学生在一堂课中从每个科目中获得的分数生成的位置

Question

我有一个学生考试成绩表，该表现在存储学生的成绩，我想根据学生的最高成绩来计算职位。

select er.student_id, first_name, class_id, subject_id, subsubject_id,
       sum(total_marks) totalmarks, sum(marks_obtained) obtainedmarks, (0) Position
from Exam_Results er inner join
     student s
     on er.student_id = s.student_id
where class_id = 272     
group by er.student_id, session_id, shift_id, class_id, subclass_id, term_id, catagory_id, exam_type_id, subject_id, subsubject_id, first_name
order by student_id desc

Answer 1

您可以尝试以下查询。

Select er.student_id, first_name, class_id, subject_id, subsubject_id,
t1.totalmarks, t1.obtainedmarks, 
DENSE_RANK() OVER (ORDER BY totalmarks) as position 
from Exam_Results er 
inner join student s on er.student_id = s.student_id
inner join(select student_id, sum(total_marks) totalmarks, 
sum(marks_obtained) obtainedmarks from Exam_Results group by student_id) t1
on t1.student_id=s.student_id
where class_id = 272

Answer 2

选择er.student_id，first_name，class_id，subject_id，subsubject_id，        sum（total_marks）个总标记，sum（marks_obtained）获得的标记，DENSE_RANK（）OVER（ORDER BY sum（marks_obtained）desc）作为位置 来自Exam_Results的内部联接      学生们      在er.student_id = s.student_id上 其中class_id = 272
按er.student_id，session_id，shift_id，class_id，subclass_id，term_id，catagory_id，exam_type_id，subject_id，subsubject_id，first_name分组 按student_id desc排序

请尝试此查询。我认为这应该可以解决您的问题，因为我本人已经尝试过以达到预期效果。

Answer 3

如果只希望基于聚合进行排名，则不需要子查询或CTE。只要做：

select er.student_id, first_name, class_id, subject_id, subsubject_id,
       sum(total_marks) as totalmarks, sum(marks_obtained) as obtainedmarks,
       dense_rank() over (order by sum(total_marks) desc) as Position
from Exam_Results er inner join
     student s
     on er.student_id = s.student_id
where class_id = 272     
group by er.student_id, session_id, shift_id, class_id, subclass_id, term_id, catagory_id, exam_type_id, subject_id, subsubject_id, first_name
order by student_id desc;