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- 1
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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)