PHP获得最接近给定时间数组的时间[重复]

Question

我有一个名为 $timeslots 的数组，其时隙如下：

array:32 [▼
  0 => "2018-12-15T12:00:00.0000000"
  1 => "2018-12-15T12:15:00.0000000"
  2 => "2018-12-15T12:30:00.0000000"
  3 => "2018-12-15T12:45:00.0000000"
  4 => "2018-12-15T13:00:00.0000000"
  5 => "2018-12-15T13:15:00.0000000"
  6 => "2018-12-15T13:45:00.0000000"
  7 => "2018-12-15T14:15:00.0000000"
  8 => "2018-12-15T14:30:00.0000000"
  9 => "2018-12-15T14:45:00.0000000"
  10 => "2018-12-15T15:00:00.0000000"
  11 => "2018-12-15T15:15:00.0000000"
  12 => "2018-12-15T15:30:00.0000000"
  13 => "2018-12-15T15:45:00.0000000"
  14 => "2018-12-15T16:15:00.0000000"
  15 => "2018-12-15T16:45:00.0000000"
  16 => "2018-12-15T17:00:00.0000000"
  17 => "2018-12-15T17:30:00.0000000"
  18 => "2018-12-15T17:45:00.0000000"
  19 => "2018-12-15T18:30:00.0000000"
  20 => "2018-12-15T18:45:00.0000000"
  21 => "2018-12-15T19:15:00.0000000"
  22 => "2018-12-15T19:45:00.0000000"
  23 => "2018-12-15T20:15:00.0000000"
  24 => "2018-12-15T20:45:00.0000000"
  25 => "2018-12-15T21:00:00.0000000"
  26 => "2018-12-15T21:15:00.0000000"
  27 => "2018-12-15T21:30:00.0000000"
  28 => "2018-12-15T21:45:00.0000000"
  29 => "2018-12-15T22:00:00.0000000"
  30 => "2018-12-15T22:15:00.0000000"
  31 => "2018-12-15T22:30:00.0000000"
]

另外，我有一个像这样的变量：

$expected_time = 2018-12-15T18:00:00.0000000; // this can be different value, so its not unique value

$expected_time 永远不会进入数组 $timeslots，但我需要找到最接近 $expected_time 的值...我该怎么做？

如何从数组$timeslots 到$expected_time 获取最接近的时隙值并计算以分钟为单位的差异？

有什么想法吗？

Answer 1

正如 Nico 在 cmets 中提到的，它非常简单。只是循环和计算时间差。

$timeslots = [...];
$expected_time = "2018-12-15T18:00:00.0000000";
$timestamp = strtotime($expected_time);
$diff = null;
$index = null;

foreach ($timeslots as $key => $time) {
    $currDiff = abs($timestamp - strtotime($time));
    if (is_null($diff) || $currDiff < $diff) {
        $index = $key;
        $diff = $currDiff;
    }
}

echo $timeslots[$index];

Answer 2

这里有一个适合您要求的解决方案：

function findClosestDate($expectedDate,$dates)
{

    $differenceInMinutes = null; 
    $expectedDate = new DateTime($expectedDate);
    $expectedDateEpoch = $expectedDate->getTimestamp();
    $returnIndex = -1;

    for($i = 0; $i<count($dates); $i++)
    {
        $dateObject = new DateTime($dates[$i]);
        $dateEpoch = $dateObject->getTimestamp();
        $difference = abs($expectedDateEpoch-$dateEpoch);
        $difference = $difference/60;
        if($differenceInMinutes === null || $difference < $differenceInMinutes)
        {
            $differenceInMinutes = $difference;
            $returnIndex = $i;
        }
    }

    return array(
        "closest" => $dates[$returnIndex],
        "difference" => $differenceInMinutes
    ) ;
}

这利用DateTime 类来创建DateTime 对象并获取相应的timestamp。然后通过expectedDate 和dates 数组中的条目之间的绝对差来计算分钟数。遍历整个数组后，最接近的匹配项和差值在一个数组中返回。

Answer 3

因为您的列表已经排序 - 您可以将元素放入数组并再次使用排序 - 并且在每次迭代中计算差异会快得多。

<?php
$timeslots = [
    ...
];

$expected_time = "2018-12-15T18:00:00.0000000";
$counter = count($timeslots);
$timeslots = array_flip($timeslots);
$timeslots[$expected_time] = $counter;
ksort($timeslots);

while (key($timeslots) !== $expected_time) {
    $prev = key($timeslots);
    next($timeslots);
}

next($timeslots);
$next = key($timeslots);

$expected_time = new \DateTime($expected_time);
$closestDiff = min(($expected_time)->diff(new \DateTime($prev)), (new \DateTime($next))->diff($expected_time));
var_dump($closestDiff->i);