我以日期2010-05-01开始,以2010-05-10结束。如何在PHP中遍历所有这些日期?


当前回答

<?php

    $start_date = '2015-01-01';
    $end_date = '2015-06-30';

    while (strtotime($start_date) <= strtotime($end_date)) {
        echo "$start_daten";
        $start_date = date ("Y-m-d", strtotime("+1 days", strtotime($start_date)));
    }

?>

其他回答

$startTime = strtotime('2010-05-01'); 
$endTime = strtotime('2010-05-10'); 

// Loop between timestamps, 1 day at a time 
$i = 1;
do {
   $newTime = strtotime('+'.$i++.' days',$startTime); 
   echo $newTime;
} while ($newTime < $endTime);

or

$startTime = strtotime('2010-05-01'); 
$endTime = strtotime('2010-05-10'); 

// Loop between timestamps, 1 day at a time 
do {
   $startTime = strtotime('+1 day',$startTime); 
   echo $startTime;
} while ($startTime < $endTime);

如果你使用Laravel并且想要使用Carbon,正确的解决方案如下:

$start_date = Carbon::createFromFormat('Y-m-d', '2020-01-01');
$end_date = Carbon::createFromFormat('Y-m-d', '2020-01-31');

$period = new CarbonPeriod($start_date, '1 day', $end_date);

foreach ($period as $dt) {
 echo $dt->format("l Y-m-d H:i:s\n");
}

记得加上:

使用碳\碳; 使用碳\ CarbonPeriod;

下面是另一个简单的实现

/**
 * Date range
 *
 * @param $first
 * @param $last
 * @param string $step
 * @param string $format
 * @return array
 */
function dateRange( $first, $last, $step = '+1 day', $format = 'Y-m-d' ) {
    $dates = [];
    $current = strtotime( $first );
    $last = strtotime( $last );

    while( $current <= $last ) {

        $dates[] = date( $format, $current );
        $current = strtotime( $step, $current );
    }

    return $dates;
}

例子:

print_r( dateRange( '2010-07-26', '2010-08-05') );

Array (
    [0] => 2010-07-26
    [1] => 2010-07-27
    [2] => 2010-07-28
    [3] => 2010-07-29
    [4] => 2010-07-30
    [5] => 2010-07-31
    [6] => 2010-08-01
    [7] => 2010-08-02
    [8] => 2010-08-03
    [9] => 2010-08-04
    [10] => 2010-08-05
)
<?php

    $start_date = '2015-01-01';
    $end_date = '2015-06-30';

    while (strtotime($start_date) <= strtotime($end_date)) {
        echo "$start_daten";
        $start_date = date ("Y-m-d", strtotime("+1 days", strtotime($start_date)));
    }

?>

这也包括最后的日期

$begin = new DateTime( "2015-07-03" );
$end   = new DateTime( "2015-07-09" );

for($i = $begin; $i <= $end; $i->modify('+1 day')){
    echo $i->format("Y-m-d");
}

如果你不需要最后的日期,只需从条件中删除=。