Question

我试图在一段时间内优化这个。这是一个巨大查询的一个小提取，但这是查询最后变慢的。我认为你可以从中获得简单的表格结构。有人知道如何做到这一优雅和苗条？

SELECT gref.gid AS gid, COUNT(o.id) AS cnt
FROM objects AS o 
RIGHT JOIN group_xref AS gx ON o.id = gx.oid 
LEFT JOIN group_xref AS gref ON o.id = gref.oid 
LEFT JOIN groups AS groups ON gref.gid = groups.id 
WHERE groups.att = 1 
GROUP BY gref.gid

好的，这是EXPLAIN： http://htmlpaste.com/ff849935d884a36076f74bd934b0212bac7658e9

Answer 1

如果不了解表结构和关系，很难说出查询在做什么。我想知道以下查询是否产生了相同的结果以及它是否运行得更快：

select gref.gid AS gid, COUNT(*) 
from group_xref gref LEFT JOIN
     groups
     ON gref.gid = groups.id 
WHERE groups.att = 1 
GROUP BY gref.gid;

至少，我认为双重加入group_xref是不必要的。以下是您的查询：

SELECT gx.gid AS gid, COUNT(o.id) AS cnt
FROM group_xref gx left join
     objects 0
     ON o.id = gx.oid LEFT JOIN
     groups groups
     ON gx.gid = groups.id 
WHERE groups.att = 1 
GROUP BY gx.gid;

如果我假设与对象的链接是1-many且所有对象id都有效，那么这简化为：

SELECT gx.gid AS gid, COUNT(gx.id) AS cnt
FROM group_xref gx left join
     groups groups
     ON gx.gid = groups.id 
WHERE groups.att = 1 
GROUP BY gx.gid;

（我注意到的是我开始的地方）。

Answer 2

确保存在以下索引：

objects.id
group_xref.oid
group_xref.gid
groups.id
groups.att

此外，与group_xref的右连接似乎是多余的，请尝试：

SELECT gref.gid AS gid, COUNT(o.id) AS cnt
FROM objects AS o 
LEFT JOIN group_xref AS gref ON o.id = gref.oid 
LEFT JOIN groups AS groups ON gref.gid = groups.id 
WHERE groups.att = 1 
GROUP BY gref.gid

丑陋和缓慢的mysql查询

2 个答案: