你可以这样做：
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\ U present是百分比present。请注意，john和matt的出席率分别为50%和66.6%。
解释
为了获得总记录，我们会这样做：

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

为了得到每个学生出席的总时间，我们会：

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

%现在是学生在场的次数除以总数。所以，这就是我在子查询中所做的。
一旦学生的数据可用，将结果与学生的信息连接起来，并从两个表中提取所需的信息。