在与学生/年级等一起开展项目时,我需要不时更新前三名的学生。我想出了以下查询。但是,我很难获得他们的排名/顺序。我知道如何在一个简单的查询中执行此操作,但是在一个更复杂的查询中,它不起作用。 我正确地获取了所有其他列,并且使用尝试获取顺序的所有方法,有时我得到0(如代码的当前状态),有时得到的值是错误的(1、11、10、10 )等。
注意:我已经检查了各种问题(包括下面的问题),但是我不知道如何将它们放入查询中。
What is the best way to generate ranks in MYSQL?
摘要:
目标:
-从marks
获取每个学生的分数总和,将其除以表中的条目数(再次为marks
)。学生来自给定的年级。
-使用sum(mark)
对这些学生进行排名。
-获得前三名。
-将该年级的前三名学生及其平均成绩(如TopStudents
和ID放在sum
表中。
表格:
学生表包含有关学生的信息,包括ID:
+-------------+---------------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+-------------+---------------------+------+-----+---------+----------------+
| id | int (20) unsigned | NO | PRI | NULL | auto_increment |
| name |varchar(20) unsigned | NO | | NULL | |
+-------------+---------------------+------+-----+---------+----------------+
分数表列出每项考试中每个学生的分数
+-------------+---------------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+-------------+---------------------+------+-----+---------+----------------+
| id |int (20) unsigned | NO | PRI | NULL | auto_increment |
| idStudent |int (20) unsigned | NO | FOR | NULL | |
| mark |tinyInt (3) unsigned | NO | | NULL | |
| idExam |int (20) unsigned | NO | FOR | NULL | |
+-------------+---------------------+------+-----+---------+----------------+
成绩表具有成绩ID和名称:
+-------------+---------------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+-------------+---------------------+------+-----+---------+----------------+
| id | int (20) unsigned | NO | PRI | NULL | auto_increment |
| name |varchar(20) unsigned | NO | | NULL | |
+-------------+---------------------+------+-----+---------+----------------+
班级表各个年级的班级。参考表
+-------------+---------------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+-------------+---------------------+------+-----+---------+----------------+
| id | int (20) unsigned | NO | PRI | NULL | auto_increment |
| name |varchar(20) unsigned | NO | | NULL | |
| idGrade | int (20) unsigned | NO | FOR | NULL | |
+-------------+---------------------+------+-----+---------+----------------+
最后是臭名昭著的 TopStudents Table 。
+-------------+---------------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+-------------+---------------------+------+-----+---------+----------------+
| id | int (20) unsigned | NO | PRI | NULL | auto_increment |
| idStudent | int (20) unsigned | NO | FOR | NULL | |
| sumMarks | int (20) unsigned | NO | | NULL | |
| rank |tinyInt (1) unsigned | NO | | NULL | |
| date |date unsigned | NO | | NULL | |
+-------------+---------------------+------+-----+---------+----------------+
尝试: 尝试1 :错误:所有等级均为0
INSERT INTO topStudents(`date`, idStudent, `sum`, `order`)
SELECT
'2018-10-10' AS DATE,
student.id AS idStudent,
AVG(marks.mark)
@n = @n + 1 AS `order`
FROM
marks
INNER JOIN student ON student.id = marks.idStudent
INNER JOIN class ON class.id = marks.idClass
INNER JOIN grade ON class.idGrade = grade.id
WHERE
grade.id = 2
GROUP BY
marks.idStudent
ORDER BY
SUM(mark)
DESC
LIMIT 3
尝试2 :返回的排名:1、11、10
SET @n := 0;
INSERT INTO topStudents(`date`, idStudent, `sum`, `rank`)
SELECT
'2018-10-10' AS DATE,
tbl.idStudent AS idStudent,
AVG(tbl.mark) AS `sum`,
rnk AS `rank`
FROM (SELECT student.id AS idStudent, SUM(mark) AS mark FROM
marks
INNER JOIN student ON student.id = marks.idStudent
INNER JOIN class ON class.id = marks.idClass
INNER JOIN grade ON class.idGrade = grade.id
WHERE
grade.id = 2
GROUP BY
marks.idStudent
ORDER BY
SUM(mark)
DESC
LIMIT 3) AS tbl, (SELECT @n = @n + 1) AS rnk
答案 0 :(得分:0)
在较新的MySQL版本中,在分配等级之前,您需要使用派生表进行排序:
INSERT INTO topStudents (`date`, idStudent, `sum`, `order`)
SELECT date, idStudent, `sum`, (@n := @n + 1) AS `order`
FROM (SELECT '2018-10-10' AS DATE, s.id AS idStudent,
SUM(m.mark) / (SELECT COUNT(*) FROM marks m2 WHERE m2.idStudent = m.idStudent) AS `sum`
FROM marks m JOIN
student s
ON s.id = m.idStudent JOIN
class c
ON c.id = m.idClass JOIN
grade g
ON c.idGrade = g.id
WHERE g.id = 2
GROUP BY m.idStudent
ORDER BY SUM(mark) DESC
LIMIT 3
) sm CROSS JOIN
(SELECT @n := 0) params;
我几乎可以肯定sum
的计算是不正确的,并且您确实打算使用avg(mark)
。但是,这就是您所提出的逻辑。