Question

好的说我在数据库中保存了两个时间戳，格式为yyyymmdd，即开始日期和结束日期。

说这些是我的开始和结束日期：20110618-20110630

如何才能让20110618,20110619,20110620等一路回到20110630？

Answer 1

// Will return the number of days between the two dates passed in
function count_days( $a, $b ) {
    // First we need to break these dates into their constituent parts:
    $gd_a = getdate( $a );
    $gd_b = getdate( $b );

    // Now recreate these timestamps, based upon noon on each day
    // The specific time doesn't matter but it must be the same each day
    $a_new = mktime( 12, 0, 0, $gd_a['mon'], $gd_a['mday'], $gd_a['year'] );
    $b_new = mktime( 12, 0, 0, $gd_b['mon'], $gd_b['mday'], $gd_b['year'] );

    // Subtract these two numbers and divide by the number of seconds in a
    //  day. Round the result since crossing over a daylight savings time
    //  barrier will cause this time to be off by an hour or two.
    return round( abs( $a_new - $b_new ) / 86400 );
}

// Prepare a few dates
$date1 = strtotime( '20110618' );
$date2 = strtotime( '20110630' );

// Calculate the differences, they should be 43 & 11353
echo "<p>There are ", count_days( $date1, $date2 ), " days.</p>\n";
$days = count_days($date1, $date2);
for ($i = 0; $i < $days; $i++) {
    echo date('Ymd', $date1+(86400*$i));
}

使用以下功能：Get number of days between two dates。

Answer 2

这应该在PHP 5.3上打印所有日期

$start = '20110618';
$end = '20110630';

$date = new DateTime($start);
$end = new DateTime($end)->getTimestamp();

while ($date->getTimestamp() <= $end) {
    echo $date->format('Ymd');
    $date->add(new DateInterval('P1D'));
}

Answer 3

使用PHP datetime对象，因此以下是persudo代码

//Convert the 2 dates to a php datetime object
//Get the number of days differance inbetween
//For each day, 
      //create a new datetime object, N days after the youngest object
      //Echo the reasult out

Answer 4

你不能只做以下事情：（编辑包含实际的php循环）

$date = new DateTime('20110618');
do {
    $d = $date->format('Ymd') . "\n";
    echo $d;
    $date->add(new DateInterval('P1D'));
} while ($d < 20110630);

两个日期戳，在每天之间回应一些东西

4 个答案: