Gor*_*don 14
我很确定之前已经回答了数千万次,但无论如何:
$start = strtotime('20-04-2010 10:00');
$end = strtotime('22-04-2010 10:00');
for($current = $start; $current <= $end; $current += 86400) {
echo date('d-m-Y', $current);
}
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该10:00部分是为了防止代码因夏令时而跳过或重复一天.
通过给出天数:
for($i = 0; $i <= 2; $i++) {
echo date('d-m-Y', strtotime("20-04-2010 +$i days"));
}
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使用PHP5.3
$period = new DatePeriod(
new DateTime('20-04-2010'),
DateInterval::createFromDateString('+1 day'),
new DateTime('23-04-2010') // or pass in just the no of days: 2
);
foreach ( $period as $dt ) {
echo $dt->format( 'd-m-Y' );
}
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你可以用mktime().
mktime()对于进行日期算术和验证很有用,因为它会自动计算超出范围输入的正确值.
如果您增加日期编号,则会获得有效的日期,即使您已超过月末:
<?php
$day= 25;
$dateEnd = mktime(0,0,0,5,3,2010);
do {
$dateCur = mktime(0,0,0,4,$day,2010);
$day++;
print date( 'd-m-y', $dateCur) .'<br>';
} while ($dateCur < $dateEnd);
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输出:
25-04-10
26-04-10
27-04-10
28-04-10
29-04-10
30-04-10
01-05-10
02-05-10
03-05-10
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