1

テーブルを考えると：

reservations (id, place_id, confirmed_at, paid_at) と places (id, name)

次のクエリを個別に表現できる集計を返す必要があります。

-- Confirmed
SELECT places.id, places.name, COUNT(reservations.*) as total_confirmed
FROM reservations
  INNER JOIN places ON places.id = reservations.place_id
WHERE
  reservations.confirmed_at IS NOT NULL
GROUP BY places.id, places.name

--  Paid
SELECT places.id, places.name, COUNT(reservations.*) as total_paid
FROM reservations
  INNER JOIN places ON places.id = reservations.place_id
WHERE
  reservations.paid_at IS NOT NULL
GROUP BY places.id, places.name

-- Paid Uncofirmed
SELECT places.id, places.name, COUNT(reservations.*) as total_paid_unconfirmed
FROM reservations
  INNER JOIN places ON places.id = reservations.place_id
WHERE
  reservations.paid_at IS NOT NULL AND reservations.confirmed_at IS NULL
GROUP BY places.id, places.name

これらのクエリを単一のクエリに書き直して、必要なものをすべて返すにはどうすればよいですか?

4

3 に答える 3

3 
SELECT places.id, places.name,
sum(case when (reservations.confirmed_at IS NOT NULL) then 1 else 0 end) as total_confirmed,
sum(case when (reservations.paid_at  IS NOT NULL) then 1 else 0 end) as total_paid,
sum(case when  (reservations.paid_at IS NOT NULL AND reservations.confirmed_at IS NULL) then 1 else 0 end) as total_confirmed_paid
FROM reservations
  INNER JOIN places ON places.id = reservations.place_id
GROUP BY places.id, places.name
于 2012-10-05T08:51:16.987 に答える
1

私はこのようなクエリを使用することを好みます。なぜなら、まだ予約されていない、またはすでに複数の予約がある可能性がある場合はどうなるからplacesですreservation。それでも安全に を計算できますSUM。

SELECT  d.*, a.total_confirmed, b.total_paid, c.total_paid_unconfirmed
FROM    places d
        LEFT JOIN
        (
            SELECT places.id, places.name, COUNT(reservations.*) as total_confirmed
            FROM reservations
              INNER JOIN places ON places.id = reservations.place_id
            WHERE
              reservations.confirmed_at IS NOT NULL
            GROUP BY places.id, places.name
        ) a ON d.id = a.id
        LEFT JOIN
        (
            SELECT places.id, places.name, COUNT(reservations.*) as total_paid
            FROM reservations
              INNER JOIN places ON places.id = reservations.place_id
            WHERE
              reservations.paid_at IS NOT NULL
            GROUP BY places.id, places.name
        ) b ON d.id = b.id
        LEFT JOIN
        (
            SELECT places.id, places.name, COUNT(reservations.*) as total_paid_unconfirmed
            FROM reservations
              INNER JOIN places ON places.id = reservations.place_id
            WHERE
              reservations.paid_at IS NOT NULL AND reservations.confirmed_at IS NULL
            GROUP BY places.id, places.name
        ) c ON d.id = c.id
于 2012-10-05T08:52:34.747 に答える
1

これらのクエリはユニオンを使用して組み合わせることができますが、ユニオンでは、すべてのクエリの出力が同じになるように、すべてのクエリが同じ結果セットを返す必要があります。これは次のように行うことができます。

SELECT places.id, places.name,'total_confirmed' as totalType, COUNT(reservations.*) as total
FROM reservations   INNER JOIN places ON places.id = reservations.place_id 
WHERE   reservations.confirmed_at IS NOT NULL 
GROUP BY places.id, places.name  
union all
SELECT places.id, places.name,'total_paid' as totalType, COUNT(reservations.*) as total
FROM reservations   INNER JOIN places ON places.id = reservations.place_id 
WHERE   reservations.paid_at IS NOT NULL 
GROUP BY places.id, places.name  
union all
SELECT places.id, places.name,'total_paid_unconfirmed' as totalType, COUNT(reservations.*) as total
FROM reservations   INNER JOIN places ON places.id = reservations.place_id 
WHERE   reservations.paid_at IS NOT NULL AND reservations.confirmed_at IS NULL 
GROUP BY places.id, places.name
于 2012-10-05T09:00:51.523 に答える