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

这很简短,而且应该在PHP4+中工作。

function getDatesFromRange($start, $end){
    $dates = array($start);
    while(end($dates) < $end){
        $dates[] = date('Y-m-d', strtotime(end($dates).' +1 day'));
    }
    return $dates;
}

其他回答

这是非常灵活的。

/**
 * Creating date collection between two dates
 *
 * <code>
 * <?php
 * # Example 1
 * date_range("2014-01-01", "2014-01-20", "+1 day", "m/d/Y");
 *
 * # Example 2. you can use even time
 * date_range("01:00:00", "23:00:00", "+1 hour", "H:i:s");
 * </code>
 *
 * @author Ali OYGUR <alioygur@gmail.com>
 * @param string since any date, time or datetime format
 * @param string until any date, time or datetime format
 * @param string step
 * @param string date of output format
 * @return array
 */
function date_range($first, $last, $step = '+1 day', $output_format = 'd/m/Y' ) {

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

    while( $current <= $last ) {

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

    return $dates;
}

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;
}

我认为这是最简短的答案

按照您的喜好编辑代码

for ($x=strtotime('2015-12-01');$x<=strtotime('2015-12-30');$x+=86400)
echo date('Y-m-d',$x);
<?
print_r(getDatesFromRange( '2010-10-01', '2010-10-05' ));

function getDatesFromRange($startDate, $endDate)
{
    $return = array($startDate);
    $start = $startDate;
    $i=1;
    if (strtotime($startDate) < strtotime($endDate))
    {
       while (strtotime($start) < strtotime($endDate))
        {
            $start = date('Y-m-d', strtotime($startDate.'+'.$i.' days'));
            $return[] = $start;
            $i++;
        }
    }

    return $return;
}