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
下面是一个函数,它将从两个方向返回日期范围,适用于PHP >=5.2.2:
function createRange($start, $end, $format = 'Y-m-d') {
$start = new DateTime($start);
$end = new DateTime($end);
$invert = $start > $end;
$dates = array();
$dates[] = $start->format($format);
while ($start != $end) {
$start->modify(($invert ? '-' : '+') . '1 day');
$dates[] = $start->format($format);
}
return $dates;
}
使用的例子:
print_r(createRange('2010-10-01', '2010-10-05'));
/*Array
(
[0] => 2010-10-01
[1] => 2010-10-02
[2] => 2010-10-03
[3] => 2010-10-04
[4] => 2010-10-05
)*/
print_r(createRange('2010-10-05', '2010-10-01', 'j M Y'));
/*Array
(
[0] => 5 Oct 2010
[1] => 4 Oct 2010
[2] => 3 Oct 2010
[3] => 2 Oct 2010
[4] => 1 Oct 2010
)*/
demo
其他回答
你也可以看看DatePeriod类:
$period = new DatePeriod(
new DateTime('2010-10-01'),
new DateInterval('P1D'),
new DateTime('2010-10-05')
);
这将为您提供一个包含DateTime对象的数组。
迭代
foreach ($period as $key => $value) {
//$value->format('Y-m-d')
}
简单却有魅力:
$period = new DatePeriod(new DateTime('2015-01-01'), new DateInterval('P1D'), new DateTime('2015-01-15 +1 day'));
foreach ($period as $date) {
$dates[] = $date->format("Y-m-d");
}
//ONLY SHOWING
echo '<pre>';
var_dump($dates);
echo '</pre>';
exit();
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 $dates_arr;
}
function GetDays($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 $aDays;
}
使用像
GetDays('2007-01-01', '2007-01-31');
我认为这是最简短的答案
按照您的喜好编辑代码
for ($x=strtotime('2015-12-01');$x<=strtotime('2015-12-30');$x+=86400)
echo date('Y-m-d',$x);