我想从这个查询中得到最好的方法
here is my tables 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
3 agz
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
here's SQL Fiddle for this
http://sqlfiddle.com/!2/3313e0/4
你可以看到她是我的SQL查询,我使用
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 => PRIMARY: ALL
3agz => DEPENDENT SUBQUERY: ref
3agz => DEPENDENT SUBQUERY: ref
这是我的问题,我可以这样做吗,什么是最好的方法
1 -我可以在select
中忽略第二个子查询并使其依赖于第一个子查询吗?2-如果第一个答案是你不能我可以忽略第一个子查询吗[has_3agz alias]在这个查询执行后,我循环显示结果[schools ids]
并像这样执行第二个查询
以为例,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
将每个计数附加到数组
中的学校希望你能理解我
SELECT schools.*, count(*) as has_3agz from schools
LEFT JOIN 3agz on 3agz.school = schools.id and fixed = 0
GROUP BY schools.id;