力扣sql中等篇练习(二十二)

1 坚定的友谊

1.1 题目内容

1.1.1 基本题目信息

1.1.2 示例输入输出

1.2 示例sql语句

# Write your MySQL query statement below
# ①求出所有朋友关系的总表
WITH T as
(
    SELECT user1_id ,user2_id
    FROM friendship
    UNION 
    SELECT user2_id user1_id ,user1_id user2_id
    FROM friendship
)

SELECT t1.user1_id,t2.user2_id,count(*) common_friend
FROM T t1
INNER JOIN T t2
ON t1.user2_id=t2.user1_id
INNER JOIN T t3
ON t1.user1_id=t3.user1_id
WHERE t2.user2_id=t3.user2_id AND t1.user1_id<t2.user2_id
GROUP BY t1.user1_id,t2.user2_id
HAVING count(*)>=3

1.3 运行截图

2 查询具有最多共同关注者的所有两两结对组

2.1 题目内容

2.1.1 基本题目信息

2.1.2 示例输入输出

2.2 示例sql语句

# 一样的关注着可能不只一个
with T as
(
    SELECT r1.user_id user1_id,r2.user_id user2_id,IFNULL(count(*),0) n
    FROM Relations r1
    INNER JOIN Relations r2
    ON r1.user_id<r2.user_id AND r1.follower_id=r2.follower_id
    GROUP BY r1.user_id,r2.user_id
)
SELECT user1_id,user2_id
FROM T
WHERE n=
(
    SELECT MAX(n)
    FROM T
)

2.3 运行截图

3 找出每所学校的最低分数要求

3.1 题目内容

3.1.1 基本题目信息1

3.1.2 基本题目信息2

3.1.3 示例输入输出

a 示例输入

b 示例输出

3.2 示例sql语句

# 只要容量大于学生招生数,且是最低分数要求就可以招生
# 先找出符合条件的,再去进行后续的筛选
SELECT s.school_id,IFNULL(MIN(e.score),-1) score
FROM Schools s
LEFT JOIN Exam e 
ON s.capacity>=e.student_count
GROUP BY s.school_id

3.3 运行截图