在PHP / Oracle日期范围内工作(Working with date ranges in P

我需要找出许多不同的情况下，在不同的一天是如何发生的，来自许多不同的范围。也许最好用一个例子解释。

18-JAN-09 to 21-JAN-09
19-JAN09 to 20-JAN-09
20-JAN-09 to 20-JAN-09

使用上面的三个例子，我需要它来收集这些信息，并显示的东西有点像...

18th Jan: 1
19th Jan: 2
20th Jan: 3
21st Jan: 1

......我会抓住从Oracle数据库FWIW的信息（因此上述^格式），而且将有数百，甚至千计的记录，所以我的跛脚尝试做各种循环和if语句将永远采取跑。

有没有比较简单，高效这样的方式？我真的不太知道从哪里开始很遗憾...

谢谢

Answer 1:

无论你的数据库引擎或你的PHP代码将不得不遍历的时间范围。

下面是一些PHP代码做的总和。当天计数由存储year-month ，以避免为广泛日期范围内一个巨大的数组。

<?php

// Get the date ranges from the database, hardcoded for example
$dateRanges[0][0] = mktime(0, 0, 0, 1, 18, 2009);
$dateRanges[0][1] = mktime(0, 0, 0, 1, 21, 2009);
$dateRanges[1][0] = mktime(0, 0, 0, 1, 19, 2009);
$dateRanges[1][1] = mktime(0, 0, 0, 1, 20, 2009);
$dateRanges[2][0] = mktime(0, 0, 0, 1, 20, 2009);
$dateRanges[2][1] = mktime(0, 0, 0, 1, 20, 2009);

for ($rangeIndex = 0; $rangeIndex < sizeof($dateRanges); $rangeIndex++)
{
  $startDate = $dateRanges[$rangeIndex][0];
  $endDate = $dateRanges[$rangeIndex][1];

  // Add 60 x 60 x 24 = 86400 seconds for each day
  for ($thisDate = $startDate; $thisDate <= $endDate; $thisDate += 86400)
  {
    $yearMonth = date("Y-m", $thisDate);
    $day = date("d", $thisDate);

    // Store the count by year-month, then by day
    $months[$yearMonth][$day]++;
  }
}

foreach ($months as $yearMonth => $dayCounts)
{
  foreach ($dayCounts as $dayNumber => $dayCount)
  {
    echo $yearMonth . "-" . $dayNumber . ": " . $dayCount . "<br>";
  }
}

?>

Answer 2:

你可以使用所描述的方法的另一个SO ：

SQL> WITH DATA AS (
  2  SELECT to_date('18-JAN-09', 'dd-mon-rr') begin_date,
  3         to_date('21-JAN-09', 'dd-mon-rr') end_date FROM dual UNION ALL
  4  SELECT to_date('19-JAN-09', 'dd-mon-rr'),
  5         to_date('20-JAN-09', 'dd-mon-rr') FROM dual UNION ALL
  6  SELECT to_date('20-JAN-09', 'dd-mon-rr'),
  7         to_date('20-JAN-09', 'dd-mon-rr') FROM dual
  8  ),calendar AS (
  9  SELECT to_date(:begin_date, 'dd-mon-rr') + ROWNUM - 1 c_date
 10    FROM dual
 11  CONNECT BY LEVEL <= to_date(:end_date, 'dd-mon-rr')
 12                      - to_date(:begin_date, 'dd-mon-rr') + 1
 13  )
 14  SELECT c.c_date, COUNT(d.begin_date)
 15    FROM calendar c
 16    LEFT JOIN DATA d ON c.c_date BETWEEN d.begin_date AND d.end_date
 17   GROUP BY c.c_date
 18  ORDER BY c.c_date;

C_DATE      COUNT(D.BEGIN_DATE)
----------- -------------------
18/01/2009                    1
19/01/2009                    2
20/01/2009                    3
21/01/2009                    1

Answer 3:

你需要一台带有一排的每一天

test_calendar：

Day
16.01.2009
17.01.2009
18.01.2009
19.01.2009
20.01.2009
21.01.2009
22.01.2009
23.01.2009
24.01.2009
25.01.2009
26.01.2009

表测试中包含bagin和inctance的处理：

DFROM   DTILL
18.01.2009  21.01.2009
19.01.2009  20.01.2009
20.01.2009  20.01.2009

以下是你需要的查询：

select day, (select count(1) 
  from test where dfrom<=t.day and dtill>=t.day) from test_calendar t
 where day>to_date('15.01.2009','DD.MM.YYYY')
 order by day