[英]Nested SQL statement with AVG()
我有三个要提取数据的表。 一个是班级表,其中存储了所提供的班级列表,包括ID,时间戳,培训师ID,开始时间,上午/下午以及举办课程的日期。 培训师ID是一个外键,将其绑定到培训师表,在该表中,我根据培训师的唯一ID拉出培训师名称。
效果很好,但我还需要显示每个班级的平均出勤率。 这涉及一个统计信息表,该表具有一个类ID外键,该类将该类ID以及该会话的参与者数放入一行。 我想返回每个班级每堂课的平均参加人数。
以下是我的选择语句:
SELECT
class_id AS id,
class_name,
class_trainerid,
class_starttime AS start,
class_ampm AS ampm,
class_days AS days,
trainer_id AS trainid,
trainer_name,
stat_class AS sclass,
AVG(stat_students) as stat_students_avg
FROM
$class_table
LEFT JOIN $trainer_table ON (class_trainerid = trainer_id)
LEFT JOIN stats ON (id = stat_students_avg)
GROUP BY
id
上面的代码可能表明,我实际上对如何执行此操作并不了解。 我已经看到了有关通过联接求平均或在select语句中使用select语句进行平均的帖子,但是我似乎无法将其翻译成我的问题。
编辑:这是类表架构:
`class_id` tinyint(8) NOT NULL AUTO_INCREMENT,
`class_datereated` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP,
`class_name` varchar(256) NOT NULL,
`class_trainerid` tinyint(8) NOT NULL,
`class_starttime` time NOT NULL,
`class_ampm` text NOT NULL,
`class_days` text NOT NULL,
PRIMARY KEY (`class_id`)
这是教练模式
`trainer_id` tinyint(8) NOT NULL AUTO_INCREMENT,
'trainer_datecreated` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP,
`trainer_name` varchar(256) NOT NULL,
`trainer_password` varchar(25) NOT NULL,
`trainer_email` varchar(256) NOT NULL,
`trainer_active` tinyint(1) NOT NULL DEFAULT '1',
PRIMARY KEY (`trainer_id`),
UNIQUE KEY `email` (`trainer_email`)
这是stats表架构:
`stat_id` int(8) NOT NULL AUTO_INCREMENT,
`stat_datecreated` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP,
`stat_class` int(8) NOT NULL,
`stat_students` int(8) NOT NULL,
`stat_trainer` tinyint(8) NOT NULL,
`stat_class_date` date NOT NULL,
PRIMARY KEY (`stat_id`)
输出将是php while语句的一部分,结果是:
echo "<table class='list'>";
echo "<tr>";
echo "<th>Class Name</th><th>Trainer</th><th>Ave</th><th>Edit</th><th>Delete</th>";
echo "</tr>";
while($row = mysql_fetch_assoc($class_result))
{
echo "<tr id='class_".$row["id"]."'>";
echo "<td>".$row["class_name"]."</td>";
echo "<td class='trainer-name'>".$row["trainer_name"]."</td>";
echo "<td class='trainer-name'>".$row["stat_students_avg"]."</td>";
echo "<td class='icon'><a href='javascript:void(0);' id='class_edit_".$row["id"]."'><span class='glyphicon glyphicon-edit'></span></a></td>";
echo "<td class='icon'><a href='javascript:void(0);' id='class_delete_".$row["id"]."'><span class='glyphicon glyphicon-remove'></span></a></td>";
echo "</tr>";
}
echo "</table>";
您的ON (id = stat_students_avg)
不对应相同的值。
可以将统计信息表联接到类表的唯一方法是类表的class_trainerid和统计表的trainer_id
你可以尝试这样的事情
SELECT
class_id ,
class_name,
class_trainerid,
class_starttime as start,
class_ampm as ampm,
class_days as days,
trainer_id as trainid,
trainer_name,
stat_class as sclass,
AVG(stat_students) as stat_students_avg
FROM
$class_table
LEFT JOIN $trainer_table ON (class_trainerid = trainer_id)
LEFT JOIN stats ON (trainer_id = class_trainerid)
GROUP BY
class_id
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