获取未通过45％考试的学生守则

Question

PostgreSQL
因此，我想返回所有学生唯一的代码，这些代码通过了超过45％的测试而未通过。

Answer 1

您可以使用此：

假定为数据- Grade (75 or above is passing) => grade < 75 is failed

SELECT code
FROM exams
GROUP BY code
HAVING SUM(CASE WHEN grade < 75 THEN 1 ELSE 0 END) :: FLOAT / COUNT(grade) > 0.45;

这是Sqlfiddle

Answer 2

我只是这样写：

SELECT e.code
FROM exams e
GROUP BY e.code
HAVING AVG( (e.grade < 75)::numeric ) > 0.45;