关系代数

Question

这是我根据我的关系构建的示例关系 table。

Assignment   aID description
             1    A1
             2    A2
             3    A3  
#where aID is the unique key

Group        gID aID
             1    1
             2    1
             3    2 
             4    2  
             5    3
#where gID is the unique key , represent groups works on assignments.

Grader       gID  Username
              1    TA1
              2    TA2
              3    TA1
              4    TA2
              5    TA1
#where gID is the unique key and represent which TA is assign to grade a  group.

Result     gID, mark
            1    50
            2    60
            3    70
            4    80 
            5    90
#where gID is the unique key and present group's mark.

找到评分每项作业的评分者，如果评分者至少给一组作业打分，我们就说评分者评分了一项作业。在我的示例中，应该报告 TA1。对这个问题有什么建议的方法吗？提前致谢。

编辑：gID 是唯一的并在所有关系中使用，意味着约束

Group[aid] /subset assignment[aid]  
grader[gid] /subset group[gid]  
result[gid] /subset group[gid]

Answer 1

不是很容易理解，我们这里没有3NF。据我了解，只有一个 Grader 可以标记一个组，而一个组只能有一个标记。在这种情况下，您可以使用此查询：

SELECT DISTINCT Username FROM Grader WHERE Username NOT IN (
    SELECT grd.Username FROM Grader grd
    LEFT JOIN Assignment a ON (1=1)
    LEFT JOIN (
        SELECT grp.aId, grd.Username FROM [Group] grp
        LEFT JOIN Grader grd ON (grp.gID = grd.gID)
        LEFT JOIN Assignment a ON (grp.aID = a.aID)
        LEFT JOIN Result r ON (grp.aID = r.gID)) innerQuery
    ON (grd.Username = innerQuery.Username and a.aId = innerQuery.aId)
    WHERE innerQuery.aId is null);

结果：

Username
TA1

Answer 2

请注意，当您清楚地命名事物时，事情会变得多么简单。评分者中的 gID 很容易成为评分者的 ID...它也以 g.

开头

这是一种实现我所理解的解决方案的方法。为了清楚起见，我重命名了一些东西。

declare @Assignment  table (AssignmentID int, Description varchar(5))
insert into @Assignment
values
    (1,'A1'),
    (2,'A2'),
    (3,'A3')

declare @Group table (GroupID int, AssignmentID int)
insert into @Group
values
    (1,1),
    (2,1),
    (3,2),
    (4,2),
    (5,3)

declare @GroupGrader table (GroupID int, GraderName varchar(5))
insert into @GroupGrader
values
    (1,'TA1'),
    (2,'TA2'),
    (3,'TA1'),
    (4,'TA2'),
    (5,'TA1')

Declare @GroupResult table (GroupID int, Mark int)
insert into @GroupResult
values
    (1,50),
    (2,60),
    (3,70),
    (4,80),
    (5,90)

select gg.GraderName
from @Group g
    --inner join @Assignment a
    --  on a.AssignmentID = g.AssignmentID
    inner join @GroupGrader gg
        on g.GroupID = gg.GroupID
    --inner join @GroupResult gr
    --  on gr.GroupID = g.GroupID
group by gg.GraderName
having count(g.AssignmentID) = (select count(*) from @Assignment)

关系代数

relational algebra

relational-algebra

relational-database