2

我有几个日期范围，形式为DateTime $begin, DateTime $end. 这些范围可以以各种可能的方式重叠：

|-------|
               |=======|
           |------|
                     |======|
           |------------|
|=======|
  |---|

等等

我要做的是获取DateInterval第一个开始到最新一个结束（在上述情况下为第四个）之间的这些范围的长度（以秒或 为单位），不包括任何范围未涵盖的区域。

只有两个范围没有问题，但我不知道如何扩展它以处理两个以上。

编辑：

class Range {
    public DateTime $begin;
    public DateTime $end;
}

$ranges = getRanges(); # function that returns array of Range objects

function getActiveHours($_ranges = array()) {
  $sum = 0;
  # this is the function I'd like to have
  return $sum;
}

仅对于两个范围，我有一个返回DateInterval对象的函数：

function addTimeRanges(DateTime $b1, DateTime $e1, DateTime $b2, DateTime $e2) {
    $res = null;
    if ($e1 < $b2 || $e2 < $b1) { # separate ranges
        $r1 = $b1->diff($e1);
        $r2 = $b2->diff($e2);
        $res = addIntervals($r1, $r2);
    } else if ($b1 <= $b2 && $e1 >= $e2) { # first range includes second
        $res = $b1->diff($e1);
    } else if ($b1 > $b2 && $e1 < $e2) { # second range includes first
        $res = $b2->diff($e2);
    } else if ($b1 < $b2 && $e1 <= $e2 && $b2 <= $e1) { # partial intersection
        $res = $b1->diff($e2);
    } else if ($b2 < $b1 && $e2 <= $e1 && $b1 <= $e2) { # partial intersection
        $res = $b2->diff($e1);
    }
    return $res;
}

whereaddIntervals是一个将两个DateInterval对象之和作为另一个DateInterval对象返回的函数。

这是一些基本版本，在我的生产代码中我使用了很多其他不相关的东西。

为了简化，假设我们只有时间部分DateTime：('06:00:00' to '08:00:00'), ('07:00:00' to '09:00:00'), ('06 :00:00', '08:00:00'), ('11:00:00' to '12:00:00') （会有很多这样的范围）。我现在想要的结果是 4 小时（从 6:00 到 9:00 + 从 11:00 到 12:00）。

4

4 回答 4

0 
$ranges = array(
    array(date_create_from_format('U', 1364654958), date_create_from_format('U', 1364655758)), //800s (intersect with 2 row, 700s) = 100s
    array(date_create_from_format('U', 1364654658), date_create_from_format('U', 1364655658)), //1000s (intersect with 1 row)
    array(date_create_from_format('U', 1364656858), date_create_from_format('U', 1364656958)), //100s
);  //total 1200s = 20m
array_multisort($ranges, SORT_ASC, array_map(function($a){return $a[0];}, $ranges));
$count = count($ranges)-1;
for ($i=0; $i < $count; $i++) {
    if ($ranges[$i+1][0] < $ranges[$i][1]) {
        $ranges[$i][1] = max($ranges[$i][1], $ranges[$i+1][1]);
        unset($ranges[$i+1]);
        $i--;
        $count--;
    }
}
$sum = date_create();
foreach ($ranges as $value) {
    date_add($sum, date_diff($value[0],$value[1]));
}
print_r(date_diff(date_create(), $sum));
于 2013-03-30T15:04:31.890 回答
0

我建议您创建一个函数，该函数返回 Range 类的一个实例，该实例的属性设置为整个周期的开始和结束。像这样的东西： -

class Range
{
    public $startDate;
    public $endDate;

    public function __construct(\DateTime $startDate, \DateTime $endDate)
    {
        $this->startDate = $startDate;
        $this->endDate = $endDate;
    }

    public function getInterval()
    {
        return $this->startDate->diff($this->endDate);
    }

    public function getSeconds()
    {
        return $this->endDate->getTimestamp() - $this->startDate->getTimestamp();
    }
}

我选择创建一个最小的工厂类，除其他外，它可以为你做这种类型的计算：

class Ranges
{
    private $ranges = array();

    public function addRange(\Range $range)
    {
        $this->ranges[] = $range;
    }

    public function getFullRange()
    {
        $fullRange = new \Range($this->ranges[0]->startDate, $this->ranges[0]->endDate);
        foreach($this->ranges as $range){
            if($range->startDate < $fullRange->startDate){
                $fullRange->startDate = $range->startDate;
            }
            if($range->endDate > $fullRange->endDate){
                $fullRange->endDate = $range->endDate;
            }
        }
        return $fullRange;
    }
}

