在mysql中计算出勤率时出现语法错误

Question

SELECT name, DISTINCT studentid, count(attendance) 
 from attendance a,students s 
 where attendance = 'p'and s.studentid=a.studentid  
having count(attendance)<3/4*sum(attendance);

我有 2 个表出勤率和学生，我想从中选择学生的姓名（来自学生表）和出勤率（来自出勤表），其中 studentid 是那些出勤率

Answer 1

你可以试试这样的：

数据准备

create table attendance (studentid int, attendance char(1));

insert into attendance values (1,'p'),(1,'a'),(2,'p'),(2,'p'),(2,'a'),(3,'p');

数据

select * from students;
+-----------+------+
| studentid | name |
+-----------+------+
|         1 | John |
|         2 | Matt |
|         3 | Mary |
+-----------+------+

select * from attendance;
+-----------+------------+
| studentid | attendance |
+-----------+------------+
|         1 | p          |
|         1 | a          |
|         2 | p          |
|         2 | p          |
|         2 | a          |
|         3 | p          |
+-----------+------------+

查询

select s.*, a.total, a.p_present
from students s
inner join (
    select studentid, count(*) as total, sum(case attendance when 'p' then 1 else 0 end) * 100/count(*) as p_present
    from attendance
    group by studentid
) a on s.studentid = a.studentid
where a.p_present < 75 ;

结果

+-----------+------+-------+-----------+
| studentid | name | total | p_present |
+-----------+------+-------+-----------+
|         1 | John |     2 |   50.0000 |
|         2 | Matt |     3 |   66.6667 |
+-----------+------+-------+-----------+

p_present 是存在百分比。请注意，John 和 Matt 的出席率分别为 50% 和 66.6%。

说明

为了获得总记录，我们会这样做：

select studentid, count(*)
from attendance
group by studentid;

为了获得每个学生在场的总时间，我们会这样做：

select studentid, sum(case attendance when 'p' then 1 else 0 end)
from attendance
group by studentid;

出席百分比将是学生出席的次数除以总数。所以，这就是我在子查询中所做的。

一旦有关学生的数据可用，将该结果与学生的信息相结合，并从两个表中提取所需的信息。