Oracle SQL 查询列出两个日期之间的周数

Question

我有一个要求

START_DATE : 03/01/2018
END_DATE : 31/01/2018

我需要一个查询，它将像这样列出这两个日期之间的所有星期开始日期和结束日期

StartWeek   EndWeek
03/01/2018  04/01/2018
07/01/2018  11/01/2018
14/01/2018  18/01/2018
21/01/2018  25/01/2018
28/01/2018  31/01/2018

这里的工作日是周日到周四。一周从星期日开始

Answer 1

试试这个，

CREATE OR REPLACE PACKAGE week_pkg
AS
    TYPE week_rec IS RECORD (start_week DATE, end_week DATE);
    TYPE week_tab IS TABLE OF week_rec;
    FUNCTION get_weeks (p_start_date DATE, p_end_date DATE) RETURN week_tab PIPELINED;
END;
/

CREATE OR REPLACE PACKAGE BODY week_pkg
AS
    FUNCTION get_weeks (p_start_date DATE, p_end_date DATE) RETURN week_tab PIPELINED
    IS
        v_date DATE;
        v_week_rec week_rec;
    BEGIN
        v_date := p_start_date - 7;

        LOOP
             v_week_rec.start_week := NEXT_DAY(v_date, 'SUNDAY');
             IF v_week_rec.start_week < p_start_date THEN
                 v_week_rec.start_week:= p_start_date;
             END IF;

             v_week_rec.end_week := NEXT_DAY(v_date, 'THURSDAY');


             IF v_week_rec.end_week >= p_end_date THEN
                 v_week_rec.end_week := p_end_date;
                 PIPE ROW (v_week_rec);
                 EXIT;
             ELSIF v_week_rec.end_week <= p_start_date THEN
                 v_week_rec.end_week := NEXT_DAY(v_week_rec.start_week, 'THURSDAY');
             END IF;

             v_date := v_week_rec.end_week;

             PIPE ROW (v_week_rec);
         END LOOP;
    END;
END;
/

SELECT *
  FROM table(week_pkg.get_weeks(to_date('03-JAN-2018', 'DD-MON-YYYY'), to_date('31-JAN-2018', 'DD-MON-YYYY')));

输出：

START_WEEK END_WEEK
---------- ---------
03-JAN-18  04-JAN-18
07-JAN-18  11-JAN-18
14-JAN-18  18-JAN-18
21-JAN-18  25-JAN-18
28-JAN-18  31-JAN-18

Answer 2

你可以使用这个 SQL。

WITH t (START_DATE, END_DATE)
     AS (SELECT TO_DATE ('03/01/2018', 'dd/mm/yyyy'),
                TO_DATE ('31/01/2018', 'dd/mm/yyyy')
           FROM DUAL)
    SELECT DISTINCT
           CASE
              WHEN LEVEL = 1
              THEN
                 START_DATE
              ELSE
                 CASE
                    WHEN TRUNC (START_DATE + LEVEL + 7, 'DAY') > END_DATE
                    THEN
                       TRUNC (START_DATE + LEVEL, 'DAY')
                    ELSE
                       TRUNC (START_DATE + LEVEL + 7, 'DAY')
                 END
           END
              START_DATE,
           CASE
              WHEN NEXT_DAY (
                      CASE
                         WHEN LEVEL = 1 THEN START_DATE
                         ELSE TRUNC (START_DATE + LEVEL + 7, 'DAY')
                      END,
                      'THURSDAY') > END_DATE
              THEN
                 END_DATE
              ELSE
                 NEXT_DAY (
                    CASE
                       WHEN LEVEL = 1
                       THEN
                          CASE
                             WHEN TRIM (TO_CHAR (START_DATE, 'DAY')) = 'THURSDAY'
                             THEN
                                START_DATE - 7
                             ELSE
                                START_DATE
                          END
                       ELSE
                          TRUNC (START_DATE + LEVEL + 7, 'DAY')
                    END,
                    'THURSDAY')
           END
              END_DATE
      FROM DUAL CROSS JOIN t
CONNECT BY LEVEL < END_DATE - START_DATE;

