内部SQL查询优化

Question

我有SQL查询的问题，有什么办法如何加快这个查询？ 表字包含14 000行查询在本地主机上花费了4.5631秒。

SELECT 
    (SELECT SUM((SELECT COUNT(*) 
        FROM word w 
        WHERE w.lecture_id = l._id AND active = 1)) 
     FROM lecture l 
     WHERE l.book_id = b._id) AS active_word_count, 
    (SELECT SUM((SELECT COUNT(*) 
       FROM word w 
       WHERE w.lecture_id = l._id)) 
     FROM lecture l 
     WHERE book_id = b._id) AS word_count, 
    (SELECT COUNT(*) 
    FROM lecture l 
    WHERE l.book_id = b._id) AS lecture_count, 
    b._id, 
    b.name, 
    b.version 
FROM book b 



CREATE TABLE book (
    _id int(11) NOT NULL AUTO_INCREMENT, 
    `name` varchar(255) NOT NULL, 
    version tinyint(3) unsigned NOT NULL, 
    lang tinyint(4) NOT NULL DEFAULT '1', 
    PRIMARY KEY (_id) 
) ENGINE=MyISAM DEFAULT CHARSET=utf8 AUTO_INCREMENT=21 ; 

CREATE TABLE lecture (
    _id int(11) NOT NULL AUTO_INCREMENT, 
    book_id int(11) NOT NULL, 
    lecture_name varchar(255) NOT NULL, 
    PRIMARY KEY (_id) 
) ENGINE=MyISAM DEFAULT CHARSET=utf8 AUTO_INCREMENT=155 ; 

CREATE TABLE word (
    _id int(11) NOT NULL AUTO_INCREMENT, 
    question varchar(255) NOT NULL, 
    answer varchar(255) NOT NULL, 
    active tinyint(1) NOT NULL, 
    lecture_id int(11) NOT NULL, 
    PRIMARY KEY (_id) 
) ENGINE=MyISAM DEFAULT CHARSET=utf8 AUTO_INCREMENT=1 ; 

查询返回的数据，女巫都可以在这里：http://pastebin.com/80KNsU7Y

谢谢你的帮助。

Answer 1

我添加了一些JOIN。所以它现在应该更快。 http://sqlfiddle.com/#!2/eec80/15

SELECT 
    (SELECT COUNT(*) 
     FROM word w 
     JOIN lecture l ON l._id = w.lecture_id 
     WHERE l.book_id = b._id 
     AND w.active = 1) AS active_word_count, 
    (SELECT COUNT(*) 
     FROM word w 
     JOIN lecture l ON w.lecture_id = l._id 
     WHERE l.book_id = b._id) AS word_count, 
    (SELECT COUNT(*) 
     FROM lecture l 
     WHERE l.book_id = b._id) AS lecture_count, 
    b._id, 
    b.name, 
    b.version 
FROM book b 

Answer 2

我已经添加了几个索引，可以加快以这种方式编写的查询。他们也可能会帮助原创。 http://sqlfiddle.com/#!2/bdcf8/1

Select 
    Sum(Case When w.Active = 1 Then 1 Else 0 End) As active_word_count, 
    Count(w._id) As word_count, 
    Count(Distinct l._id) As lecture_count, 
    b._id, 
    b.name, 
    b.version 
From 
    Book b 
    Left Outer Join 
    Lecture l 
    On l.book_id = b._id 
    Left Outer Join 
    Word w 
    On w.lecture_id = l._id 
Group By 
    b._id, 
    b.name, 
    b.version