Question

表MyTable的定义如下：

Student_id(int), course_id(int), pass(bool)

如何通过student_ids获得pass_rate > 95%？还有course_ids和pass_rate > 95%吗？

谢谢

Answer 1

按组划分并具有101类

SELECT student_id, SUM(CAST(pass AS INT)) * 100 / COUNT(course_id) as pass_rate
FROM MyTable
GROUP BY student_id
HAVING(sum(cast(pass as int)) * 100 / count(course_id)) > 95

Answer 2

在两种情况下，您都需要有条件的avg：

select sudent_id, avg(case when pass = 1 then 1 else 0 end) average
from mytable
group by sudent_id
having avg(case when pass = 1 then 1 else 0 end) > 0.95;

select course_id, avg(case when pass = 1 then 1 else 0 end) average
from mytable
group by course_id
having avg(case when pass = 1 then 1 else 0 end) > 0.95;

如果您的dbms支持，则可以用pass = 1更改条件pass = true。

Answer 3

实际上，要这样做，假设您有一栏显示学生的分数，您可以这样做；

SELECT * FROM records WHERE scores > 95

SQL查询百分比

3 个答案: