MYSQL查詢優化，多個查詢或一個大型查詢

Question

我有一個查詢有一些子查詢（內部選擇），我試着找出哪個更好的性能，更大的查詢或很多較小的查詢，當我的服務器上的所有時間都發生變化時，我發現很難嘗試和計算差異。

我使用下面的查詢一次返回10個結果，以使用分頁（偏移和限制）在我的網站上顯示。

SELECT adverts.*, breed.breed, breed.type, sellers.profile_name, sellers.logo, users.user_level , 
round(sqrt((((adverts.latitude - '51.558430') * (adverts.latitude - '51.558430')) * 69.1 * 69.1) + ((adverts.longitude - '-0.0069345') * (adverts.longitude - '-0.0069345') * 53 * 53)), 1) as distance, 
(SELECT advert_images.image_name FROM advert_images WHERE advert_images.advert_id = adverts.advert_id AND advert_images.main = 1 LIMIT 1) as imagename, 
(SELECT count(advert_images.advert_id) from advert_images WHERE advert_images.advert_id = adverts.advert_id) AS num_photos 
FROM adverts 
LEFT JOIN breed ON adverts.breed_id = breed.breed_id 
LEFT JOIN sellers ON (adverts.user_id = sellers.user_id) 
LEFT JOIN users ON (adverts.user_id = users.user_id) 
WHERE (adverts.status = 1) AND (adverts.approved = 1) 
AND (adverts.latitude BETWEEN 51.2692837281 AND 51.8475762719) AND (adverts.longitude BETWEEN -0.472015213613 AND 0.458146213613) 
having (distance <= '20') 
ORDER BY distance ASC 
LIMIT 0,10 

它會更好，從主查詢中刪除低於2個內選擇，然後在我的PHP循環，調用2選擇10次，一次在迴路中的每個記錄？

(SELECT advert_images.image_name FROM advert_images WHERE advert_images.advert_id = adverts.advert_id AND advert_images.main = 1 LIMIT 1) as imagename, 
(SELECT count(advert_images.advert_id) from advert_images WHERE advert_images.advert_id = adverts.advert_id) AS num_photos 

Answer 1

避免子查詢

據我瞭解你內心的選擇，他們有兩個目的：找到任何名稱相關的圖片，並計數相關圖像的數量。你可能會實現雙方使用左連接，而不是內部的選擇：

SELECT …, 
     advert_images.image_name AS imagename, 
     COUNT(advert_images.advert_id) AS num_photos, 
     … 
FROM … 
    LEFT JOIN advert_images ON advert_images.advert_id = adverts.advert_id 
… 
GROUP BY adverts.advert_id 
… 
LIMIT 0,10 

我沒有試過，但也許是MySQL的發動機是足夠聰明，只進行查詢部分的行你」實際上返回。

請注意，根本沒有任何保證哪個圖像名稱此查詢將返回給定的一組圖像。如果你想得到可重現的結果，你應該在那裏使用一些聚合函數，例如MIN(advert_images.image_name)選擇字典中的第一個圖像。

單獨的選擇，但沒有環

如果上述方法無效，即查詢仍然會檢查advert_images表所有行計算的結果，那麼你很可能真的被執行更好第二個查詢。然而，你可以嘗試避免for循環，而是在一個查詢中獲取所有這些行：

SELECT advert_images.image_name AS imagename, 
     COUNT(advert_images.advert_id) AS num_photos 
FROM advert_images 
WHERE advert_images.advert_id IN (?, ?, ?, ?, ?, ?, ?, ?, ?, ?) 
GROUP BY advert_images.advert_id 

這個查詢中的十個參數對應於十行您當前生成結果。請注意，廣告沒有的相關照片將不包含在該結果中。因此，請確保在您的代碼中將num_photos設爲零，並將imagename設爲NULL。

臨時表

另一種方式來實現你試圖做什麼是使用一個明確的臨時內存表：第一，你選擇你感興趣的結果，然後檢索所有相關信息。

CREATE TEMPORARY TABLE tmp 
SELECT adverts.advert_id, round(…) as distance 
FROM adverts 
WHERE (adverts.status = 1) AND (adverts.approved = 1) 
    AND (adverts.latitude BETWEEN 51.2692837281 AND 51.8475762719) 
    AND (adverts.longitude BETWEEN -0.472015213613 AND 0.458146213613) 
HAVING (distance <= 20) 
ORDER BY distance ASC 
LIMIT 0,10; 

SELECT tmp.distance, adverts.*, … 
     advert_images.image_name AS imagename, 
     COUNT(advert_images.advert_id) AS num_photos, 
     … 
FROM tmp 
    INNER JOIN adverts ON tmp.advert_id = adverts.advert_id 
    LEFT JOIN breed ON adverts.breed_id = breed.breed_id 
    LEFT JOIN sellers ON adverts.user_id = sellers.user_id 
    LEFT JOIN users ON adverts.user_id = users.user_id 
    LEFT JOIN advert_images ON advert_images.advert_id = adverts.advert_id 
GROUP BY adverts.advert_id 
ORDER BY tmp.distance ASC; 

DROP TABLE tmp; 

這將確保所有其他表格僅針對您當前正在處理的結果進行查詢。畢竟，關於advert_images表幾乎沒有什麼魔力，除了你可能需要多行。

子查詢作爲從前款的方式加入因子

大廈，你甚至可以避免管理的臨時表，代替它使用子查詢：

SELECT sub.distance, adverts.*, … 
     advert_images.image_name AS imagename, 
     COUNT(advert_images.advert_id) AS num_photos, 
     … 
FROM (SELECT adverts.advert_id, round(…) as distance 
     FROM adverts 
     WHERE (adverts.status = 1) AND (adverts.approved = 1) 
      AND (adverts.latitude BETWEEN 51.2692837281 AND 51.8475762719) 
      AND (adverts.longitude BETWEEN -0.472015213613 AND 0.458146213613) 
     HAVING (distance <= 20) 
     ORDER BY distance ASC 
     LIMIT 0,10; 
    ) AS sub 
    INNER JOIN adverts ON sub.advert_id = adverts.advert_id 
    LEFT JOIN breed ON adverts.breed_id = breed.breed_id 
    LEFT JOIN sellers ON (adverts.user_id = sellers.user_id) 
    LEFT JOIN users ON (adverts.user_id = users.user_id) 
    LEFT JOIN advert_images ON advert_images.advert_id = adverts.advert_id 
GROUP BY adverts.advert_id 
ORDER BY sub.distance ASC 

同樣，你確定相關行僅使用adverts表中的數據，並且僅連接其他表中的必需行。很可能，該中間結果將在內部存儲在一個臨時表中，但這取決於SQL服務器的決定。

Answer 2

我認爲MySQL使用文件排序+臨時表來執行您的查詢。這就是爲什麼在大餐桌上你的建議會帶來更好的結果。一般來說，你最好執行較小的查詢，然後是1大。