mysql查询优化步骤或如何优化查询

我对查询优化知之甚少，但我知道查询的执行顺序

FROM子句

WHERE子句

分组依据子句

有从句

SELECT子句

按条款订货

这就是我写的问题

SELECT 
    `main_table`.forum_id,
     my_topics.topic_id,
     (
        SELECT MAX(my_posts.post_id) FROM my_posts WHERE my_topics.topic_id = my_posts.topic_id 
     ) AS `maxpostid`,
     (
       SELECT my_posts.admin_user_id FROM my_posts WHERE my_topics.topic_id = my_posts.topic_id ORDER BY my_posts.post_id DESC LIMIT 1
     ) AS `admin_user_id`, 
     (
       SELECT my_posts.user_id FROM my_posts WHERE my_topics.topic_id = my_posts.topic_id ORDER BY my_posts.post_id DESC LIMIT 1
     ) AS `user_id`,
     (
      SELECT COUNT(my_topics.topic_id) FROM my_topics WHERE my_topics.forum_id = main_table.forum_id ORDER BY my_topics.forum_id DESC LIMIT 1
     ) AS `topicscount`,
     (
      SELECT COUNT(my_posts.post_id) FROM my_posts WHERE my_topics.topic_id = my_posts.topic_id ORDER BY my_topics.topic_id DESC LIMIT 1
     ) AS `postcount`, 
     (
      SELECT CONCAT(admin_user.firstname,' ',admin_user.lastname) FROM admin_user INNER JOIN my_posts ON my_posts.admin_user_id = admin_user.user_id WHERE my_posts.post_id = maxpostid ORDER BY my_posts.post_id DESC LIMIT 1
     ) AS `adminname`, 
    (
     SELECT forum_user.nick_name FROM forum_user INNER JOIN my_posts ON my_posts.user_id = forum_user.user_id WHERE my_posts.post_id = maxpostid ORDER BY my_posts.post_id DESC LIMIT 1
    ) AS `nickname`, 
    (
      SELECT CONCAT(ce1.value,' ',ce2.value) AS fullname FROM my_posts INNER JOIN customer_entity_varchar AS ce1 ON ce1.entity_id = my_posts.user_id INNER JOIN customer_entity_varchar AS ce2 ON ce2.entity_id=my_posts.user_id WHERE (ce1.attribute_id = 1) AND (ce2.attribute_id = 2) AND my_posts.post_id = maxpostid ORDER BY my_posts.post_id DESC LIMIT 1
    ) AS `fullname`
    FROM `my_forums` AS `main_table`
       LEFT JOIN `my_topics` ON main_table.forum_id = my_topics.forum_id 
    WHERE  (forum_status = '1')

现在我想知道是否有任何方法可以优化它？因为所有的逻辑都写在

Select

section而不是

From

，但是我不知道如何在查询的

From

部分中写相同的逻辑

这有什么区别吗？或者两者都一样

---

多亏了

相关子查询真的应该是最后的手段，它们通常会在RBAR中执行，并且考虑到您的许多子查询非常相似，尝试使用联接获得相同的结果将导致更少的表扫描

我注意到的第一件事是，您的所有子查询都包括表

my_posts

，并且大多数都包含按my_posts排序的

ORDER.post_id DESC LIMIT 1

，这些查询没有计数，没有分组依据，因此顺序和限制是多余的，因此我的第一步是加入my_posts：

SELECT  *
FROM    my_forums AS f
        LEFT JOIN my_topics AS t
            ON f.forum_id = t.forum_id 
        LEFT JOIN
        (   SELECT  topic_id, MAX(post_id) AS post_id
            FROM my_posts 
            GROUP BY topic_id
        ) AS Maxp
            ON Maxp.topic_id = t.topic_id 
        LEFT JOIN my_posts AS p
            ON p.post_id = Maxp.post_id
WHERE   forum_status = '1';

