0

特定の日付とその日付の価格を取得するために使用するクエリがありますが、特定の曜日の平均価格を取得するために同様のものを使用したいと考えています。

availables というテーブルから特定の日付を取得するために機能する現在のクエリを次に示します。

SELECT rooms.name, rooms.roomtype, rooms.id, max(availables.updated_at),
MAX(IF(to_days(availables.bookdate) - to_days('2009-12-10') = 0, (availables.price*0.66795805223432), '')) AS day1,
MAX(IF(to_days(availables.bookdate) - to_days('2009-12-10') = 1, (availables.price*0.66795805223432), '')) AS day2,
MAX(IF(to_days(availables.bookdate) - to_days('2009-12-10') = 2, (availables.price*0.66795805223432), '')) AS day3,
MAX(IF(to_days(availables.bookdate) - to_days('2009-12-10') = 3, (availables.price*0.66795805223432), '')) AS day4,
MAX(IF(to_days(availables.bookdate) - to_days('2009-12-10') = 4, (availables.price*0.66795805223432), '')) AS day5,
MAX(IF(to_days(availables.bookdate) - to_days('2009-12-10') = 5, (availables.price*0.66795805223432), '')) AS day6,
MAX(IF(to_days(availables.bookdate) - to_days('2009-12-10') = 6, (availables.price*0.66795805223432), '')) AS day7,
MIN(spots) as spots
     FROM `availables`
     INNER JOIN rooms
     ON availables.room_id=rooms.id
     WHERE rooms.hotel_id = '5064' AND bookdate
     BETWEEN '2009-12-10' AND DATE_ADD('2009-12-10', INTERVAL 6 DAY)
     GROUP BY rooms.name
     ORDER BY rooms.ppl

おそらくDAYSOFWEEK関数がto_days関数と大きく異なるため、機能しない私の最初の刺し傷...

SELECT rooms.id, rooms.name,
MAX(IF(DAYOFWEEK(availables.bookdate) - DAYOFWEEK('2009-12-10') = 0, (availables.price*0.66795805223432), '')) AS day1,
MAX(IF(DAYOFWEEK(availables.bookdate) - DAYOFWEEK('2009-12-10') = 1, (availables.price*0.66795805223432), '')) AS day2,
MAX(IF(DAYOFWEEK(availables.bookdate) - DAYOFWEEK('2009-12-10') = 2, (availables.price*0.66795805223432), '')) AS day3,
MAX(IF(DAYOFWEEK(availables.bookdate) - DAYOFWEEK('2009-12-10') = 3, (availables.price*0.66795805223432), '')) AS day4,
MAX(IF(DAYOFWEEK(availables.bookdate) - DAYOFWEEK('2009-12-10') = 4, (availables.price*0.66795805223432), '')) AS day5,
MAX(IF(DAYOFWEEK(availables.bookdate) - DAYOFWEEK('2009-12-10') = 5, (availables.price*0.66795805223432), '')) AS day6,
MAX(IF(DAYOFWEEK(availables.bookdate) - DAYOFWEEK('2009-12-10') = 6, (availables.price*0.66795805223432), '')) AS day7,rooms.ppl AS spots FROM `availables` 
 INNER JOIN `rooms` ON `rooms`.id = `availables`.room_id 
 WHERE (rooms.hotel_id = 5064 AND rooms.ppl > 3 AND availables.price > 0 AND availables.spots > 1) 
 GROUP BY rooms.name
 ORDER BY rooms.ppl

多分私はこれを非常に難しくしていて、誰かがもっと簡単な方法を知っています。

このようなデータが必要です

#Availables
id    room_id   price    spots    bookdate
1     26        $5       5        2009-10-20
2     26        $6       5        2009-10-21

に：

+----+-------+--------------------+---------------------+---------------------+---------------------+------+------+------+------+
| id | spots | name               | day1                | day2                | day3                | day4 | day5 | day6 | day7 |
+----+-------+--------------------+---------------------+---------------------+---------------------+------+------+------+------+
| 25 | 4     | Blue Room          | 14.9889786921381408 | 14.9889786921381408 | 14.9889786921381408 |      |      |      |      |
| 26 | 6     | Whatever           | 13.7398971344599624 | 13.7398971344599624 | 13.7398971344599624 |      |      |      |      |
| 27 | 8     | Some name          | 11.2417340191036056 | 11.2417340191036056 | 11.2417340191036056 |      |      |      |      |
| 28 | 8     | Another            | 9.9926524614254272  | 9.9926524614254272  | 9.9926524614254272  |      |      |      |      |
| 29 | 10    | Stuff              | 7.4944893460690704  | 7.4944893460690704  | 7.4944893460690704  |      |      |      |      |
+----+-------+--------------------+---------------------+---------------------+---------------------+------+------+------+---
4

1 に答える 1

0

私の理解が正しければ、「 - DAYOFWEEK('2009-12-10')」 DAYOFWEEK(availables.bookdate) はすでに曜日を表す数値を返しているので、取り出す必要があるようです。

また、DAYOFWEEK は 0 ～ 6 ではなく、1 ～ 7 の数値を返すため、それに応じて調整する必要があります。

"GROUP BY room_id, DAYOFWEEK(availables.bookdate)" を実行して、内側のサブクエリで部屋と日ごとに平均価格をグループ化し、外側のクエリでピボットを実行する方が効率的な場合があります。

于 2009-12-11T02:51:38.787 に答える