力扣sql中等篇练习(二十)

力扣sql中等篇练习(二十)

1 寻找面试候选人

1.1 题目内容

1.1.1 基本题目信息1

1.1.2 基本题目信息2

1.1.3 示例输入输出

a 示例输入

b 示例输出

1.2 示例sql语句

# 分为以下两者情况,分别考虑,然后union进行处理(有可能同时满足,需要去进行去重)
# ①该用户在 三场及更多不同的 比赛中赢得 金牌（这些比赛可以不是连续的）分组然后数量大于3即可
#连续三年以及更多,意味着只需要满足三年的情况就行
#使用with创建临时表(查出每年用户是否获奖),inner join后面可以直接接上union
WITH T AS
(SELECT u.user_id,c.contest_idFROM Contests cINNER JOIN Users uON c.gold_medal=u.user_idOR c.silver_medal=u.user_idOR c.bronze_medal=u.user_idORDER BY u.user_id,c.contest_id
)
SELECT name,mail
FROM Users
WHERE user_id IN
(SELECT distinct t1.user_idFROM T t1INNER JOIN T t2ON t1.user_id=t2.user_id AND t1.contest_id+1=t2.contest_idINNER JOIN T t3ON t2.user_id=t3.user_id AND t2.contest_id+1=t3.contest_idUNIONSELECT gold_medal user_idFROM ContestsGROUP BY gold_medalHAVING count(contest_id)>=3
)

1.3 运行截图

2 每天的最大交易

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

# Write your MySQL query statement below
SELECT t1.transaction_id
FROM
(SELECT transaction_id,date_format(day,'%Y-%m-%d') d,amountFROM Transactions 
)t1
INNER JOIN
(SELECT t2.d,MAX(t2.amount) nFROM(SELECT transaction_id,date_format(day,'%Y-%m-%d') d,amountFROM Transactions )t2GROUP BY t2.d
)t3
ON t1.d=t3.d AND t1.amount=t3.n
ORDER BY t1.transaction_id asc

2.3 运行截图

3 联赛信息统计

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

3.2 示例sql语句

# 注意表中是进球数,不是得分数
# 分别求出客场时得分和主场时得分,两者合并后再分组即可
# 按照字典顺序默认就是升序
SELECT (SELECT team_name FROM Teams WHERE team_id=t1.team_id) team_name,count(*) matches_played,sum(t1.score) points,SUM(t1.n1) goal_for,SUM(t1.n2) goal_against,SUM(t1.n1)-SUM(t1.n2) goal_diff
FROM
(SELECT home_team_id team_id,IF(home_team_goals>away_team_goals,3,IF(home_team_goals=away_team_goals,1,0)) score,home_team_goals n1,away_team_goals n2FROM MatchesUNION ALLSELECT away_team_id team_id,IF(home_team_goals<away_team_goals,3,IF(home_team_goals=away_team_goals,1,0)) score,away_team_goals n1,home_team_goals n2FROM Matches
)t1
GROUP BY t1.team_id
ORDER BY points desc,goal_diff desc,team_name asc

3.3 运行截图