对2列求和并用desc对其排序,只显示1条记录,并首先插入优先级(最小的id)

ovfsdjhp  于 2021-06-21  发布在  Mysql
关注(0)|答案(1)|浏览(232)

你好,我知道我的问题,看起来有点复杂,
所以首先我有一个学生

+------------+------- +---------+
| id_student | score_a|score_b  |
+------------+--------+---------+
| 1          | 70     |80       |
+------------+--------+---------+
| 2          | 70     |90       |
+------------+--------+---------+
| 3          | 80     |70       |
+------------+--------+---------+
| 4          | 80     |70       |
+------------+--------+---------+
| 5          | 80     |90       |
+------------+--------+---------+
| 6          | 60     |70       |
+------------+--------+---------+
| 7          | 60     |80       |
+------------+--------+---------+
| 8          |40      |70       |
+------------+--------+---------+

然后上课

+----------+--------------+
| id_class | limit_people |
+----------+--------------+
| 1        | 2            |
+----------+--------------+
| 2        | 2            |
+----------+--------------+
| 3        | 1            |
+----------+--------------+

最后让我们说注册。在这个表格中,学生可以选择2个班级,但他们只能选择1个班级(插入的第一个班级是优先顺序),当使用2个分数之和排序时( student.score_a 以及 student.score_b )基于限制的描述 class.limit_people field(这可能吗?),我会给你一张这张表,但我不知道这是否正确。我希望你能给出一个更好的字段/栏
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+-------------+-------------+----------+
| id_register | id_students | id_class |
+-------------+-------------+----------+
| 1           | 1           | 1        |
+-------------+-------------+----------+
| 2           | 1           | 2        |
+-------------+-------------+----------+
| 3           | 2           | 2        |
+-------------+-------------+----------+
| 4           | 2           | 3        |
+-------------+-------------+----------+
| 5           | 3           | 1        |
+-------------+-------------+----------+
| 6           | 3           | 3        |
+-------------+-------------+----------+
| 7           | 4           | 3        |
+-------------+-------------+----------+
| 8           | 4           | 2        |
+-------------+-------------+----------+

我使用的当前查询是

SELECT register.id_register, register.id_students, SUM((student.score_a*50/100)+ (student.score_b*50/100)) as total,register.id_class 
FROM `register` LEFT JOIN `student` ON register.id_students=student.id_student 
GROUP by register.id_register ORDER BY total DESC

这就是结果

+-------------+-------------+-------+----------+
| id_register | id_students | total | id_class |
 +-------------+-------------+-------+----------+
| 3           | 2           | 80    | 2        |
+-------------+-------------+-------+----------+
| 4           | 2           | 80    | 3        |
+-------------+-------------+-------+----------+
| 2           | 1           | 75    | 2        |
+-------------+-------------+-------+----------+
| 5           | 3           | 75    | 1        |
+-------------+-------------+-------+----------+
| 6           | 3           | 75    | 3        |
+-------------+-------------+-------+----------+
| 7           | 4           | 75    | 3        |
+-------------+-------------+-------+----------+
| 8           | 4           | 75    | 2        |
+-------------+-------------+-------+----------+
| 1           | 1           | 75    | 1        |
+-------------+-------------+-------+----------+

正如你可以看到所有的学生都展示了2次,如何使它只有1而不求和所有 total 具有相同id的学生,如果有多个具有相同id的记录,如何优先显示第一个插入的id register.id_students ?
这是我想要的table

+-------------+-------------+-------+----------+
| id_register | id_students | total | id_class |
+-------------+-------------+-------+----------+
| 3           | 2           | 80    | 2        |
+-------------+-------------+-------+----------+
| 1           | 1           | 75    | 1        |
+-------------+-------------+-------+----------+
| 5           | 3           | 75    | 1        |
+-------------+-------------+-------+----------+
| 7           | 4           | 75    | 3        |
+-------------+-------------+-------+----------+
xytpbqjk

xytpbqjk1#

使用 GROUP BY register.id_students 而不是 GROUP by register.id_register 获取 id_students 现场。因此,查询将如下所示:

SELECT 
  register.id_register, 
  register.id_students, 
  (student.score_a + student.score_b)*50/100 AS total,
  register.id_class 
FROM 
  `register` 
LEFT JOIN 
  `student` 
ON (register.id_students = student.id_student)
GROUP BY register.id_students 
ORDER BY total DESC

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