小编Alo*_*tal的帖子

我想得到每个学生的平均资格.

例如,有30名学生拥有:

Pupil 1: Qualification 1: 5,6 Qualification 2: 3,2 Qualification 3: 9,1
Pupil 2: Qualification 1: 5,1 Qualification 2: 8,6 Qualification 3: 3,1
Pupil 3: Qualification 1: 1,9 Qualification 2: 7,2 Qualification 3: 5,1
Pupil 4: Qualification 1: 4,6 Qualification 2: 5,2 Qualification 3: 9,5

等等...

我想得到:

我试过:

select distinct pupils.name, pupils.surname, qualifications.id_trimester, round(avg(qualifications.qualification),2), count(qualifications.qualification) from pupils, qualifications where pupils.level='1' and pupils.class='A'  and qualifications.id_trimester=1 and qualifications.type_qualification='class' group by pupils.surname.

但它显示了所有行的相同平均资格和计数.它显示了所有行中重复的第一个学生的平均资格...

表格是:

pupils:

id_pupil
name
surname
email
user
pass
level …