Sqlalchemy (Справочная информация об этой ошибке: https://sqlalche.me/e/20/e3q8) – таймаут соединения ⇐ Python

[Anonymous]

Сейчас я пытаюсь вернуть первый результат этой команды SQL:
SELECT backend.campsites.campsite_id AS backend_campsites_campsite_id, backend.campsites.campground_id AS backend_campsites_campground_id, backend.campsites.available_from AS backend_campsites_available_from, backend.campsites.available_to AS backend_campsites_available_to, backend.campsites.site_number AS backend_campsites_site_number, backend.campsites.size AS backend_campsites_size, backend.campsites.amenities AS backend_campsites_amenities, backend.campsites.availability AS backend_campsites_availability
FROM backend.campsites
WHERE backend.campsites.size = 'Medium' AND backend.campsites.campground_id = 2

Но когда я вызываю это в этой функции, я получаю ошибку соединения:
def assign_campsites(Campsites):
campsite_session = SessionLocal()
retrieve_all_bookings = bookings.find()
retrieve_available_campsites = campsite_session.query(Campsites).filter(Campsites.availability == True).all()
for booking in retrieve_all_bookings:
for campsite in retrieve_available_campsites:
if campsite.availability == True:
departure_date = datetime.strptime(booking['arrival_date'], "%Y-%m-%d") + timedelta(days=7)
existing_booking = campsites.find_one({
"booking_id": booking['booking_id'],
})
if not existing_booking:
if booking['campsite_size']

Подробнее здесь: https://stackoverflow.com/questions/791 ... 3q8-connec