Question

学生

+----+-----------+--------------+
| id | First Name| Last Name    |
+----+-----------+--------------+
| 1  | John      | A            |
+----+-----------+--------------+
| 2  | Jane      | B            |
+----+-----------+--------------+

认证

+----+------------------+
| id | Name             |
+----+------------------+
| 1  | Certification 1  |
+----+------------------+
| 2  | Certification 2  |
+----+------------------+
| 3  | Certification 3  |
+----+------------------+

StudentCertifications

+----+------------+-----------------+
| id | StudentID  | CertificationID |
+----+------------+-----------------+
| 1  | 1          | 1               |
+----+------------+-----------------+
| 2  | 1          | 2               |
+----+------------+-----------------+
| 3  | 1          | 3               |
+----+------------+-----------------+

我想通过SQL查询找到的内容：

StudentMissingCertifications

+----+------------+-----------------+
| id | StudentID  | CertificationID |
+----+------------+-----------------+
| 1  | 2          | 1               |
+----+------------+-----------------+
| 2  | 2          | 2               |
+----+------------+-----------------+
| 3  | 2          | 3               |
+----+------------+-----------------+

Answer 1

使用NOT EXISTS

SELECT * FROM Studients S CROSS JOIN Certifications C
WHERE NOT EXISTS
 (
  SELECT 1 FROM StudentCertifications SC WHERE SC.StudentId=S.Id AND SC.CertificationID=C.Id
 )

或使用EXCEPT

SELECT S.ID,C.ID FROM Studients S CROSS JOIN Certifications C
EXCEPT 
SELECT StudentId, CertificationID FROM StudentCertifications

Answer 2

通过CROSS JOIN将所有学生与所有认证相关联，然后使用NOT EXISTS过滤已有的认证。

SELECT
    StudentID = S.ID,
    MissingCertificationID = C.ID
FROM
    Student AS S
    CROSS JOIN Certifications AS C
WHERE
    NOT EXISTS (
        SELECT
            'student does not have the certification'
        FROM
            StudentCertifications X
        WHERE
            X.StudentID = S.ID AND
            X.CertificationID = C.ID)

Answer 3

交叉加入以获得所有可能的学生/证书，然后将其加入学生证书表，其中结果为空。

$ docker run --name artifactory -d -p 8081:8081 docker.bintray.io/jfrog/artifactory-oss:latest

Answer 4

解决您的问题：

select Student.id,[First Name],[Last Name],Certifications.id,name
from Student cross join Certifications
left join StudentCertifications
on 
Student.id=StudentCertifications.Studentid
WHERE StudentCertifications.Studentid IS NULL

<强>输出：

id  First_Name  Last_Name   id     name
2   Jane         B          1    Certification 1
2   Jane         B          2    Certification 2
2   Jane         B          3    Certification 3

请点击演示链接：

http://sqlfiddle.com/#!18/866ce/16

Answer 5

执行join s（即LEFT JOIN）查找尚未获得任何证书的学生

select 
       s.id as StudentID, c.id as CertificationID  
from Student s
left join StudentCertifications sc on sc.StudentID = s.id
cross join (select id from Certifications) c
where sc.StudentID is null

Answer 6

如果我理解了这个问题，那么这个查询就是你要找的

SELECT First Name, Last Name FROM Student
WHERE Student.id NOT IN (SELECT StudentID From StudentCertifications)

没有认证的学生

6 个答案: