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
// Specify the start date. This date can be any English textual format
$date_from = "2018-02-03";
$date_from = strtotime($date_from); // Convert date to a UNIX timestamp
// Specify the end date. This date can be any English textual format
$date_to = "2018-09-10";
$date_to = strtotime($date_to); // Convert date to a UNIX timestamp
// Loop from the start date to end date and output all dates inbetween
for ($i=$date_from; $i<=$date_to; $i+=86400) {
echo date("Y-m-d", $i).'<br />';
}
其他回答
为了让穆斯塔法的回答更完整,这绝对是最简单和最有效的方法:
function getDatesFromRange($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 $dates_array;
}
// see the dates in the array
print_r( getDatesFromRange('2017-02-09', '2017-02-19') );
如果在调用函数时添加第三个参数,您甚至可以更改默认的输出日期格式,否则它将使用默认格式,即设置为'Y-m-d'。
我希望它能帮助你:)
public static function countDays($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 $dates;
}
这很简短,而且应该在PHP4+中工作。
function getDatesFromRange($start, $end){
$dates = array($start);
while(end($dates) < $end){
$dates[] = date('Y-m-d', strtotime(end($dates).' +1 day'));
}
return $dates;
}
使用DateTime对象的PHP 5.2解决方案。但是startDate必须在endDate之前。
function createRange($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');
$日期包含:
Array( '2010-10-01', '2010-10-02', '2010-10-03', '2010-10-04', '2010-10-05' )
我认为这是最简短的答案
按照您的喜好编辑代码
for ($x=strtotime('2015-12-01');$x<=strtotime('2015-12-30');$x+=86400)
echo date('Y-m-d',$x);