優化複雜的SQL表達式

Question

這個超出了我在SQL中的能力。我有一個查詢選擇具有特定半徑的對象的所有對象，完美地工作。它首先創建一個邊界框以獲取所有潛在候選項，然後從該邊界框內計算半徑以選擇結果。

 SELECT * 
    FROM (
    SELECT b.*, pr.postcode, pr.prize, pr.title, pr.collection, pr.redeemed, pr.delivery, pr.archived, bt.category, b.id as objectid, b.updated as changed, 
      p.radius, 
      p.distance_unit 
        * DEGREES(ACOS(COS(RADIANS(p.latpoint)) 
        * COS(RADIANS(b.lat)) 
        * COS(RADIANS(p.longpoint - b.lng)) 
        + SIN(RADIANS(p.latpoint)) 
        * SIN(RADIANS(b.lat)))) 
      AS distance 
     FROM bubbles AS b, bubble_prizes AS pr, bubble_types AS bt 
     JOIN ( 
      SELECT ? AS latpoint, ? AS longpoint, 
        ? AS radius,  ? AS distance_unit 
     ) AS p 
     WHERE pr.bubble = b.id 
     AND b.deleted = 0 
     AND b.type IN ($placeholders) 
     AND b.type = bt.type 
     AND b.updated > $since 
     AND b.lat 
     BETWEEN p.latpoint - (p.radius/p.distance_unit) 
      AND p.latpoint + (p.radius/p.distance_unit) 
     AND b.lng 
     BETWEEN p.longpoint - (p.radius/(p.distance_unit * COS(RADIANS(p.latpoint)))) 
      AND p.longpoint + (p.radius/(p.distance_unit * COS(RADIANS(p.latpoint)))) 

    ) AS d 
    WHERE distance <= radius 
    ORDER BY distance"; 

我的問題是，現在我想修改這個（這個查詢是從這篇文章http://www.movable-type.co.uk/scripts/latlong-db.html派生），因此它選擇的所有對象，其中

b.created =用戶ID或者b .catch_held = userid OR [物體在某點的某個半徑內（如上）]

它也需要高效。也就是說，在b.created = userid或b.catch_held = userid的情況下，我不想計算DISTANCE或執行任何邊界框計算，並且我不確定如何重構查詢來實現此目的。

請注意，無論創建還是catct_held，無距離相關條件都需要保留在原位。那是

 pr.bubble = b.id 
     AND b.deleted = 0 
     AND b.type IN ($placeholders) 
     AND b.type = bt.type 
     AND b.updated > $since 

有人可以幫忙嗎？

Answer 1

首先，您應該只使用明確的join語法重寫您的查詢。一個簡單的規則：從不在from子句中使用逗號。

您正在尋找效率，所以基本上有兩種方法。一種是保持查詢原樣並使用union來引入其他行。另一種是修改查詢。爲此，只需在內部查詢和外部查詢中添加條件即可。這些都是簡單的條件，所以它不那麼難。

我對「距離」相關條件的含義有點模糊。以下假定所有的非加盟條件是距離相關：

SELECT * 
FROM (SELECT b.*, pr.postcode, pr.prize, pr.title, pr.collection, pr.redeemed, pr.delivery, 
      pr.archived, bt.category, b.id as objectid, b.updated as changed, 
      p.radius, 
      (p.distance_unit 
         * DEGREES(ACOS(COS(RADIANS(p.latpoint)) 
         * COS(RADIANS(b.lat)) 
         * COS(RADIANS(p.longpoint - b.lng)) 
         + SIN(RADIANS(p.latpoint)) 
         * SIN(RADIANS(b.lat)))) 
      ) AS distance 
     FROM bubbles b JOIN 
      bubble_prizes pr 
      ON pr.bubble = b.id JOIN 
      bubble_types bt 
      ON b.type = bt.type CROSS JOIN 
      (SELECT ? AS latpoint, ? AS longpoint, ? AS radius, ? AS distance_unit 
      ) p 
     WHERE (b.deleted = 0 AND 
      b.type IN ($placeholders) AND 
      b.updated > $since AND 
      b.lat BETWEEN p.latpoint - (p.radius/p.distance_unit) AND p.latpoint + (p.radius/p.distance_unit) AND 
      b.lng BETWEEN p.longpoint - (p.radius/(p.distance_unit * COS(RADIANS(p.latpoint)))) AND p.longpoint + (p.radius/(p.distance_unit * COS(RADIANS(p.latpoint)))) 
      ) or 
      (b.created = userid OR b.catch_held = userid) 
    ) b 
WHERE distance <= radius or b.created = userid OR b.catch_held = userid 
ORDER BY distance; 

如果這不是你想要的東西，那麼它應該是足夠接近。