mysql学生成绩排名,分组取前 N 条记录

转载  https://blog.csdn.net/jslcylcy/article/details/72627762

score表:

CREATE TABLE `score` (
  `student_id` int(10) DEFAULT NULL,
  `class_id` int(10) DEFAULT NULL,
  `score` int(5) DEFAULT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4 COLLATE=utf8mb4_unicode_ci

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字段 student_id 学生 id ,class_id:班级 id ,score:分数 
数据准备:

insert into score values(1,1,100),(2,1,93),(3,1,89),(4,1,96),(5,2,98),(6,2,97),(7,2,90),(8,2,88),(9,1,96);

表结构如下:

mysql> select * from score;
+------------+----------+-------+
| student_id | class_id | score |
+------------+----------+-------+
|          1 |        1 |   100 |
|          2 |        1 |    93 |
|          3 |        1 |    89 |
|          4 |        1 |    96 |
|          5 |        2 |    98 |
|          6 |        2 |    97 |
|          7 |        2 |    90 |
|          8 |        2 |    88 |
|          9 |        1 |    96 |
+------------+----------+-------+
9 rows in set (0.00 sec)

1.取每个班级前两名的学生(包含并列第二名)

mysql> select * from score s1 where (select count(0) from score s2 where s1.class_id = s2.class_id and s1.score < s2.score) < 2;
+------------+----------+-------+
| student_id | class_id | score |
+------------+----------+-------+
|          1 |        1 |   100 |
|          4 |        1 |    96 |
|          5 |        2 |    98 |
|          6 |        2 |    97 |
|          9 |        1 |    96 |
+------------+----------+-------+
5 rows in set (0.00 sec)

sql 解释:取表 s1的数据,这些数据中 class_id 和 s2 class_id相同的数据下,比 s1的 score 分数大的 s2的数据条目必须小于2

或者使用 left join 的方式:

mysql> select s1.* from score s1 left join score s2 on s1.class_id = s2.class_id and s1.score<s2.score group by s1.class_id,s1.student_id,s1.score having count(s2.student_id)<2;
+------------+----------+-------+
| student_id | class_id | score |
+------------+----------+-------+
|          1 |        1 |   100 |
|          4 |        1 |    96 |
|          9 |        1 |    96 |
|          5 |        2 |    98 |
|          6 |        2 |    97 |
+------------+----------+-------+
5 rows in set (0.00 sec)

2.取学生分数数据且表示排名

mysql> select s1.*,(select count(0) + 1 from score s2 where s2.score > s1.score)rank from score s1;
+------------+----------+-------+------+
| student_id | class_id | score | rank |
+------------+----------+-------+------+
|          1 |        1 |   100 |    1 |
|          2 |        1 |    93 |    6 |
|          3 |        1 |    89 |    8 |
|          4 |        1 |    96 |    4 |
|          5 |        2 |    98 |    2 |
|          6 |        2 |    97 |    3 |
|          7 |        2 |    90 |    7 |
|          8 |        2 |    88 |    9 |
|          9 |        1 |    96 |    4 |
+------------+----------+-------+------+
9 rows in set (0.00 sec)

sql解释:将 s2中比s1中分数大的条目显示出来就行了(count 时需要加1)

3.取学生成绩数据,表示班级排名

mysql> select s1.*,(select count(0) + 1 from score s2 where s1.class_id = s2.class_id and s2.score > s1.score)rank from score s1 order by class_id,rank;
+------------+----------+-------+------+
| student_id | class_id | score | rank |
+------------+----------+-------+------+
|          1 |        1 |   100 |    1 |
|          4 |        1 |    96 |    2 |
|          9 |        1 |    96 |    2 |
|          2 |        1 |    93 |    4 |
|          3 |        1 |    89 |    5 |
|          5 |        2 |    98 |    1 |
|          6 |        2 |    97 |    2 |
|          7 |        2 |    90 |    3 |
|          8 |        2 |    88 |    4 |
+------------+----------+-------+------+
9 rows in set (0.00 sec)

与之前一样,但过滤条件中只需要计算班级相同的数据条目

4.取每个班级前两名(并列的只取前面的数据)