用于考试（测试测验）的MySQL数据库查询分数

我有一个数据库，其中存储了一个包含用户信息的表，一个包含测试（答案和分数）的表，以及一个包含每个问题的用户答案的表。每个问题总共得1分，可以有一个或多个正确答案。如果所有答案都是正确的，并且用户只检查了一个，那么他只会得到0.25分。 我想做一个查询来检查每个用户的总分，但我没有找到一个好方法

用户表：

+--------+------------+-----------+-------------------+------------+--------+
| userID | first_name | last_name | email             | password   | points |
+--------+------------+-----------+-------------------+------------+--------+
|      1 | Jhon       | Jhonny    | jhon@yahoo.com    | secretPass |      0 |
|      2 | Dan        | Dan       | dan@yahoo.com     | 1234       |      0 |
|      3 | Dick       | Pop       | dd@yahoo.com      | 123456     |      0 |
|      4 | Mihaela    | Micky     | mihaela@yahoo.com | pass12     |      0 |
+--------+------------+-----------+-------------------+------------+--------+

问题表： （1表示答案是好的-我们可以有多个正确答案）

用户答案表：（1表示用户选择该答案，但可能不正确）

试试这个

SELECT 
  a.*,
  u.name,
  q.*,
  # (a.a & q.a) + (a.b & q.b) + (a.c & q.c) + (a.d & q.d) userCorrects,
  # (a.a + a.b + a.c + a.d) questionCorrects,
  ((a.a & q.a) + (a.b & q.b) + (a.c & q.c) + (a.d & q.d)) / (a.a + a.b + a.c + a.d) as userGrade
FROM
 answer a
    INNER JOIN
 user u ON a.userID = u.id
    INNER JOIN
 question q ON a.questionID = q.id

---

你要把问题和答案联系起来，把正确的答案加起来。每个答案的公式很简单：正确答案的数量除以5。您可以按用户ID和总和分组，以获得每个用户的结果

select u.userid, u.first_name, u.last_name, counted.points
from
(
  select
    a.userid,
    sum(((a.a = q.a) + (a.b = q.b) + (a.c = q.c) + (a.d = q.d) + (a.e = q.e)) / 5)
      as points
  from question q
  join answer a on a.questionid = q.questionid
  group by a.userid
) counted
join users u on u.userid = counted.userid;

在MySQL中

true=1

和

false=0

，因此

（a.a=q.a）

正确时为1，错误时为0。

问得好

我认为此查询将帮助您处理此场景中可能发生的所有情况

SELECT 
u.userID, u.first_name, u.last_name, u.email, 
SUM(1 - ((!(answer.a = question.a)) + (!(answer.b = question.b)) + (!(answer.c = question.c)) + (!(answer.d = question.d))) * 0.25) AS Score 
FROM answer 
INNER JOIN 
user u ON answer.userID = u.userID 
INNER JOIN 
question ON answer.questionID = question.questionID 
GROUP BY answer.userID

---

我猜你的意思是5个答案中的1个正确=0.2分（即1的20%），而不是0.25分？你的问题在哪里还不清楚。你当然知道你应该加入问题和答案。您还知道如何比较值，并且可能知道如何聚合每个用户的数据。那么你到底被困在哪里了？谢谢，我喜欢它！：）谢谢我试试看。

select u.userid, u.first_name, u.last_name, counted.points
from
(
  select
    a.userid,
    sum(((a.a = q.a) + (a.b = q.b) + (a.c = q.c) + (a.d = q.d) + (a.e = q.e)) / 5)
      as points
  from question q
  join answer a on a.questionid = q.questionid
  group by a.userid
) counted
join users u on u.userid = counted.userid;

SELECT 
u.userID, u.first_name, u.last_name, u.email, 
SUM(1 - ((!(answer.a = question.a)) + (!(answer.b = question.b)) + (!(answer.c = question.c)) + (!(answer.d = question.d))) * 0.25) AS Score 
FROM answer 
INNER JOIN 
user u ON answer.userID = u.userID 
INNER JOIN 
question ON answer.questionID = question.questionID 
GROUP BY answer.userID