Question

只有当他没有成功完成X，或者他已经成功完成了构建X课程的所有课程时，Mike才能参加CS课程。例如，迈克不能拿IE2900，因为他还没有成功完成IE2700。（在查询中，只应考虑直接构建课程的课程。例如，IE3200仅直接构建在IE2201上，而不是IE1100上。）

简而言之，找到迈克可以在下学期学习CS课程代码的查询。

结果应如下所示：

Code
----
IE2700
IE3200

以下是包含示例数据的数据库表：

学生

Sid         Name      
----------  ----------
2           Kathlyn     
6           Mike       
7           Nick       
9           Anny

课程

Code        Title        Dept        Credits   
----------  -----------  ----------  ----------
IE1100      programming  CS          10        
IE2201      operating s  CS          20        
IE2700      database     CS          10        
IE2900      software en  CS          10        
IE3200      distributed  CS          10        
MAT1001     calculus     Math        30        
DOK1012     database     Hum         10

CoursesBuiltOn

Code        BuiltOn   
----------  ----------
IE2201      IE1100   
IE2700      IE1100   
IE2900      IE2201   
IE2900      IE2700   
IE3200      IE2201

等级

Sid         Code        Year        Grade     
----------  ----------  ----------  ----------
2           MAT1001     2010        A         
2           IE1100      2011        A         
2           IE2201      2012        A         
2           IE2700      2013        A         
2           IE2900      2014        B         
2           IE3200      2014        B         
6           MAT1001     2011        D         
6           IE1100      2012        C         
6           IE2201      2013        B         
6           IE2700      2013        F         
7           MAT1001     2013        B         
7           DOK1012     2013        A         
9           MAT1001     2013        F

数据库结构：

CREATE TABLE Students (
Sid TEXT PRIMARY KEY,
Name TEXT
);
CREATE TABLE Courses (
Code TEXT PRIMARY KEY,
Title TEXT,
Dept TEXT,
Credits INT
);
CREATE TABLE Grades (
Sid INT,
Code TEXT,
Year INT,
Grade TEXT,
PRIMARY KEY (Sid, Code),
FOREIGN KEY (Sid) REFERENCES Students (Sid),
FOREIGN KEY (Code) REFERENCES Courses (Code)
);
CREATE TABLE CoursesBuiltOn (
Code TEXT,
BuiltOn TEXT,
PRIMARY KEY (Code, BuiltOn),
FOREIGN KEY (Code) REFERENCES Courses (Code),
FOREIGN KEY (Code) REFERENCES Courses (BuiltOn)
);

我尝试将所有表格连接在一起以获取外键并获取信息，但我找不到任何方法来获得正确的答案。

编辑：

这是我到目前为止所得到的：

with NotAvailable as (select COB.Code
from Courses C
left join CoursesBuiltOn COB on COB.BuiltOn = C.Code
left join Grades Gr on Gr.Code = C.Code
inner join Students S on S.Sid = Gr.Sid
where Gr.Sid = S.Sid and Gr.Grade = 'F'
), AlreadyTakenAndPassed as (select Gr.Code 
from Grades Gr where Gr.Sid = S.Sid and Gr.Grade <> 'F'
), TakenButNotYetPassed as (select Gr.Code 
from Grades Gr where Gr.Sid = S.Sid and Gr.Grade = 'F'
)
select COB.Code
from Courses C
left join CoursesBuiltOn COB on COB.BuiltOn = C.Code
left join Grades G on G.Code = C.Code
inner join Students S on G.Sid = S.Sid
where C.Dept = 'CS' and S.Name = 'Mike'
and (COB.Code in TakenButNotYetPassed or COB.Code not in NotAvailable or COB.Code not in AlreadyTakenAndPassed)
;

Code      
----------
IE2700   
IE2900   
IE3200   
IE2900   


select COB.Code
from Courses C
left join CoursesBuiltOn COB on COB.BuiltOn = C.Code
left join Grades G on G.Code = C.Code
inner join Students S on S.Sid = G.Sid
where S.Name = 'Mike' and G.Grade = 'F';

Code      
----------
IE2900

正如您在上面的查询中所看到的，我想创建子查询，从整个课程集中排除课程。但是，NotAvailable子查询不会排除IE2900。这是为什么？我以为我是在正确的方向......这可能不是最有效的方式触摸...

Answer 1

我自己做了，将整个查询切入案例：

with 
TakenButNotYetPassed as (select Gr.Code 
from Grades Gr 
inner join Students St on St.Sid = Gr.Sid
where St.Name = 'Mike' and Gr.Grade = 'F'
), 
NotAvailable as (select Code
from CoursesBuiltOn
where BuiltOn in TakenButNotYetPassed
), 
Passed as (select Gr.Code
from Grades Gr
inner join Students St on St.Sid = Gr.Sid
where St.Name = 'Mike' and Gr.Grade <> 'F'
)

select *
from Courses C
where C.Dept = 'CS' 
and C.Code not in Passed
and C.Code not in NotAvailable
;

SQL：Mike下学期可以学习的课程名称

编辑：

1 个答案: