我想我差不多了 - 尝试获取当前时间然后再添加一个小时然后将选项输出到15分钟间隔 - 所以如果它是14:12,则第一个可用插槽将是14:15然后每个15分钟,直到我的结束时间
$start = new DateTime("Now");
$start->add(new DateInterval('PT1H'));
$end = new DateTime("2015-05-16 17:00");
$interval = new DateInterval('PT15M');
$period = new DatePeriod($start, $interval, $end);
$current_date = date('d-M-Y g:i:s A');
$current_time = strtotime($current_date);
foreach ($period as $dt)
{
echo $dt->format('H:i')."<br>";
}
答案 0 :(得分:2)
您可以将数字钳位到15的下一个倍数,如下所示:
ceil($num / 15) * 15
如上所述调整$start
日期。例如:
$start = new DateTime("today 14:12"); // 2015-05-16 14:12:00
$hour = (int) $start->format("H"); // 14
$minute = (int) $start->format("i"); // 12
$clamped = ceil($minute / 15) * 15; // 15
$start->setTime($hour, $clamped); // 2015-05-16 14:15:00
setTime
注意:超出范围的值会添加到其父值中。
答案 1 :(得分:0)
我假设你的问题是如何在第15分钟到达正确的位置?
这可以解决您的问题吗?:
<?php
$start = new DateTime("Now");
$currentMinutes = $start->format('i');
if ($currentMinutes > 15) {
$start->setTime(($start->format('H') + 1), 15);
} elseif($currentMinutes < 15) {
$start->setTime($start->format('H'), 15);
}
$end = new DateTime("2015-05-16 17:00");
$interval = new DateInterval('PT15M');
$period = new DatePeriod($start, $interval, $end);
$current_date = date('d-M-Y g:i:s A');
$current_time = strtotime($current_date);
foreach ($period as $dt)
{
echo $dt->format('H:i')."<br>";
}
或者如果您喜欢简短:
$start = new DateTime("2015-05-16 13:15");
$currentMinutes = $start->format('i');
$currentHour = $start->format('H');
$start->setTime(($currentMinutes > 15 ? ++$currentHour : $currentHour), 15);
[...]
答案 2 :(得分:-1)
$min15InSecs = 15*60;
$min15 = time()-(time()%$min15InSecs)+$min15InSecs;
$start = new DateTime(date("Y-m-d H:i", $min15));
$start->add(new DateInterval('PT1H'));
$end = new DateTime("2015-05-16 19:00");
$interval = new DateInterval('PT15M');
$period = new DatePeriod($start, $interval, $end);
$current_date = date('d-M-Y g:i:s A');
$current_time = strtotime($current_date);
foreach ($period as $dt)
{
echo $dt->format('H:i')."<br>";
}