總結取決於多個條件：轉換爲單個查詢

Question

鑑於表：

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

我需要返回可單獨表達瞭如下的查詢彙總：

-- 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 

我如何重寫這些查詢到一個單一的並返回所有必要的？

Answer 1

我希望有這樣的查詢，因爲什麼，如果有一些places有沒有reservation又或者可能有多個預訂了。仍然可以安全地計算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 

Answer 2

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 

Answer 3

這些查詢可以通過工會結合在一起，但在工會的所有查詢必須返回相同的結果集，對於所有查詢的輸出必須做出相同。這是可以做到如下：

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