预期的输入:

getDatesFromRange( '2010-10-01', '2010-10-05' );

预期的输出:

Array( '2010-10-01', '2010-10-02', '2010-10-03', '2010-10-04', '2010-10-05' )

当前回答

为了让穆斯塔法的回答更完整,这绝对是最简单和最有效的方法:

function getDatesFromRange($start_date, $end_date, $date_format = 'Y-m-d')
   {
      $dates_array = array();
      for ($x = strtotime($start_date); $x <= strtotime($end_date); $x += 86400) {
         array_push($dates_array, date($date_format, $x));
      }

      return $dates_array;
   }

   // see the dates in the array
   print_r( getDatesFromRange('2017-02-09', '2017-02-19') );

如果在调用函数时添加第三个参数,您甚至可以更改默认的输出日期格式,否则它将使用默认格式,即设置为'Y-m-d'。

我希望它能帮助你:)

其他回答

function datesbetween ($date1,$date2)
{
    $dates= array();
    for ($i = $date1
       ; $i<= $date1
       ; $i=date_add($i, date_interval_create_from_date_string('1 days')) ) 
    {            
       $dates[] = clone $i;
    }
    return $dates;
}
function getWeekdayDatesFrom($format, $start_date_epoch, $end_date_epoch, $range) {

    $dates_arr = array();

    if( ! $range) {
        $range = round(abs($start_date_epoch-$end_date_epoch)/86400) + 1;
    } else {
        $range = $range + 1; //end date inclusive
    }

    $current_date_epoch = $start_date_epoch;

    for($i = 1; $i <= $range; $i+1) {

        $d = date('N',  $current_date_epoch);

        if($d <= 5) { // not sat or sun
            $dates_arr[] = "'".date($format, $current_date_epoch)."'";
        }

        $next_day_epoch = strtotime('+'.$i.'day', $start_date_epoch);
        $i++;
        $current_date_epoch = $next_day_epoch;

    }

    return $dates_arr;
} 

为了让穆斯塔法的回答更完整,这绝对是最简单和最有效的方法:

function getDatesFromRange($start_date, $end_date, $date_format = 'Y-m-d')
   {
      $dates_array = array();
      for ($x = strtotime($start_date); $x <= strtotime($end_date); $x += 86400) {
         array_push($dates_array, date($date_format, $x));
      }

      return $dates_array;
   }

   // see the dates in the array
   print_r( getDatesFromRange('2017-02-09', '2017-02-19') );

如果在调用函数时添加第三个参数,您甚至可以更改默认的输出日期格式,否则它将使用默认格式,即设置为'Y-m-d'。

我希望它能帮助你:)

public static function countDays($date1,$date2)
{
    $date1 = strtotime($date1); // or your date as well
    $date2 = strtotime($date2);
    $datediff = $date1 - $date2;
    return floor($datediff/(60*60*24));
}

public static function dateRange($date1,$date2)
{
    $count = static::countDays($date1,$date2) + 1;
    $dates = array();
    for($i=0;$i<$count;$i++)
    {
        $dates[] = date("Y-m-d",strtotime($date2.'+'.$i.' days'));
    }
    return $dates;
}

有很多方法可以做到这一点,但最后这都取决于您使用的PHP版本。以下是所有解决方案的总结:

获取PHP版本:

echo phpinfo();

PHP 5。+。

$period = new DatePeriod(
     new DateTime('2010-10-01'),
     new DateInterval('P1D'),
     new DateTime('2010-10-05')
);

PHP 4 +

/**
 * creating between two date
 * @param string since
 * @param string until
 * @param string step
 * @param string date format
 * @return array
 * @author Ali OYGUR <alioygur@gmail.com>
 */
function dateRange($first, $last, $step = '+1 day', $format = 'd/m/Y' ) { 

    $dates = array();
    $current = strtotime($first);
    $last = strtotime($last);

    while( $current <= $last ) { 

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

    return $dates;
}

PHP < 4

你应该升级:)