Question

我有一个数组，每个元素都包含一个在＆＃39; dd.mm.yy＆＃39;中形成的startdate和enddate。

 Array
    (
        [0] =
            (
                [date_start] = 23.07.2014
                [date_end] = 02.08.2014
            )
     )

现在我需要构建一个数组，该数组将包含date_start和date_end之间的每一天，以便稍后验证它们。

现在我的问题是如何计算这些日期之间的每一天并将其保存到数组？

Answer 1

如果您只是想要差异

<?php
        $arr=array
        (
            array
                (
                    'date_start' =>'23.07.2014',
                    'date_end' =>'02.08.2014'
                ),
            array
            (
                'date_start' =>'15.07.2014',
                'date_end' =>'02.08.2014'
            )
         );

         foreach($arr as $key=>$val){
            $diff=strtotime($val['date_end'])-strtotime($val['date_start']);
            $new_arr[]=array(
                                'day_start'=>$val['date_start'],
                                'day_end'=>$val['date_end'],
                                'difference'=>$diff/(24*60*60)-1
                            );  
         }
         print_r($new_arr);
    ?>

输出

Array
(
    [0] => Array
        (
            [day_start] => 23.07.2014
            [day_end] => 02.08.2014
            [difference] => 9
        )

    [1] => Array
        (
            [day_start] => 15.07.2014
            [day_end] => 02.08.2014
            [difference] => 17
        )

)

如果你想要在阵列之间的所有日期

<?php
    $arr=array
    (
        array
            (
                'date_start' =>'23.07.2014',
                'date_end' =>'02.08.2014'
            ),
        array
        (
            'date_start' =>'15.07.2014',
            'date_end' =>'02.08.2014'
        )
     );

     foreach($arr as $key=>$val){
        $diff=strtotime($val['date_end'])-strtotime($val['date_start']);
        $day_diff=$diff/(24*60*60);
        $alldiff=array();
        for($i=1;$i<=$day_diff-1;$i++){

            $alldiff[]=date('d.m.Y',(strtotime($val['date_start'])+$i*24*60*60));
        }
            $new_arr[]=array(
                                'day_start'=>$val['date_start'],
                                'day_end'=>$val['date_end'],
                                'difference'=>$alldiff
                            );  

     }
     echo "<pre>";
     print_r($new_arr);
?>

输出

Array
(
    [0] => Array
        (
            [day_start] => 23.07.2014
            [day_end] => 02.08.2014
            [difference] => Array
                (
                    [0] => 24.07.2014
                    [1] => 25.07.2014
                    [2] => 26.07.2014
                    [3] => 27.07.2014
                    [4] => 28.07.2014
                    [5] => 29.07.2014
                    [6] => 30.07.2014
                    [7] => 31.07.2014
                    [8] => 01.08.2014
                )

        )

    [1] => Array
        (
            [day_start] => 15.07.2014
            [day_end] => 02.08.2014
            [difference] => Array
                (
                    [0] => 16.07.2014
                    [1] => 17.07.2014
                    [2] => 18.07.2014
                    [3] => 19.07.2014
                    [4] => 20.07.2014
                    [5] => 21.07.2014
                    [6] => 22.07.2014
                    [7] => 23.07.2014
                    [8] => 24.07.2014
                    [9] => 25.07.2014
                    [10] => 26.07.2014
                    [11] => 27.07.2014
                    [12] => 28.07.2014
                    [13] => 29.07.2014
                    [14] => 30.07.2014
                    [15] => 31.07.2014
                    [16] => 01.08.2014
                )

        )

)

Answer 2

这应该有效：

$startDate = new DateTime('2014-05-01');
$endDate  = new DateTime('2014-06-02');
 // calculate diff
$diff = $startDate->diff($endDate);
$daysDiff = $diff->days;


for ($i = 0; $i < $daysDiff; $i++) {
    // clone so the original date will not be changed
    $currentDate = clone $startDate;
    $currentDate->modify('+'.$i.' days');
    echo $currentDate->format('Y-m-d H:i:s')."\n";
}

现在你只需要为每个数组条目创建它并将其保存到数组

Answer 3

为什么不将日期转换为时间戳，然后从timestamp_start循环到timestamp_end，按天数秒，然后将每个步骤转换为新日期以便稍后检查。

Answer 4

这样的东西？

// Convert start and end dates to unix timestamp
$start = strptime($date_array[0]['date_start'], '%d.%m.%Y');
$end = strptime($date_array[0]['date_end'], '%d.%m.%Y');

$dates = array();
while ($start < $end){
  $dates[] = date('d.m.Y', $start);
  $start += 86400 // One day equals 86400 seconds
}

Answer 5

试试这个，因为它适合我

function getDates($startTime, $endTime) {
$day = 86400;
$format = 'd.m.Y';
$startTime = strtotime($startTime);
$endTime = strtotime($endTime);
$numDays = round(($endTime - $startTime) / $day);  

$days = array();

for ($i = 1; $i < $numDays; $i++) { 
    $days[] = date($format, ($startTime + ($i * $day)));
}

return $days;

}

$ days = getDates（'23 .07.2014'，'02 .08.2014'）;

PHP日期差异，在数组之间保存每一天

5 个答案: