MySQL - 特定の日付に利用可能なホテルを見つける

Question

私は多くの同様の質問を読んだが、それは自分のために働くことができませんでした。私はホテルの予約システムを作成しており、予約された予約日に少なくとも1つの部屋があるホテルをリストしたいと思います。ここで

は、私のテーブル構造が

だとここで私はこれまで

SELECT 
    ac.id, 
    ac.name, 
    ac.link, 
    ac.type, 
    ac.country, 
    co.name AS countryName, 
    ac.state, 
    st.statename, 
    ac.cityid AS cityId, 
    ci.name AS cityName, 
    ac.addr, 
    ac.featuredimage, 
    ac.amenities, 
    ac.starrating, 
    ac.bookings_received, 
    ro.roomprice, 
    SUM(ro.numberofrooms) AS totalRooms, 
    roomsBooked = (
    SELECT 
    *, 
    SUM(numberofroomsbooked) AS roomsBooked 
    FROM 
    bookings 
    WHERE 
    reservationto >= '2016-07-07' AND reservationfrom <= '2016-07-09' 
    AND hotelid = ro.hotel 
) 
FROM 
    accomodations ac 
JOIN countries co ON ac.country = co.id 
JOIN states st ON ac.state = st.id 
JOIN city ci ON ci.id = ac.cityid 
JOIN room ro ON ac.id = ro.hotel 
WHERE 
    ac.active = 1 AND ac.delete = 0 
GROUP BY 
    ac.id 

を持っているクエリは、私は、MySQLへの新たなんだだと、これは本当に手に負えなくなってきています。アイデアは、ホテルの総客室数を計算し、指定された期間内に予約された客室の総数を差し引くことです。私はそれについていくつかの助けを得ることができますか？

サブクエリ完全にそれ自身ではなく、この1で

 SELECT 
     *, 
     SUM(numberofroomsbooked) AS roomsBooked 
     FROM 
     bookings 
     WHERE 
     reservationto >= '2016-07-07' AND reservationfrom <= '2016-07-09' 
     AND hotelid = ro.hotel 

作品。

Answer 1

編集：前のクエリ

SELECT 
    ac.id, 
    ac.name, 
    ac.link, 
    ac.type, 
    ac.country, 
    co.name AS countryName, 
    ac.state, 
    st.statename, 
    ac.cityid AS cityId, 
    ci.name AS cityName, 
    ac.addr, 
    ac.featuredimage, 
    ac.amenities, 
    ac.starrating, 
    ac.bookings_received, 
    ro.roomprice, 
    ro.totalRooms 
FROM 
    accomodations ac 
JOIN countries co ON ac.country = co.id 
JOIN states st ON ac.state = st.id 
JOIN city ci ON ci.id = ac.cityid 
JOIN (
    SELECT 
    hotel, 
    roomprice, 
    SUM(numberofrooms) AS totalRooms 
    FROM 
    room 
    GROUP BY 
    hotel 
) ro ON ac.id = ro.hotel 
JOIN (
    SELECT 
    hotelid, 
    SUM(numberofroomsbooked) ASroomsBooked 
    FROM 
    bookings 
    WHERE 
    reservationto >= '2016-07-07' AND reservationfrom <= '2016-07-09' 
    GROUP BY 
    hotelid 
) bo ON ac.id = bo.hotelid 
WHERE 
    ac.active = 1 AND ac.delete = 0 
GROUP BY 
    ac.id 

更新クエリは、selectインナーを取り出し、インナーとして、それを追加すると内蔵は、このクエリを

SELECT 
    ac.id, 
    ac.name, 
    ac.link, 
    ac.type, 
    ac.country, 
    co.name AS countryName, 
    ac.state, 
    st.statename, 
    ac.cityid AS cityId, 
    ci.name AS cityName, 
    ac.addr, 
    ac.featuredimage, 
    ac.amenities, 
    ac.starrating, 
    ac.bookings_received, 
    ro.roomprice, 
    SUM(ro.numberofrooms) AS totalRooms, 
    SUM(tm.roomsBooked) as roomsBooked, 
FROM 
    accomodations ac 
JOIN countries co ON ac.country = co.id 
JOIN states st ON ac.state = st.id 
JOIN city ci ON ci.id = ac.cityid 
JOIN room ro ON ac.id = ro.hotel 
(
    SELECT 
    hotelid, 
    SUM(numberofroomsbooked) AS roomsBooked 
    FROM 
    bookings 
    WHERE 
    reservationto >= '2016-07-07' AND reservationfrom <= '2016-07-09' 
    group by hotelid 
) tm on tm.hotelid=ro.hotel 
WHERE 
    ac.active = 1 AND ac.delete = 0 
GROUP BY 
    ac.id 

Answer 2

てみましたでしょうか？あなたが宿泊施設のホテルでグループ化しているので、予約された部屋で機能SUMを使用すると、予約された部屋の合計が返されます。私はそれを試していない。

私はそれがあなたの問題を解決することを願っています。

SELECT 
    ac.id, 
    ac.name, 
    ac.link, 
    ac.type, 
    ac.country, 
    co.name AS countryName, 
    ac.state, 
    st.statename, 
    ac.cityid AS cityId, 
    ci.name AS cityName, 
    ac.addr, 
    ac.featuredimage, 
    ac.amenities, 
    ac.starrating, 
    ac.bookings_received, 
    ro.roomprice, 
    SUM(ro.numberofrooms) AS totalRooms, 
    SUM(numberofroomsbooked) AS bookedRooms 
FROM 
    accomodations ac 
JOIN bookings b ON ac.id = b.hotelid 
JOIN countries co ON ac.country = co.id 
JOIN states st ON ac.state = st.id 
JOIN city ci ON ci.id = ac.cityid 
JOIN room ro ON ac.id = ro.hotel 
WHERE 
    ac.active = 1 AND ac.delete = 0 
    AND reservationto >= '2016-07-07' AND reservationfrom <= '2016-07-09' 
GROUP BY 
    ac.id;