WHEN TO_CHAR(TRUNC(MYDATE,'YYYY'),'DAY') = 'SUN' THEN 1 ELSE 0 END AS NR_OF_SUNDAYS
FROM
( SELECT TRUNC (SYSDATE, 'YY') - 1 + LEVEL AS MYDATE
FROM DUAL
CONNECT BY LEVEL <= TRUNC (ADD_MONTHS (SYSDATE, 24), 'YY') -
TRUNC (SYSDATE, 'YY')
)
)

我需要一列指定以下情况：

1) MYDATE
2) 如果我的当前 =
3) 如果我当前的一周 + 一周然后 “下周”（不包括逾期）
4) 否则未来

非常感谢您的帮助。

【问题讨论】：

如果您编写了所显示的查询，您应该能够轻松找到其余问题的解决方案，不是吗？你被困在哪里了？
MYDATE 是一个 DATE 值，您不应该将它与 CHAR 进行比较！写CASE WHEN MYDATE < TRUNC(SYSDATE) THEN 'PAST DUE'

标签： sql oracle calendar oracle-sqldeveloper

【解决方案1】：

因此，在我的回答中，我尝试保留您的周数计算背后的逻辑。

但是请记住，您可以使用 oracle to_char(date,'WW')、to_char(date,'IW')、to_char(date,'W') 函数计算周数，然后您的生活会更轻松。

WW  Week of year (1-53) where week 1 starts on the first day of the year and continues to the seventh day of the year.
W   Week of month (1-5) where week 1 starts on the first day of the month and ends on the seventh.
IW  Week of year (1-52 or 1-53) based on the ISO standard.

说了这么多，我的解决方案只使用了 sql（请注意，定义和使用函数会容易得多），基于您的计算方法。

with date_table as (
SELECT MYDATE, to_number(TO_CHAR(NR_OF_SUNDAYS + 1,'FM09')) WEEK_NUM,  to_number(to_char(MYDATE+1,'IW')) as nu
FROM
(
SELECT MYDATE,
( (TRUNC(MYDATE,'DAY') - TRUNC(TRUNC(MYDATE,'YYYY'),'DAY')) / 7 ) +
CASE WHEN TO_CHAR(TRUNC(MYDATE,'YYYY'),'DY', 'NLS_DATE_LANGUAGE = american') = 'SUN' THEN 1 ELSE 0 END AS NR_OF_SUNDAYS
FROM
( SELECT TRUNC (SYSDATE, 'YY') - 1 + LEVEL AS MYDATE
FROM DUAL
CONNECT BY LEVEL <= TRUNC (ADD_MONTHS (SYSDATE, 24), 'YY') -TRUNC (SYSDATE,'YY')
)
)
),
todays_week as 
(
select distinct WEEK_NUM from date_table 
where trunc(sysdate)=trunc(mydate)
),
pre_final as (
select MYDATE,WEEK_NUM, (select  WEEK_NUM from todays_week) as todaysweek from date_table)
select MYDATE,sysdate,WEEK_NUM,todaysweek,
case when trunc(MYDATE) < trunc(sysdate) then 'PAST DUE' 
     when todaysweek = WEEK_NUM and abs(MYDATE-sysdate)<=7 then 'CURRENT WEEK'
     when todaysweek +1 = WEEK_NUM and abs(MYDATE-sysdate)<=14 then 'Next Week'
     else 'Future' end as description
     from pre_final;

主要思想是找到今天的周数，然后使用case when。

这是我与结果的小提琴链接。 http://sqlfiddle.com/#!4/3149e4/148

编辑 1：现在，类似的结果可以通过以下方式实现：

select res.*,
case when trunc(MYDATE) < trunc(sysdate) then 'PAST DUE' 
     when todaysweek = WEEK_NUM and abs(MYDATE-sysdate)<=7 then 'CURRENT WEEK'
     when todaysweek +1 = WEEK_NUM and abs(MYDATE-sysdate)<=14 then 'Next Week'
     else 'Future' end as description
 from (
SELECT MYDATE, to_number(to_char(MYDATE,'IW')) as WEEK_NUM,to_number(to_char(sysdate,'IW')) as todaysweek 
FROM
( SELECT TRUNC (SYSDATE, 'YY') - 1 + LEVEL AS MYDATE
FROM DUAL
CONNECT BY LEVEL <= TRUNC (ADD_MONTHS (SYSDATE, 24), 'YY') -TRUNC (SYSDATE,'YY')
)) res

【讨论】：

非常感谢您的帮助。
@Deluq 不客气 :) 请注意，我编辑了我的答案是为了处理下一年的相同周数......否则你会得到两组 current week 和 next week .
是的，幸运的是，我使用这张表只是为了提前 1 个月提出计划订单。
TO_CHAR(TRUNC(MYDATE,'YYYY'),'DAY') 将返回 "SUNDAY " - 不是 "SUN"，使用 TO_CHAR(TRUNC(MYDATE,'YYYY'),'DY', 'NLS_DATE_LANGUAGE = american') 我想知道这个查询如何提供所需的结果。
@WernfriedDomscheit apperanlty 这部分CASE WHEN TO_CHAR(TRUNC(MYDATE,'YYYY'),'DAY') = 'SUN' THEN 1 ELSE 0 END 始终是0 并且不参与计算。我留下了 OPs 查询，没有进一步调查。但是，既然您指出了这一点-显然您是正确的。