在PHP中如何从两个日期时间计算小时间隔？

我正在创建一个在线预订系统。当用户单击日历中的某个日期时，它返回该日期的两个日期时间开始和结束。我试图计算我能够计算的从开始到结束的所有小时数，但我需要以间隔显示小时数

假设用户添加了明天10.00-14.00的可用时间，那么我需要如下显示时间：

10.00-11.00

11.00-12.00

12.00-13.00

13.00-14.00

具体日期

到目前为止我所拥有的

public function getTimes()
{

  $user_id = Input::get("id"); //get the user id
  $selectedDay = Input::get('selectedDay');   // We get the data from AJAX for the day selected, then we get all available times for that day
  $availableTimes = Nanny_availability::where('user_id', $user_id)->get();

  // We will now create an array of all booking datetimes that belong to the selected day
  // WE WILL NOT filter this in the query because we want to maintain compatibility with every database (ideally)

  // For each available time...
  foreach($availableTimes as $t => $value) {
    $startTime = new DateTime($value->booking_datetime);

    if ($startTime->format("Y-m-d") == $selectedDay) {
      $endTime = new DateTime($value->booking_datetime);

      date_add($endTime, DateInterval::createFromDateString('3600 seconds'));

      // Try to grab any appointments between the start time and end time
      $result = Nanny_bookings::timeBetween($startTime->format("Y-m-d H:i"), $endTime->format("Y-m-d H:i"));

      // If no records are returned, the time is okay, if not, we must remove it from the array
      if($result->first()) {
        unset($availableTimes[$t]);
      }

    } else {
      unset($availableTimes[$t]);
    }
  }

  return response()->json($availableTimes);
}

---

如何获取时间间隔？

根据您的问题，假设开始和结束之间的时差为1，您可以使用和来迭代时间，如：

$startDate = new DateTime( '2017-07-18 10:15:00' );
$endDate = new DateTime( '2017-07-18 14:15:00' );
$interval = new DateInterval('PT1H'); //interval of 1 hour
$daterange = new DatePeriod($startDate, $interval ,$endDate);

$times = [];
foreach($daterange as $date){
    $times[] = $date->format("H:i") . " -- " 
        . $date->add(new DateInterval("PT1H"))->format("H:i");
}
echo "<pre>"; print_r($times);
//gives
Array
(
    [0] => 10:15 -- 11:15
    [1] => 11:15 -- 12:15
    [2] => 12:15 -- 13:15
    [3] => 13:15 -- 14:15
)

---

根据您的问题，假设开始和结束之间的小时差为1，您可以使用和来迭代时间，如下所示：

$startDate = new DateTime( '2017-07-18 10:15:00' );
$endDate = new DateTime( '2017-07-18 14:15:00' );
$interval = new DateInterval('PT1H'); //interval of 1 hour
$daterange = new DatePeriod($startDate, $interval ,$endDate);

$times = [];
foreach($daterange as $date){
    $times[] = $date->format("H:i") . " -- " 
        . $date->add(new DateInterval("PT1H"))->format("H:i");
}
echo "<pre>"; print_r($times);
//gives
Array
(
    [0] => 10:15 -- 11:15
    [1] => 11:15 -- 12:15
    [2] => 12:15 -- 13:15
    [3] => 13:15 -- 14:15
)

---

它使用的是普通php。您能给我一个如何以json形式返回日期的示例吗？@raqulka您可以使用json_encode for$times数组返回json数据，因为更新后的answerbounty是您的，先生。谢谢。它正在使用普通php。您能给我一个如何以json形式返回日期的示例吗？@raqulka您可以使用json_encode for$times数组返回json数据，因为更新后的answerbounty是您的，先生。谢谢