Question

我正在尝试创建一个学生数据库，但我遇到了给定的要求。

我得到了创建2个表的表，一个表是学生的注册号，科目总和，另一个表是Student_name，总和的排名。我可以理解，这里我们需要单独使用学生姓名，以后再组合使用，但是让我对学生科目感到震惊的部分可能会有所不同，这意味着如果学生编号：1包含3个科目，则意味着学生编号：2可能包含2个，而学生否：3个可能有5个，因此我们必须将排名排序

CREATE TABLE Student
(StudentID int, StudentName varchar(6), Details varchar(1));    


INSERT INTO Student
(StudentID, StudentName, Details)
  VALUES
(1, 'John', 'X'),
(2, 'Paul', 'X'),
(3, 'George', 'X'),
(4, 'Paul', 'X');

CREATE TABLE Subject
(SubjectID varchar(1), SubjectName varchar(7));    

INSERT INTO Subject
  (SubjectID, SubjectName)
VALUES
('M', 'Math'),
('E', 'English'),
('H', 'History');

CREATE TABLE Mark
(StudentID int, SubjectID varchar(1), MarkRate int);    

INSERT INTO Mark
 (StudentID, SubjectID, MarkRate)
VALUES
(1, 'M', 90),
(1, 'E', 100),
(2, 'M', 95),
(2, 'E', 70),
(3, 'E', 95),
(3, 'H', 98),
(4, 'H', 90),
(4, 'E', 100);

我需要2个输出第一个是

ID  |subjects |marks
----------------------------------------------------------
1     maths     98
1     science   87
1     social    88
2     maths     87
2     english   99
3     maths     96
3     evs       100
3     social     88
3     history    90

第二个表为

NO  |name |total|rank
----------------------------------------------------------
1   xxx     123     1
2   yyy     456     2
3   zzz     789      3

我需要这样的输出才能输入n个条目

Answer 1

1 st one
SELECT 
A.StudentID AS [ID],
C.SubjectName AS [subjects],
B.MarkRate AS [marks] 
FROM STUDENT A JOIN Mark B ON A.StudentID=B.StudentID JOIN Subject C ON C.SubjectID=B.SubjectID

2nd one
SELECT 
A.StudentID AS [NO],
A.StudentName AS [name],
SUM(B.MarkRate) AS [total],
ROW_NUMBER() OVER(ORDER BY SUM(B.MarkRate) DESC) AS [rank]
FROM STUDENT A JOIN Mark B ON A.StudentID=B.StudentID JOIN Subject C ON C.SubjectID=B.SubjectID
GROUP BY A.StudentID,A.StudentName 
ORDER BY [total] DESC

Answer 2

我知道第一个查询列出了每个学生的分数

SELECT
    m.StudentID as ID,
    s.SubjectName as subjects,
    m.MarkRate as marks
FROM 
    Mark m
    INNER JOIN Subject s on m.SubjectID = m.SubjectID 
ORDER BY 
    m.StudentID,
    s.SubjectName

第二个查询给出学生的总得分及其排名。

SELECT 
    X.StudentID,
    X.StudentName,
    ROWNUMBER() OVER ( ORDER BY X.TotalMark desc) as Rank
FROM (
    SELECT 
        m.StudentID,
        s.StudentName,
        sum(m.MarkRate) TotalMark
    FROM 
        Mark m
        INNER JOIN Student s on s.StudentID = m.StudentID 
    GROUP BY 
        m.StudentID,
        s.StudentName
) X
ORDER BY X.TotalMark desc

Answer 3

select StudentID ,SubjectName ,MarkRate from
 #Mark a join #Subject b on a.SubjectID=b.SubjectID

输出

StudentID   SubjectName MarkRate
1              Math       90
1              English    100
2              Math       95
2              English    70
3              English    95
3              History    98
4              History    90
4              English    100

第二次查询

  with cte as
    (
    select a.StudentID,StudentName,sum(MarkRate)MarkRate  from #Student a join #Mark B on a.StudentID=b.StudentID
    group by a.StudentID,StudentName
    )
    select *,rank() over( order by MarkRate desc) as rn  from cte

输出

StudentID   StudentName MarkRate    rn
3             George    193         1
4             Paul      190         2
1             John      190         2
2             Paul      165         4

Answer 4

要获得带有主题名称的标记列表，很简单：您需要使用JOIN：

SELECT M.StudentId
        ,SU.SubjectName
        ,M.MarkRate
FROM Mark M
INNER JOIN  Subject SU ON M.SubjectID = SU.SubjectID

要获得排名的总分，您需要使用GROUP BY和RANK()函数：

SELECT M.StudentId
        ,ST.StudentName
        ,SUM(MarkRate) Total
        ,RANK() OVER(ORDER BY SUM(MarkRate) ) Rank
        --,RANK() OVER(ORDER BY SUM(MarkRate) DESC) Rank
FROM Mark M
INNER JOIN  Student ST ON M.StudentId = ST.StudentId
GROUP BY M.StudentId
        ,ST.StudentName

Answer 5

您可以尝试对两个输出使用以下查询。

SELECT
    Student.StudentID as ID,
    [Subject].SubjectName as subjects,
    Mark.MarkRate as marks
FROM 
    Student
    INNER JOIN Mark on Student.StudentID = Mark.StudentID
    INNER JOIN [Subject] on [Subject].SubjectID = Mark.SubjectID 
ORDER BY 
    Student.StudentID,
    SubjectName

在下面查询第二个输出。正如您所说的，那些得分较低的人获得了等级1，否则您可以在下面的查询中将Order By Total Desc放在得分较高的人中。

SELECT ID, StudentName, Total, Row_number() Over (Order BY Total) Ranks FROM(
SELECT Id, StudentName, SUM(marks) as Total FROM (
SELECT
    Student.StudentID as ID,
    [Subject].SubjectName as subjects,
    Mark.MarkRate as marks,
    Student.StudentName
FROM 
    Student
    INNER JOIN Mark on Student.StudentID = Mark.StudentID
    INNER JOIN [Subject] on [Subject].SubjectID = Mark.SubjectID 
)Tot
Group By Id, StudentName
)Ranks

如何将一列的结果合并到另一列

5 个答案: