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

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;
}

其他回答

  function GetDays($sStartDate, $sEndDate){  
      // Firstly, format the provided dates.  
      // This function works best with YYYY-MM-DD  
      // but other date formats will work thanks  
      // to strtotime().  
      $sStartDate = gmdate("Y-m-d", strtotime($sStartDate));  
      $sEndDate = gmdate("Y-m-d", strtotime($sEndDate));  

      // Start the variable off with the start date  
     $aDays[] = $sStartDate;  

     // Set a 'temp' variable, sCurrentDate, with  
     // the start date - before beginning the loop  
     $sCurrentDate = $sStartDate;  

     // While the current date is less than the end date  
     while($sCurrentDate < $sEndDate){  
       // Add a day to the current date  
       $sCurrentDate = gmdate("Y-m-d", strtotime("+1 day", strtotime($sCurrentDate)));  

       // Add this new day to the aDays array  
       $aDays[] = $sCurrentDate;  
     }  

     // Once the loop has finished, return the  
     // array of days.  
     return $aDays;  
   }  

使用像

GetDays('2007-01-01', '2007-01-31'); 
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;
}
$report_starting_date=date('2014-09-16');
$report_ending_date=date('2014-09-26');
$report_starting_date1=date('Y-m-d',strtotime($report_starting_date.'-1 day'));
while (strtotime($report_starting_date1)<strtotime($report_ending_date))
{

    $report_starting_date1=date('Y-m-d',strtotime($report_starting_date1.'+1 day'));
    $dates[]=$report_starting_date1;
  } 
  print_r($dates);

 // dates    ('2014-09-16', '2014-09-26')


 //print result    Array
(
[0] => 2014-09-16
[1] => 2014-09-17
[2] => 2014-09-18
[3] => 2014-09-19
[4] => 2014-09-20
[5] => 2014-09-21
[6] => 2014-09-22
[7] => 2014-09-23
[8] => 2014-09-24
[9] => 2014-09-25
[10] => 2014-09-26
)
<?
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;
}

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:)