Answer 3

您可以使用递归子查询分解子句：

WITH input_dates ( start_date, end_date ) AS (
  SELECT DATE '2018-01-03', DATE '2018-01-31'
  FROM   DUAL
),
valid_start_date ( start_date, end_date ) AS (
  SELECT CASE
         WHEN start_date - TRUNC( start_date, 'IW' )
                IN (
                  0, -- Monday
                  1, -- Tuesday
                  2, -- Wednesday
                  3, -- Thursday
                  6  -- Sunday
                )
         THEN start_date
         ELSE NEXT_DAY( start_date, 'SUNDAY' )
         END,
         end_date
  FROM   input_dates
),
dates ( start_week, end_week, end_date ) AS (
  SELECT start_date,
         LEAST( NEXT_DAY( start_date, 'THURSDAY' ), end_date ),
         end_date
  FROM   valid_start_date
  WHERE  start_date <= end_date
UNION ALL
  SELECT NEXT_DAY( start_week, 'SUNDAY' ),
         LEAST( end_week + INTERVAL '7' DAY, end_date ),
         end_date
  FROM   dates
  WHERE  NEXT_DAY( start_week, 'SUNDAY' ) <= end_date
)
SELECT start_week, end_week
FROM   dates;

Oracle Live SQL

输出：

START_DATE END_DATE
---------- ----------
2018-01-03 2018-01-04
2018-01-07 2018-01-11
2018-01-14 2018-01-18
2018-01-21 2018-01-25
2018-01-28 2018-01-31

Answer 4

类似的东西：Oracle 11 及更高版本（也许还有 Oracle 9 和 10...）

如果你的一周从星期日开始，也许你应该在内部减去 1 天 sql。（"select trunc(dt, 'IW')-1 as beg_period" 表示从星期日开始的一周 - 而不是 "select trunc(dt, 'IW') as beg_period" 表示从星期一开始的一周）

with periods as
(
    select greatest(to_date(&date_from, 'dd.mm.yyyy'), beg_period) beg_period -- date_from or first day of week
           , least(to_date(&date_to, 'dd.mm.yyyy'), lead(beg_period) over (order by beg_period)-1 )  end_period -- date_to or last day of week (and =NULL if week begins AFTER date_to)
    from 
    (select trunc(dt, 'IW')-1 as beg_period -- calc the beginning of week for every day generated, taking week beginning in Sunday ("-1")
     from 
        (-- generate all the days between &date_from and FIRST DAY OF NEXT WEEK JUST AFTER &date_to (including)
         select to_date(&date_from, 'dd.mm.yyyy') + (level-1) as dt 
         from dual 
         connect by level <= (trunc(to_date(&date_to, 'dd.mm.yyyy'), 'IW') + 7 -- also calc the first day of next week just AFTER date_to; it needed 
                                                                               -- for lead(...) in further (just to calc end-of-week for date_to, by lead(...))
                               - to_date(&date_from, 'dd.mm.yyyy') 
                               + 1
                             )   
        )                              
     group by trunc(dt, 'IW') -- take only 1 beg_period for every week (may use "group by", or "distinct")
   )
)     
select beg_period
  , end_period - case to_char(end_period, 'DAY','NLS_DATE_LANGUAGE=''numeric date language''') when '7' then 2 when '6' then 1 else 0 end
   as end_period -- correction for weekend (Friday and Saturday) for week beginning in Sunday
from periods 
where end_period is not null -- exclude next week just after date_to (which has end_period=NULL)
order by 1
;

看最后"case"：这取决于NLS。在我的表达中，我的意思是 Saturday='7' 和 Friday='6'（我想如果一周从星期日开始是正确的）。

Answer 5

Elegant solution

  SELECT DISTINCT  TO_CHAR(z.START_DATE,'MMRRRR') MM  
            ,TO_CHAR(z.START_DATE,'DD-MON-RRRR') START_DATE 
            ,TO_CHAR(z.END_DATE,'DD-MON-RRRR') END_DATE
    FROM  (WITH C_RANGE AS   
    (SELECT TRUNC(:STARTDATE) + LEVEL - 1 FECHA_INI
                ,ADD_MONTHS(TRUNC(:STARTDATE),LEVEL) - 1 FECHA_FIN
        FROM DUAL 
    CONNECT BY LEVEL <= (TRUNC(:ENDDATE) - TRUNC(:STARTDATE)))
    SELECT DISTINCT
               (CASE WHEN L.FECHA_INI = TRUNC(:STARTDATE) THEN TRUNC(:STARTDATE) ELSE ADD_MONTHS(L.FECHA_FIN,-1) +1 END) START_DATE
              ,(CASE WHEN L.FECHA_FIN +1 >= TRUNC(:ENDDATE)  THEN  TRUNC(:ENDDATE) ELSE L.FECHA_FIN  END)  END_DATE  
      FROM  C_RANGE L
     WHERE ADD_MONTHS(L.FECHA_FIN,-1) <= TRUNC(:ENDDATE)
         AND  TRUNC(:ENDDATE) >= TRUNC(:STARTDATE)) z
   ORDER BY MM ASC ;

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