MySQL超过5名学生的课
- SQL架构
- 题目描述
- 题解
- 方法一：使用 GROUP BY 子句和子查询【通过】
- 方法二：使用 GROUP BY 和 HAVING 条件【通过】
SQL架构
Create table If Not Exists courses (student varchar(255), class varchar(255));
insert into courses (student, class) values ('A', 'Math');
insert into courses (student, class) values ('B', 'English');
insert into courses (student, class) values ('C', 'Math');
insert into courses (student, class) values ('D', 'Biology');
insert into courses (student, class) values ('E', 'Math');
insert into courses (student, class) values ('F', 'Computer');
insert into courses (student, class) values ('G', 'Math');
insert into courses (student, class) values ('H', 'Math');
insert into courses (student, class) values ('I', 'Math');
题目描述
请列出所有超过或等于5名学生的课。
例如，表：
+---------+------------+
| student | class |
+---------+------------+
| A | Math |
| B | English |
| C | Math |
| D | Biology |
| E | Math |
| F | Computer |
| G | Math |
| H | Math |
| I | Math |
+---------+------------+
应该输出:
+---------+
| class |
+---------+
| Math |
+---------+
提示：
学生在每个课中不应被重复计算。
题解
方法一：使用 GROUP BY 子句和子查询【通过】
思路
先统计每门课程的学生数量，再从中选择超过 5 名学生的课程。
算法
使用 GROUP BY 和 COUNT 获得每门课程的学生数量。
SELECT
class, COUNT(DISTINCT student)
FROM
courses
GROUP BY class;
注：使用 DISTINCT 防止在同一门课中学生被重复计算。
+----------+-------------------------+
| class | COUNT(DISTINCT student) |
+----------+-------------------------+
| Biology | 1 |
| Computer | 1 |
| English | 1 |
| Math | 6 |
+----------+-------------------------+
4 rows in set (0.00 sec)
使用上面查询结果的临时表进行子查询，筛选学生数量超过 5 的课程。
SELECT
class
FROM
(SELECT
class, COUNT(DISTINCT student) AS num
FROM
courses
GROUP BY class) AS temp_table
WHERE
num >= 5;
+-------+
| class |
+-------+
| Math |
+-------+
1 row in set (0.00 sec)
方法二：使用 GROUP BY 和 HAVING 条件【通过】
算法
在 GROUP BY 子句后使用 HAVING 条件是实现子查询的一种更加简单直接的方法。
SELECT
class
FROM
courses
GROUP BY class
HAVING COUNT(DISTINCT student) >= 5;
+-------+
| class |
+-------+
| Math |
+-------+
1 row in set (0.00 sec)
