MYSQL查询优化，多个查询或一个大查询

我有一个查询，它有一些子查询（内部选择），我正在尝试找出哪一个对性能更好，一个更大的查询或许多更小的查询，我发现很难尝试和计时差异，因为它在我的服务器上一直在变化

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

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我认为MySQL使用文件排序+临时表来执行查询。这就是为什么在大桌子上，你的建议会有更好的效果。通常，执行较小的查询比执行较大的查询更好

避免子查询 据我所知，你的内心选择有两个目的：找到关联图像的任何名称，并计算关联图像的数量。您可能会使用左连接而不是内部选择来实现这两个目标：

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

我还没有尝试过，但也许MySQL引擎足够聪明，可以只对实际返回的行执行这部分查询

请注意，对于给定的一组图像，根本无法保证此查询将返回哪个图像名称。如果你想得到可复制的结果，你应该在那里使用一些聚合函数，例如

MIN（adverd\u images.image\u name）

来选择按字典顺序排列的第一幅图像

单独选择但不循环 如果上述操作不起作用，即查询仍将检查

advert\u images

表中计算结果的所有行，那么执行第二次查询可能会更好。但是，您可以尝试避免

for

循环，而是在单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

此查询中的十个参数对应于当前生成的结果的十行。请注意，没有相关照片的广告根本不会包含在结果中。因此，请确保在代码中将

num_photos

默认为零，并将

imagename

默认为

NULL

临时桌 另一种方法是使用显式的临时内存表：首先选择感兴趣的结果，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;

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同样，您可以仅使用

广告

表中的数据来确定相关行，并仅连接其他表中所需的行。最有可能的是，中间结果将在内部存储在一个临时表中，但这取决于SQL server。

Hi，因为我按距离排序是一个计算字段，所以它使用文件排序，当表较大时会减慢排序速度。因此，是否会对主查询表中的每个记录运行2个内部选择？如果是这样的话，我会认为在结果集上运行2个内部选择会更快。是的，内部选择将在每个rowHi上执行，感谢您的详细回答。我尝试了你提到的第一种方法，通过group by和删除内部选择，但是查询速度比原来慢了很多。临时表听起来确实是个好主意，但是当查询每秒在服务器上运行大约10次时，它能正常工作吗，因为网站非常繁忙？@user1052096，只要临时表方法的两个查询足够接近，影响应该很小。临时表是连接的本地表，因此不会有任何名称冲突。与第二个查询结果的许多列相比，

tmp

表的内存消耗应该很小，因此组合解决方案可能比原始查询使用更少的内存。但是我有另一个想法，我会马上编辑到我的答案中。嗨，MvG，谢谢你使用子查询进行的更新，它确实有效，但我不知道它是否更快。如果我只运行子查询，它将在0.02秒内运行，但是如果我在没有选择广告id的情况下运行子查询，它将在0.01秒的速度下运行两倍。

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

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;

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