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' )
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 $dates;
}
其他回答
这是非常灵活的。
/**
* Creating date collection between two dates
*
* <code>
* <?php
* # Example 1
* date_range("2014-01-01", "2014-01-20", "+1 day", "m/d/Y");
*
* # Example 2. you can use even time
* date_range("01:00:00", "23:00:00", "+1 hour", "H:i:s");
* </code>
*
* @author Ali OYGUR <alioygur@gmail.com>
* @param string since any date, time or datetime format
* @param string until any date, time or datetime format
* @param string step
* @param string date of output format
* @return array
*/
function date_range($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 $dates;
}
使用DateTime对象的PHP 5.2解决方案。但是startDate必须在endDate之前。
function createRange($startDate, $endDate) {
$tmpDate = new DateTime($startDate);
$tmpEndDate = new DateTime($endDate);
$outArray = array();
do {
$outArray[] = $tmpDate->format('Y-m-d');
} while ($tmpDate->modify('+1 day') <= $tmpEndDate);
return $outArray;
}
使用:
$dates = createRange('2010-10-01', '2010-10-05');
$日期包含:
Array( '2010-10-01', '2010-10-02', '2010-10-03', '2010-10-04', '2010-10-05' )
function GetDays($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');
$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($dates);
// dates ('2014-09-16', '2014-09-26')
//print result Array
(
[0] => 2014-09-16
[1] => 2014-09-17
[2] => 2014-09-18
[3] => 2014-09-19
[4] => 2014-09-20
[5] => 2014-09-21
[6] => 2014-09-22
[7] => 2014-09-23
[8] => 2014-09-24
[9] => 2014-09-25
[10] => 2014-09-26
)
我认为这是最简短的答案
按照您的喜好编辑代码
for ($x=strtotime('2015-12-01');$x<=strtotime('2015-12-30');$x+=86400)
echo date('Y-m-d',$x);