MYSQL查询优化，多个查询或一个大型查询

Question

我有一个查询有一些子查询（内部选择），我试着找出哪个更好的性能，更大的查询或很多较小的查询，当我的服务器上的所有时间都发生变化时，我发现很难尝试和计算差异。

我使用下面的查询一次返回10个结果，以使用分页（偏移和限制）在我的网站上显示。

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 

它会更好，从主查询中删除低于2个内选择，然后在我的PHP循环，调用2选择10次，一次在回路中的每个记录？

(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

避免子查询

据我了解你内心的选择，他们有两个目的：找到任何名称相关的图片，并计数相关图像的数量。你可能会实现双方使用左连接，而不是内部的选择：

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 

我没有试过，但也许是MySQL的发动机是足够聪明，只进行查询部分的行你”实际上返回。

请注意，根本没有任何保证哪个图像名称此查询将返回给定的一组图像。如果你想得到可重现的结果，你应该在那里使用一些聚合函数，例如MIN(advert_images.image_name)选择字典中的第一个图像。

单独的选择，但没有环

如果上述方法无效，即查询仍然会检查advert_images表所有行计算的结果，那么你很可能真的被执行更好第二个查询。然而，你可以尝试避免for循环，而是在一个查询中获取所有这些行：

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 

这个查询中的十个参数对应于十行您当前生成结果。请注意，广告没有的相关照片将不包含在该结果中。因此，请确保在您的代码中将num_photos设为零，并将imagename设为NULL。

临时表

另一种方式来实现你试图做什么是使用一个明确的临时内存表：第一，你选择你感兴趣的结果，然后检索所有相关信息。

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; 

这将确保所有其他表格仅针对您当前正在处理的结果进行查询。毕竟，关于advert_images表几乎没有什么魔力，除了你可能需要多行。

子查询作为从前款的方式加入因子

大厦，你甚至可以避免管理的临时表，代替它使用子查询：

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 

同样，你确定相关行仅使用adverts表中的数据，并且仅连接其他表中的必需行。很可能，该中间结果将在内部存储在一个临时表中，但这取决于SQL服务器的决定。

Answer 2

我认为MySQL使用文件排序+临时表来执行您的查询。这就是为什么在大餐桌上你的建议会带来更好的结果。一般来说，你最好执行较小的查询，然后是1大。