Php 减少mySQL查询和时间

我目前正在加快一个网站的速度，这是从一个查询返回300000多行。虽然我不认为DB服务器上的负载太大，但根据用户拥有的“库”数量，这个查询是在While循环中进行的

例如，乔的帐户中有10个画廊。每个图库都有x个图像，这些图像上有x个评论。因此当前正在运行的查询

SELECT count(*) as total 
FROM galleryimage a 
INNER JOIN imagecomments b ON a.id=b.imgId 
WHERE a.galleryId='".$row['id']."' 
AND b.note <> ''

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您可以合并查询并消除循环的需要：

SELECT 
    a.id, 
    a.created, 
    a.name, 
    b.clientName, 
    a.isFeatured, 
    a.views, 
    a.clientId,
    COALESCE(c.img_cnt, 0) AS gallery_image_count,
    COALESCE(c.comment_cnt, 0) AS gallery_comment_count
FROM 
    gallery a 
INNER JOIN 
    client b ON a.clientId = b.id 
LEFT JOIN
(
    SELECT aa.galleryId, 
           COUNT(DISTINCT aa.id) AS img_cnt, 
           COUNT(1) AS comment_cnt
    FROM galleryimage aa
    INNER JOIN imagecomments bb ON aa.id = bb.imgId
    WHERE bb.note <> ''
    GROUP BY aa.galleryId
) c ON a.id = c.galleryId
WHERE 
    a.isTemp = 0 AND
    a.clientRef = '{$clientRef}' AND
    a.finish = 1 AND
    a.isArchive = 0 
ORDER BY 
    a.created DESC

---

在创建查询时，您是否拥有所有库ID？@AlexandreP.Levasseur-是的，我有-更新的问题您说您不能更改架构，但可以添加表吗？是的，添加表是一种可能性…首先，如果您在创建查询时拥有所有的ID，您可以将WHERE子句修改为a.galleryId所在的类型。。。你在哪里替换。。。使用您拥有的ID，用逗号分隔。此外，如果您可以索引galleryId，则检索过程会快得多。

SELECT 
    a.id, 
    a.created, 
    a.name, 
    b.clientName, 
    a.isFeatured, 
    a.views, 
    a.clientId,
    COALESCE(c.img_cnt, 0) AS gallery_image_count,
    COALESCE(c.comment_cnt, 0) AS gallery_comment_count
FROM 
    gallery a 
INNER JOIN 
    client b ON a.clientId = b.id 
LEFT JOIN
(
    SELECT aa.galleryId, 
           COUNT(DISTINCT aa.id) AS img_cnt, 
           COUNT(1) AS comment_cnt
    FROM galleryimage aa
    INNER JOIN imagecomments bb ON aa.id = bb.imgId
    WHERE bb.note <> ''
    GROUP BY aa.galleryId
) c ON a.id = c.galleryId
WHERE 
    a.isTemp = 0 AND
    a.clientRef = '{$clientRef}' AND
    a.finish = 1 AND
    a.isArchive = 0 
ORDER BY 
    a.created DESC