一些代码来证明它的工作原理： -

$ranges = new \Ranges();
$ranges->addRange(new \Range(new \DateTime(), new \DateTime('+ 2 hours')));
$ranges->addRange(new \Range(new \DateTime('1st Jan 2012'), new \DateTime('3rd Jan 2012')));
$ranges->addRange(new \Range(new \DateTime('- 4 days'), new \DateTime('+ 30 days')));
$fullRange = $ranges->getFullRange();
var_dump($fullRange);
var_dump($fullRange->getInterval());
var_dump($fullRange->getSeconds());

在我运行它时，我得到了以下结果：-

object(Range)[11]
  public 'startDate' => 
    object(DateTime)[6]
      public 'date' => string '2012-01-01 00:00:00' (length=19)
      public 'timezone_type' => int 3
      public 'timezone' => string 'Europe/London' (length=13)
  public 'endDate' => 
    object(DateTime)[10]
      public 'date' => string '2013-05-14 19:36:06' (length=19)
      public 'timezone_type' => int 3
      public 'timezone' => string 'Europe/London' (length=13)

object(DateInterval)[12]
  public 'y' => int 1
  public 'm' => int 4
  public 'd' => int 13
  public 'h' => int 19
  public 'i' => int 36
  public 's' => int 6
  public 'invert' => int 0
  public 'days' => int 499

int 43180566

这将以任何顺序处理任意数量的 Range 对象，并始终返回一个 Range 对象，该对象为您提供所提供的所有范围所跨越的最早和最晚日期。

我还添加了一些方法，允许您以 DateInterval 实例或秒数的形式获取结果。

于 2013-04-14T16:55:21.470 回答
0

给定任务的示例

class DateRange
{
    private $startDate;
    private $endDate;

    public function getStart(){
        return clone $this->startDate;
    }

    public function getEnd(){
        return clone $this->endDate;
    }

    public function __construct(\DateTime $startDate, \DateTime $endDate = null)
    {
        $this->startDate = $startDate;
        if (is_null($endDate)) {
            $this->endDate = new \DateTime();
        } else {
            $this->endDate = $endDate;
        }
    }
}

class DateRanges
{
    private $ranges = array();

    public function addRange(\DateRange $range)
    {
        $this->ranges[] = $range;
    }

    private function _RageToArray(\DateRange $_in)
    {
        $_r = array();
        $start = $_in->getStart();
        $end = $_in->getEnd();
        while($start<$end){
            $_r[$start->format('Y-m-d')] = null;
            $start->modify('+1 days');
        }
        return $_r;
    }

    public function getDaysCount()
    {
        $_r = array();

        foreach($this->ranges as $range){
            $_r += $this->_RageToArray($range);
        }
        return count($_r);
    }
}

$today = new DateTime();
$ranges = new DateRanges();

$x = new stdClass();
$x->start = (clone $today);
$x->start->modify('-3 years');
$x->end = (clone $x->start);
$x->end->modify('+1 month');
$ranges->addRange(new DateRange($x->start, $x->end));

$x = new stdClass();
$x->start = (clone $today);
$x->start->modify('-3 years');
$x->end = (clone $x->start);
$x->end->modify('+15 days');
$ranges->addRange(new DateRange($x->start, $x->end));

$x = new stdClass();
$x->start = (clone $today);
$x->start->modify('-4 years');
$x->end = (clone $x->start);
$x->end->modify('+15 days');
$ranges->addRange(new DateRange($x->start, $x->end));

echo $ranges->getDaysCount() . ' must be near ' . (31 + 15) . PHP_EOL;
于 2013-07-12T11:49:24.850 回答
0

将日期转换为时间戳后，以下代码可用作解决方案的一部分： https ://stackoverflow.com/a/3631016/1414555

一旦 $data 是带有时间戳的数组，您就可以使用它：

usort($data, function($a, $b) { return $a[0] - $b[0]; });

$n = 0; $len = count($data);
for ($i = 1; $i < $len; ++$i) {
    if ($data[$i][0] > $data[$n][1] + 1)
        $n = $i;
    else {
        if ($data[$n][1] < $data[$i][1])
            $data[$n][1] = $data[$i][1];
        unset($data[$i]);
    }
}

$duration = 0; //Duration in seconds
foreach ($data as $range)
    $duration += ($range[1] - $range[0]);
于 2020-10-08T17:19:43.297 回答