我正在创建一个在线预订系统。当用户点击日历中的日期时,它会返回该日期的两个日期时间(开始和结束)。我试图计算从头到尾的所有小时数,但我需要按时间间隔显示小时数。
假设用户明天从10.00-14.00增加了可用时间,那么我需要显示这样的时间:
10.00-11.00
11.00-12.00
12.00-13.00
13.00-14.00
特定日期。
到目前为止我有什么。
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);
}
我如何获得间隔?
答案 0 :(得分:4)
假设根据您的问题,开始和结束之间的小时差异为1,您可以使用DateInterval和DatePeriod来迭代以下时间:
$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
)
<强>更新强>
您可以使用json_encode()来返回json次数据,如下:
$jsonTimes = json_encode($times);