Question

我想获得从此查询获得结果的最佳方式

这是我的表instructrue

学校

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

这里的SQL小提琴

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

你可以看到她使用我的SQL查询

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

解释是

schools =＆gt;主要：所有

3agz =＆gt;相关提示：参考

这里我问我能做到这一点以及最好的方式

1 - 我可以忽略第二个子查询在哪里，并依赖于select

中的第一个子查询

2-如果1号答案是你不能的话我可以忽略第一个子查询[has_3agz别名] 执行此查询后，我循环trow结果[学校ids]

并像这样进行第二次查询

例如第一个查询返回学校ID 1,2,3,4

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

将每个计数附加到阵列

的学校希望你了解我

祝你好运

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;

mysql：优化这个SQL查询的最佳方法

1 个答案: