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
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;
}