获得所有分数高于该 class 平均分的学生

Question

我在下面有两个名为tbl_student的表，我们在其中存储学生的数据，例如 id、name、class、roll no。 和年龄，以及tbl_marks我们存储考试得分的地方。 -

CREATE TABLE IF NOT EXISTS public.tbl_student
(
    student_id integer NOT NULL GENERATED ALWAYS AS IDENTITY,
    student_name varchar(255) NOT NULL,
    student_class integer,
    student_roll_no integer,
    age integer,
    CONSTRAINT tbl_student_pkey PRIMARY KEY (student_id)
)

和

CREATE TABLE IF NOT EXISTS public.tbl_marks
(
    marks_id integer NOT NULL GENERATED ALWAYS AS IDENTITY,
    student_id integer NOT NULL,
    subject varchar(255),
    marks integer,
    CONSTRAINT tbl_marks_pkey PRIMARY KEY (marks_id)
)

这些是各个表的一些数据：

INSERT INTO
    public.tbl_student (student_name, student_class, student_roll_no, age)
VALUES
    ('John Doe',5,1, 12),
    ('Clark Keny', 5,2, 10),
    ('Ross Barkley', 5, 3, 11),
    ('Frank Lampard', 5, 4, 9),
    ('John Terry', 5, 5, 10),
    ('Peter Parker',6,1, 13),
    ('Tony Stark', 6,2,11),
    ('Bruce Wayne', 6, 3,14),
    ('Johnny Depp', 6, 4,11),
    ('Jackie Chan',7,1,15),
    ('John Wick', 7,2,14),
    ('Indiana Jones', 7, 3,11),
    ('Halley Berry', 7, 4,12),
    ('Jane Doe', 7, 5,15),
    ('Martha Kent', 7, 6,13);

和

INSERT INTO
    public.tbl_marks (student_id, subject, marks)
VALUES
    (1, 'Math' , 79),
    (1, 'Science' , 64),
    (2, 'Math' , 69),
    (2, 'Science' , 72),
    (3, 'Math' , 30),
    (3, 'Science' , 50),
    (4, 'Math' , 77),
    (4, 'Science' , 72),
    (5, 'Math' , 84),
    (5, 'Science' , 88),
    (6, 'Math' , 36),
    (6, 'Science' , 54),
    (7, 'Math' , 55),
    (7, 'Science' , 54),
    (8, 'Math' , 66),
    (8, 'Science' , 78),
    (9, 'Math' , 43),
    (9, 'Science' , 20),
    (10, 'Math' ,87 ),
    (10, 'Science' , 92),
    (11, 'Math' , 68),
    (11, 'Science' , 82),
    (12, 'Math' , 78),
    (12, 'Science' , 93),
    (13, 'Math' , 20),
    (13, 'Science' , 22),
    (14, 'Math' , 40),
    (14, 'Science' , 42),
    (15, 'Math' , 52),
    (15, 'Science' , 62);

现在，我需要让每个 class 中所有分数高于各自班级平均分数的学生。

我使用带有PARTITION BY子句的RANK() function 进行了尝试。

Answer 1

尝试这个：

SELECT a.*
  FROM
     ( SELECT s.*
            , m.subject
            , m.marks
            , avg(m.marks) OVER (PARTITION BY m.subject, s.student_class) AS marks_avg
         FROM tbl_marks AS m
        INNER JOIN tbl_student AS s
           ON s.student_id = m.student_id
    ) AS a
WHERE a.marks > a.marks_avg

在dbfiddle中查看结果。

请参阅有关 window 功能的手册：

https://www.postgresql.org/docs/current/tutorial-window.html https://www.postgresql.org/docs/current/queries-table-expressions.html#QUERIES-WINDOW https://www. postgresql.org/docs/current/functions-window.html

还有关于可以用作 window 函数的聚合函数：

https://www.postgresql.org/docs/current/sql-expressions.html#SYNTAX-AGGREGATES https://www.postgresql.org/docs/current/functions-aggregate.html

Answer 2

好的，所以找到平均值是这样的：

    SELECT s.student_class, m.subject, AVG(m.marks) AS average
      FROM tbl_marks m INNER JOIN tbl_student s USING (student_id) 
      GROUP by s.student_class, m.subject

所以，我们只需要将它用作连接中的另一个表：

SELECT DISTINCT s.student_name, s.student_class, m.subject, m.marks
  FROM tbl_marks m 
  INNER JOIN tbl_student s USING (student_id) 
  INNER JOIN (
    SELECT s.student_class, m.subject, AVG(m.marks) AS average
      FROM tbl_marks m INNER JOIN tbl_student s USING (student_id) 
      GROUP by s.student_class, m.subject) x1 
    USING (student_class, subject)
  WHERE m.marks >= x1.average;

Output：

John Doe              5              Math        79        
Clark Keny            5              Math        69        
Clark Keny            5              Science     72        
Frank Lampard         5              Math        77        
Frank Lampard         5              Science     72        
John Terry            5              Math        84        
John Terry            5              Science     88        
Peter Parker          6              Science     54        
Tony Stark            6              Math        55        
Tony Stark            6              Science     54        
Bruce Wayne           6              Math        66        
Bruce Wayne           6              Science     78        
Jackie Chan           7              Math        87        
Jackie Chan           7              Science     92        
John Wick             7              Math        68        
John Wick             7              Science     82        
Indiana Jones         7              Math        78        
Indiana Jones         7              Science     93