Question

我有这个代码及其临时表，因此您可以运行它。

create table #student(
    id int identity(1,1),
    name varchar(50)
)

create table #quiz(
    id int identity(1,1),
    name varchar(50),
    points_worth int
)

create table #exam(
    id int identity(1,1),
    sequence int,
    question varchar(50),
    answer varchar(50),
    quiz_id int
)

create table #student_taken(
    id int identity(1,1),
    sequence int,
    answer varchar(50),
    student_id int,
    quiz_id int
)

insert into #student(name)
values('Uzumaki Naruto'),('Uchiha Sasuke'),('Haruno Sakura')

insert into #quiz(name,points_worth)
values('Chunin Exam',2)

insert into #exam(sequence,question,answer,quiz_id)
values(1,'Hinata and Neji are siblings','True',1),
(2,'Uchiha Sasuke is part of the Akatsuki','False',1),
(3,'Tsunade and Jiraiya has a son','False',1)

insert into #student_taken(sequence,answer,quiz_id,student_id)
values(1,'True',1,1),(2,'True',1,1),(3,'True',1,1),(1,'True',1,2),(2,'False',1,2),(3,'False',1,2),
(1,'True',1,3),(2,'False',1,3),(3,'False',1,3)

drop table #student
drop table #exam
drop table #quiz
drop table #student_taken

所以你可以看到我Uzumaki Naruto只有1分，因为他只有1个正确的答案，Sakura和Sasuke都有3分。

现在我希望它像这样：

  id     name              score
  1      Uzumaki Naruto     2
  2      Uchiha Sasuke      6
  3      Haruno Sakura      6

他得到了6，因为在我的#quiz表中我添加了值（这是每个项目的价值）。

所以现在我想知道是否需要group by子句呢？什么是正确的求和，我希望如果True = True则加1点，False = False，False = True不计算。

这是我的尝试

select 
  ST.student_id,
  SUM(1 * Q.points_worth) 'sum'
from #student_taken ST
inner join #exam E
on E.quiz_id = ST.quiz_id
inner join #quiz Q
on Q.id = E.quiz_id
group by ST.student_id

Answer 1

我不确定你的问题在这里。 @JorgeCampos不太正确，因为只有在同一数据集中返回加重和非聚合字段时才需要GROUP BY（不使用OVER子句）。

至于获得结果集，我不太确定你是如何得出结论的。 points_worth的值在你的测验表中，而不是考试表中，所以我假设每个问题的测验值都相同？如果是这样，这是您的查询的一个想法：

SELECT q.[name] AS QuizName,
       s.[name] As StudentName,
       COUNT(CASE WHEN st.answer = e.answer THEN 1 END) * q.points_worth AS Score,
       COUNT(CASE WHEN st.answer = e.answer THEN 1 END) AS Correct,
       COUNT(CASE WHEN st.answer != e.answer THEN 1 END) AS Incorrect
FROM #student s
     JOIN #student_taken st ON s.id = st.student_id
     JOIN #exam e ON st.[sequence] = e.id
     JOIN #quiz q ON e.quiz_id = q.id
GROUP BY q.[name], s.[name],
         q.points_worth;

然而，我们可以进一步看看学生是否真的回答了所有问题（并排除那些没有回答的问题），因此我们得到：

INSERT INTO #quiz([name],points_worth)
VALUES('Basic Maths',1);

INSERT INTO #exam([sequence],question,answer,quiz_id)
VALUES(1,'5 + 2 * 3 = 21','False',2),
      (2,'9 - 1 * 2 = 7','True',2);

INSERT INTO #student ([name])
VALUES ('Jane Smith'),('Sally Bloggs');

INSERT INTO #student_taken ([sequence],answer,quiz_id,student_id)
VALUES (1, 'false', 1, 4),
       (1, 'false', 2, 4),
       (2, 'true', 2, 4),
       (1, 'true', 2, 5);
GO

SELECT q.[name] AS QuizName,
       s.[name] As StudentName,
       COUNT(CASE WHEN st.answer = e.answer THEN 1 END) * q.points_worth AS Score,
       COUNT(CASE WHEN st.answer = e.answer THEN 1 END) AS Correct,
       COUNT(CASE WHEN st.answer != e.answer THEN 1 END) AS Incorrect,
       COUNT(CASE WHEN st.answer IS NULL THEN 1 END) AS Unanswered
FROM #quiz q
     JOIN #exam e ON q.id = e.quiz_id
     CROSS JOIN #student s
     LEFT JOIN #student_taken st ON s.id = st.student_id
                                AND e.[sequence] = st.[sequence]
                                AND q.id = st.quiz_id
WHERE EXISTS (SELECT 1 FROM #student_taken sq WHERE sq.student_id = s.id AND sq.quiz_id = q.id)
GROUP BY q.[name], s.[name],
         q.points_worth;

希望有所帮助。

Answer 2

在sum内添加条件。我还注意到从#student_taken到#exam的加入并未正常工作，因为您＆＃39 ;只需加入quiz_id，您还需要加入sequence。

所以，您尝试进行这些修改：

select 
    ST.student_id,
    SUM(IIF(ST.answer = E.answer, Q.points_worth, 0)) 'sum'
from #student_taken ST
    inner join #exam E
    on E.quiz_id = ST.quiz_id and ST.sequence = E.sequence
    inner join #quiz Q
    on Q.id = st.quiz_id
group by ST.student_id

IIF函数将第一个参数计算为条件，如果为true则返回第二个参数，如果为false则返回第三个参数。因此，如果学生回答了考试期望的内容（ST.answer = E.answer），则会返回值，否则为0。

如果您不想使用IIF，则可以使用案例陈述：case when ST.answer = E.answer then Q.points_worth else 0 end。我个人认为IIF看起来更干净，如果你弄乱一个案例陈述，SQL Server Management Studio会因为语法提示而有些疯狂。

Answer 3

你可以试试这个方法：找出多少正确的分数/学生（在CTE内部查询），然后取结果并加入#quiz表，通过应用points_worth来计算最终分数

;with cte as (
select
    st.student_id
    ,st.quiz_id
    ,COUNT(e.id) as points
from #student_taken st
left join #exam e
    on st.quiz_id = e.quiz_id
    and st.answer = e.answer
    and st.sequence = e.sequence
group by st.student_id, st.quiz_id
) select
    student_id
    ,s.name
    --,quiz_id
    ,points * q.points_worth
from cte
inner join #quiz q
    on quiz_id = q.id
inner join #student s
    on student_id = s.id

如果答案与group by子句的正确答案相同，则SQL Server会添加分数

3 个答案: