0
我试图将函数的返回结果用于另一个函数。这是我的第一个功能:从其他函数访问变量LARAVEL
public function availableRooms(){
if(isset($_POST['selectedDate'])){
$checkInDateReceived = str_replace('/', '-', $_POST['selectedDate']);
$checkInDate = date('Y-m-d', strtotime($checkInDateReceived));
$availableRooms = DB::table('rooms')
->leftJoin('reservations', function ($secondJoin) use($checkInDate) {
$secondJoin->where('reservations.checkout_date', '<', $checkInDate);
})
->leftJoin('reservation_room_rel', function ($firstJoin) {
$firstJoin
->on('reservation_room_rel.room_id', '=', 'rooms.id')
->on('reservation_room_rel.reservation_id', '=', 'reservations.id');
})
->where([
['rooms.status', '=', 1]
])
->whereNull('reservation_room_rel.room_id')
->get();
return $availableRooms;
}
}
这里是我的第二个功能:
public function returnAddReservation(){
$availableRoomsRet = $this->availableRooms();
$mealSupplement = new Meal;
$accommodationSupplement = new AccommodationSupplement;
$accommodationSupplements = $accommodationSupplement::where('status', 1)
->orderBy('name', 'asc')
->get();
$mealSupplements = $mealSupplement::where('status', 1)
->orderBy('name', 'asc')
->get();
return view('addReservation', [
'availableRooms' => $availableRoomsRet,
'accommodationSupplements' => $accommodationSupplements,
'mealSupplements' => $mealSupplements
]);
}
首先函数返回正确的结果,如果它独立运行,但是,当我dd($availableRoomsRet)
返回null。 任何想法如何正确访问$availableRooms
从public function availableRooms()
到public function returnAddReservation()
? 谢谢!
什么是'$这个 - > availableRooms()'的'var_dump'结果试试吗?它是NULL吗? – Chester
是的,它为空 –
然后可能你的查询失败,请确保你有一个查询结果。 – Chester