Question

假设我有一张表格如下：

Class   |   Subject | Student   | Marks
----------------------------------------
1       |   Maths   |   A       |   70   
1       |   Eng     |   B       |   80
1       |   IT      |   A       |   90 
1       |   IT      |   C       |   80 
2       |   Maths   |   D       |   60   
2       |   Eng     |   E       |   75
2       |   Maths   |   E       |   90 
2       |   IT      |   F       |   80 
3       |   Maths   |   A       |   160   
3       |   Eng     |   B       |   165
3       |   IT      |   G       |   90

我希望输出为

Class   |   Student     | Marks
----------------------------------------
1       |       A       |   160   
2       |       E       |   165
3       |       B       |   165

即。结果包含具有最大标记聚合的类别，学生姓名。如何为此编写SQL查询？例如对于1级，学生A有70 + 90 = 160，在B和C都达到最大值80。

Answer 1

一种解决方案是计算学生每班的最高分数，并将其用作过滤联接：

select  ClassStudentSum.*
from    (
        select  class
        ,       student
        ,       sum(Marks) as SumMarks
        from    YourTable
        group by
                class
        ,       student
        ) as ClassStudentSum
join    (
        select  class
        ,       max(SumMarks) as MaxSumMarks
        from    (
                select  class
                ,       student
                ,       sum(Marks) as SumMarks
                from    YourTable
                group by
                        class
                ,       student
                ) ClassStudentSum2
        group by
                class
        ) MaxPerClass
on      MaxPerClass.class = ClassStudentSum.class
        and MaxPerClass.MaxSumMarks = ClassStudentSum.SumMarks

Live example at SQL Fiddle.

Answer 2

更传统（也更慢）的方法......

SELECT x.*
  FROM
     ( SELECT class
            , student
            , SUM(marks) ttl_marks
         FROM yourtable 
        GROUP
           BY class
            , student
     ) x
  LEFT 
  JOIN
     ( SELECT class
            , student
            , SUM(marks) ttl_marks
         FROM yourtable 
        GROUP
           BY class
            , student
     ) y
    ON y.class = x.class
   AND y.ttl_marks > x.ttl_marks
 WHERE y.class IS NULL;

Answer 3

尝试此查询

查询1 ：

select a.*, if(@prv=class, 1, 0) as flag, @prv:=class from 
(select class,student, sum(marks) as total from table1 
group by class, student
order by class, total desc)a join (select @prv:=0)tmp
where if(@prv=class, 1, 0) = 0

的 SQL FIDDLE ：

| CLASS | STUDENT | TOTAL | FLAG | @PRV:=CLASS | ------------------------------------------------ | 1 | A | 160 | 0 | 1 | | 2 | E | 165 | 0 | 2 | | 3 | B | 165 | 0 | 3 |

希望这有帮助

Answer 4

正确的查询是：

select class, student, sums.mark
from (select class, student, sum(marks) as mark
      from student
      group by class, student
      order by mark desc) sums
group by class

Answer 5

使用此Statement.it正常工作

   select Class,Student,Marks 
from(
    select Class,Student,Marks,Dense_rank() over( partition by class order by marks desc) Rank
      from(Select Class, Student,Max(Marks) Marks 
        from(select Class, Student,Sum(Marks) Marks from Temp14
          group by Class,Student
          order by 1)
        group by Class, Student
        order by 1)
    )
where Rank=1

试试这个。

Answer 6

我想这是一个更简单的查询。它工作正常。不需要加入。

SELECT class, ( SELECT student FROM yourtable WHERE class = YT.class GROUP BY student ORDER BY SUM(marks) DESC LIMIT 1 ) AS student, ( SELECT SUM(marks) FROM yourtable WHERE class = YT.class GROUP BY student ORDER BY SUM(marks) DESC LIMIT 1 ) AS marks FROM yourtable AS YT GROUP BY class

Answer 7

    SQL> with cte as
      2  (select class, student, sum(marks) marks
      3  from engineer
      4  group by class, student
      5  order by class)
      6  select class, student, marks
      7  from (select class, student, marks, dense_rank() over(partition by class order by marks desc) rank
      8  from cte)
      9  where rank=1;

         CLASS S      MARKS
    ---------- - ----------
             1 A        160
             2 E        165
             3 B        165

    SQL>

Here the most important inner query is for cte table. Resulr set created for this one will be as below.

SQL> select class, student, sum(marks)
  2  from engineer
  3  group by class, student
  4  order by class;

     CLASS S SUM(MARKS)
---------- - ----------
         1 A        160
         1 B         80
         1 C         80
         2 D         60
         2 E        165
         2 F         80
         3 A        160
         3 B        165
         3 G         90

9 rows selected.

类别明智，学生姓名具有最大标记总数

7 个答案:

的 SQL FIDDLE ：