在这里，子查询只是确保您获得每个主题id的最新帖子。为了方便起见，我在这里缩短了您的表别名，我不确定您为什么要使用比实际表名长的表别名

现在您有了大量的查询，您可以开始在列中添加，为了获得post计数，我在子查询

Maxp

中添加了一个计数，我还必须添加一些连接以获得一些细节，例如名称：

SELECT  f.forum_id,
        t.topic_id,
        p.post_id AS `maxpostid`,
        p.admin_user_id,
        p.user_id,
        t2.topicscount,
        maxp.postcount,
        CONCAT(au.firstname,' ',au.lastname) AS adminname, 
        fu.nick_name AS nickname
        CONCAT(ce1.value,' ',ce2.value) AS fullname 
FROM    my_forums AS f
        LEFT JOIN my_topics AS t
            ON f.forum_id = t.forum_id 
        LEFT JOIN
        (   SELECT  topic_id, 
                    MAX(post_id) AS post_id,
                    COUNT(*) AS postcount
            FROM my_posts 
            GROUP BY topic_id
        ) AS Maxp
            ON Maxp.topic_id = t.topic_id 
        LEFT JOIN my_posts AS p
            ON p.post_id = Maxp.post_id
        LEFT JOIN admin_user AS au
            ON au.admin_user_id = p.admin_user_id
        LEFT JOIN forum_user AS fu
            ON fu.user_id = p.user_id
        LEFT JOIN customer_entity_varchar AS ce1
            ON ce1.entity_id = p.user_id
            AND ce1.attribute_id = 1
        LEFT JOIN customer_entity_varchar AS ce2
            ON ce2.entity_id = p.user_id
            AND ce2.attribute_id = 2
        LEFT JOIN
        (   SELECT  forum_id, COUNT(*) AS topicscount
            FROM    my_topics
            GROUP BY forum_id
        ) AS t2
            ON t2.forum_id = f.forum_id
WHERE   forum_status = '1';

我不熟悉您的模式，因此上面的内容可能需要一些调整，但主要内容仍然是-在子选择上使用连接

我要做的下一个优化阶段是去掉

customer\u entity\u varchar

表，或者至少停止使用它来存储诸如名字和姓氏之类的基本内容。实体属性值模型是一个SQL反模式，如果在

论坛用户

表中添加两列，

FirstName

和

LastName

，则会立即丢失查询中的两个联接。我不会参与太多的讨论，因为这已经被广泛讨论了很多次，我没有更多的补充

最后一个阶段是添加适当的索引，您最好决定什么是适当的，我建议您可能至少需要每个表中外键的索引，可能更多

---

编辑

要为每个

论坛id

获取一行，您需要使用以下命令：

SELECT  f.forum_id,
        t.topic_id,
        p.post_id AS `maxpostid`,
        p.admin_user_id,
        p.user_id,
        MaxT.topicscount,
        maxp.postcount,
        CONCAT(au.firstname,' ',au.lastname) AS adminname, 
        fu.nick_name AS nickname
        CONCAT(ce1.value,' ',ce2.value) AS fullname 
FROM    my_forums AS f
        LEFT JOIN
        (   SELECT  t.forum_id, 
                    COUNT(DISTINCT t.topic_id) AS topicscount,
                    COUNT(*) AS postCount,
                    MAX(t.topic_ID) AS topic_id
            FROM    my_topics AS t
                    INNER JOIN my_posts AS p
                        ON p.topic_id = p.topic_id
            GROUP BY t.forum_id
        ) AS MaxT
            ON MaxT.forum_id = f.forum_id
        LEFT JOIN my_topics AS t
            ON t.topic_ID = Maxt.topic_ID 
        LEFT JOIN
        (   SELECT  topic_id, MAX(post_id) AS post_id
            FROM    my_posts 
            GROUP BY topic_id
        ) AS Maxp
            ON Maxp.topic_id = t.topic_id 
        LEFT JOIN my_posts AS p
            ON p.post_id = Maxp.post_id
        LEFT JOIN admin_user AS au
            ON au.admin_user_id = p.admin_user_id
        LEFT JOIN forum_user AS fu
            ON fu.user_id = p.user_id
        LEFT JOIN customer_entity_varchar AS ce1
            ON ce1.entity_id = p.user_id
            AND ce1.attribute_id = 1
        LEFT JOIN customer_entity_varchar AS ce2
            ON ce2.entity_id = p.user_id
            AND ce2.attribute_id = 2
WHERE   forum_status = '1';

---

据我所知，MySQL用户在性能方面更喜欢连接，而不是相关的子选择。有没有可能重写为join with GROUP BY？@jarlh我想你错过了问题“但我不知道如何在查询的From部分编写相同的逻辑？”；）