Question

以下是我所说的简化示例：

Table: students      exam_results
_____________       ____________________________________
| id | name |       | id | student_id | score |   date |
|----+------|       |----+------------+-------+--------|
|  1 | Jim  |       |  1 |          1 |    73 | 8/1/09 | 
|  2 | Joe  |       |  2 |          1 |    67 | 9/2/09 |
|  3 | Jay  |       |  3 |          1 |    93 | 1/3/09 |
|____|______|       |  4 |          2 |    27 | 4/9/09 |
                    |  5 |          2 |    17 | 8/9/09 |
                    |  6 |          3 |   100 | 1/6/09 |
                    |____|____________|_______|________|

为了这个问题，假设每个学生都至少录制了一个考试成绩。

您如何选择每位学生及其最高分？ 编辑：...以及该记录中的其他字段？

预期产出：

_________________________
| name | score |   date |
|------+-------|--------|
|  Jim |    93 | 1/3/09 |
|  Joe |    27 | 4/9/09 |
|  Jay |   100 | 1/6/09 |
|______|_______|________|

欢迎使用所有类型的DBMS的答案。

Answer 1

回答EDITED问题（即获取相关列）。

在Sql Server 2005+中，最好的方法是将ranking/window function与CTE结合使用，如下所示：

with exam_data as
(
    select  r.student_id, r.score, r.date,
            row_number() over(partition by r.student_id order by r.score desc) as rn
    from    exam_results r
)
select  s.name, d.score, d.date, d.student_id
from    students s
join    exam_data d
on      s.id = d.student_id
where   d.rn = 1;

对于符合ANSI-SQL的解决方案，子查询和自联接将起作用，如下所示：

select  s.name, r.student_id, r.score, r.date
from    (
            select  r.student_id, max(r.score) as max_score
            from    exam_results r
            group by r.student_id
        ) d
join    exam_results r
on      r.student_id = d.student_id
and     r.score = d.max_score
join    students s
on      s.id = r.student_id;

最后一个假设没有重复的student_id / max_score组合，如果有和/或你想要计算去重复它们，你需要使用另一个子查询加入一个确定性的东西来决定哪个记录拉。例如，假设您不能为具有相同日期的特定学生提供多条记录，如果您想根据最新的max_score打破平局，您可以执行以下操作：

select  s.name, r3.student_id, r3.score, r3.date, r3.other_column_a, ...
from    (
            select  r2.student_id, r2.score as max_score, max(r2.date) as max_score_max_date
            from    (
                        select  r1.student_id, max(r1.score) as max_score
                        from    exam_results r1
                        group by r1.student_id
                    ) d
            join    exam_results r2
            on      r2.student_id = d.student_id
            and     r2.score = d.max_score
            group by r2.student_id, r2.score
        ) r
join    exam_results r3
on      r3.student_id = r.student_id
and     r3.score = r.max_score
and     r3.date = r.max_score_max_date
join    students s
on      s.id = r3.student_id;

编辑：感谢Mark在评论中的好评，添加了正确的重复数据删除查询

Answer 2

SELECT s.name,
    COALESCE(MAX(er.score), 0) AS high_score
FROM STUDENTS s
    LEFT JOIN EXAM_RESULTS er ON er.student_id = s.id
GROUP BY s.name

Answer 3

试试这个，

Select student.name, max(result.score) As Score from Student
        INNER JOIN
    result
        ON student.ID = result.student_id
GROUP BY
    student.name

Answer 4

使用Oracle的分析功能很容易：

SELECT DISTINCT
       students.name
      ,FIRST_VALUE(exam_results.score)
       OVER (PARTITION BY students.id
             ORDER BY exam_results.score DESC) AS score
      ,FIRST_VALUE(exam_results.date)
       OVER (PARTITION BY students.id
             ORDER BY exam_results.score DESC) AS date
FROM   students, exam_results
WHERE  students.id = exam_results.student_id;

Answer 5

使用MS SQL Server：

SELECT name, score, date FROM exam_results
JOIN students ON student_id = students.id
JOIN (SELECT DISTINCT student_id FROM exam_results) T1
ON exam_results.student_id = T1.student_id
WHERE exam_results.id = (
    SELECT TOP(1) id FROM exam_results T2
    WHERE exam_results.student_id = T2.student_id
    ORDER BY score DESC, date ASC)

如果存在并列分数，则返回最早的日期（将date ASC更改为date DESC以返回最新的日期。）

输出：

Jim 93  2009-01-03 00:00:00.000
Joe 27  2009-04-09 00:00:00.000
Jay 100 2009-01-06 00:00:00.000

试验台：

CREATE TABLE students(id int , name nvarchar(20) );

CREATE TABLE exam_results(id int , student_id int , score int, date datetime);

INSERT INTO students
VALUES
(1,'Jim'),(2,'Joe'),(3,'Jay')

INSERT INTO exam_results VALUES
(1, 1, 73, '8/1/09'), 
(2, 1, 93, '9/2/09'),
(3, 1, 93, '1/3/09'),
(4, 2, 27, '4/9/09'),
(5, 2, 17, '8/9/09'),
(6, 3, 100, '1/6/09')

SELECT name, score, date FROM exam_results
JOIN students ON student_id = students.id
JOIN (SELECT DISTINCT student_id FROM exam_results) T1
ON exam_results.student_id = T1.student_id
WHERE exam_results.id = (
    SELECT TOP(1) id FROM exam_results T2
    WHERE exam_results.student_id = T2.student_id
    ORDER BY score DESC, date ASC)

在MySQL上，我认为您可以在语句结尾处将TOP（1）更改为LIMIT 1。我没有测试过这个。

Answer 6

Select Name, T.Score, er. date 
from Students S inner join
          (Select Student_ID,Max(Score) as Score from Exam_Results
           Group by Student_ID) T 
On S.id=T.Student_ID inner join Exam_Result er
On er.Student_ID = T.Student_ID And er.Score=T.Score

根据最大值连接表

6 个答案: