如何在PL/SQL中迭代日期范围

Question

我需要编写一个报告,针对每个记录的日期范围的表生成摘要总计.

table data:
option   start_date   end_date
opt1     6/12/2009    6/19/2009
opt1     6/3/2009     6/13/2009
opt2     6/5/2009     6/6/2009

我想要的基本上是这样的:

date       option    count
6/1/2009   opt1      0
6/1/2009   opt2      0
6/2/2009   opt1      0
6/2/2009   opt2      0
6/3/2009   opt1      0
6/3/2009   opt2      1

我很难弄清楚如何迭代日期范围.我确信这是一个可以为此创建的简单游标,但我不知所措.最好是PL/SQL

更新:

我最终使用这里的例子来完成我想要做的事情.这将创建一个生成日期表的函数.

Answer 1

您将需要某种日历来循环一系列日期.我已经使用逐级连接技巧构建了一个.然后,您可以将日历与您的数据一起加入(即使当天没有选项,您也想要一行,因此可以交叉加入):

SQL> WITH calendar AS (
  2     SELECT to_date(:begin_date, 'mm/dd/yyyy') + ROWNUM - 1 c_date
  3       FROM dual
  4      CONNECT BY LEVEL <= to_date(:end_date, 'mm/dd/yyyy') 
                             - to_date(:begin_date, 'mm/dd/yyyy') + 1
  5  )
  6  SELECT c_date "date", d_option "option", COUNT(one_day)
  7    FROM (SELECT c.c_date, d.d_option,
  8                  CASE
  9                     WHEN c.c_date BETWEEN d.start_date AND d.end_date THEN
 10                      1
 11                  END one_day
 12             FROM DATA d, calendar c)
 13   GROUP BY c_date, d_option
 14  ORDER BY 1,2;

date        option COUNT(ONE_DAY)
----------- ------ --------------
01/06/2009  opt1                0
01/06/2009  opt2                0
02/06/2009  opt1                0
02/06/2009  opt2                0
03/06/2009  opt1                1
03/06/2009  opt2                0
04/06/2009  opt1                1
04/06/2009  opt2                0
05/06/2009  opt1                1
05/06/2009  opt2                1
06/06/2009  opt1                1
06/06/2009  opt2                1

12 rows selected

Answer 2

我使用的一个解决方案是将日期范围转换为可以在for循环中使用的整数范围,然后转换回日期以使用它执行操作.你不能以任何方式进行任何连接或任何方式,但它是一个比已经发布的小得多的解决方案:

declare
  start_date number;
  end_date number;
  business_date varchar2(8);
begin
  start_date := to_number(to_char(to_date('2013-04-25', 'yyyy-MM-dd'), 'j'));
  end_date := to_number(to_char(to_date('2013-05-31', 'yyyy-MM-dd'), 'j'));
  for cur_r in start_date..end_date loop
    business_date := to_char(to_date(cur_r, 'j'), 'yyyy-MM-dd');
    dbms_output.put_line(business_date);
  end loop;
end;

Answer 3

Using while loop (better)

declare
 dfrom date;
 dtill date;
 day date;
begin
    dfrom := TO_DATE('09.09.1988', 'dd.mm.yyyy');
    dtill := TO_DATE('19.09.1988', 'dd.mm.yyyy');
    day := dfrom;

WHILE day <= dtill
LOOP
   DBMS_OUTPUT.PUT_LINE(day);
   day := day + 1;
END LOOP;
    
end;
/

//using cursor
declare
 dfrom date;
 dtill date;
begin
    dfrom := TO_DATE('09.09.1988', 'dd.mm.yyyy');
    dtill := TO_DATE('19.09.1988', 'dd.mm.yyyy');
FOR cur IN (
      SELECT dfrom + LEVEL - 1 AS today
        FROM dual
        CONNECT BY LEVEL <= dtill - dfrom + 1
    ) LOOP
    DBMS_OUTPUT.PUT_LINE(cur.today);
    END LOOP;
    
end;
/

Answer 4

作为对其他技术的补充，我迭代日期的一种方法如下：

/* List of days for the past year, starting with today at midnight */
SELECT TRUNC(SYSDATE) + 1 - LEVEL AS today,
       TRUNC(SYSDATE) + 2 - LEVEL AS tomorrow
FROM DUAL
CONNECT BY LEVEL <= 365