mysql：優化此sql查詢的最佳方法

Question

我想以最好的方式從此查詢中獲取結果

這是我的桌子指示

學校

CREATE TABLE `schools` (
 `id` int(11) NOT NULL AUTO_INCREMENT,
 `user` mediumint(5) NOT NULL,
 `gender` tinyint(1) NOT NULL,
 `time` int(11) NOT NULL,
 `status` tinyint(2) NOT NULL,
 `number` mediumint(6) NOT NULL,
 `name` varchar(75) NOT NULL,
 `address` varchar(75) NOT NULL,
 `admin` varchar(50) NOT NULL,
 `admin_phone` varchar(20) NOT NULL,
 `admin_email` varchar(30) NOT NULL,
 `school_phone` varchar(20) NOT NULL,
 `learn` tinyint(2) NOT NULL,
 `mr7la` tinyint(2) NOT NULL,
 `sfof` smallint(3) NOT NULL,
 `fswl` smallint(3) NOT NULL,
 `json` text NOT NULL,
 PRIMARY KEY (`id`),
 UNIQUE KEY `user` (`user`),
 KEY `status` (`status`),
 KEY `learn` (`learn`),
 KEY `mr7la` (`mr7la`),
 KEY `number` (`number`)
) ENGINE=MyISAM AUTO_INCREMENT=20 DEFAULT CHARSET=utf8

3agz

CREATE TABLE `3agz` (
 `id` bigint(20) NOT NULL AUTO_INCREMENT,
 `school` int(11) NOT NULL,
 `tkss` int(11) NOT NULL,
 `teacher_7ess` int(11) NOT NULL,
 `teacher_master_7ess` int(11) NOT NULL,
 `time_added` int(11) NOT NULL,
 `reported` int(11) NOT NULL DEFAULT '0',
 `fixed` int(11) NOT NULL,
 `info` text NOT NULL,
 PRIMARY KEY (`id`),
 UNIQUE KEY `school` (`school`,`tkss`,`teacher_7ess`,`fixed`),
 KEY `school_2` (`school`),
 KEY `tkss` (`tkss`),
 KEY `reported` (`reported`),
 KEY `time_added` (`time_added`),
 KEY `school_3` (`school`,`time_added`),
 KEY `school_4` (`school`,`fixed`)
) ENGINE=InnoDB AUTO_INCREMENT=85 DEFAULT CHARSET=utf8

這是為此的SQL小提琴

http://sqlfiddle.com/#!2/3313e0/4

---

你可以看到她是我使用的SQL查詢

SELECT
    schools.* , ( select count(id) from 3agz where 3agz.school = schools.id and fixed = 0 ) as has_3agz
           FROM
                schools
           WHERE
                ( select count(id) from 3agz where 3agz.school = schools.id and fixed = 0 ) > 0

limit 10

的解釋是

學校=>小學：全部

3agz => DEPENDENT子查詢：ref

3agz => DEPENDENT子查詢：ref

這是我問我可以這樣做嗎，什么是最好的方法

1-我可以忽略第二個子查詢的位置並取決於選擇中的第一個子查詢嗎

2-如果數字1的答案是您不能執行此查詢后，我不能忽略第一個子查詢[has_3agz別名]，我循環拋出結果[schools ids]

然后像這樣進行第二次查詢

例如，第一個查詢返回的學校ID為1、2、3、4

select school , count(id) from 3agz where school in ( 1 , 2 , 3 , 4 ) and fixed = 0

將每個計數附加到數組中的學校

希望你能理解我

最好的祝福

Answer 1

SELECT schools.*, count(*) as has_3agz from schools
LEFT JOIN 3agz on 3agz.school = schools.id and fixed = 0
GROUP BY schools.id;