MySQL高阶练习题1- 寻找面试候选人

news2024/11/26 1:48:02

题目

准备数据

分析数据

实现代码

总结

题目

返回 所有面试候选人 的姓名 name 和邮件 mail 。当用户满足以下两个要求中的 任意一条 ，其成为 面试候选人 :

该用户在 连续三场及更多 比赛中赢得任意奖牌。
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）

准备数据

## 创建库
create database db;
use db;

## 创建Contests表
Create table If Not Exists Contests (contest_id int, gold_medal int, silver_medal int, bronze_medal int);

## 创建Users表
Create table If Not Exists Users (user_id int, mail varchar(50), name varchar(30));

## 向表中插入数据
Truncate table Contests;
insert into Contests (contest_id, gold_medal, silver_medal, bronze_medal) values ('190', '1', '5', '2');
insert into Contests (contest_id, gold_medal, silver_medal, bronze_medal) values ('191', '2', '3', '5');
insert into Contests (contest_id, gold_medal, silver_medal, bronze_medal) values ('192', '5', '2', '3');
insert into Contests (contest_id, gold_medal, silver_medal, bronze_medal) values ('193', '1', '3', '5');
insert into Contests (contest_id, gold_medal, silver_medal, bronze_medal) values ('194', '4', '5', '2');
insert into Contests (contest_id, gold_medal, silver_medal, bronze_medal) values ('195', '4', '2', '1');
insert into Contests (contest_id, gold_medal, silver_medal, bronze_medal) values ('196', '1', '5', '2');
Truncate table Users;
insert into Users (user_id, mail, name) values ('1', 'sarah@leetcode.com', 'Sarah');
insert into Users (user_id, mail, name) values ('2', 'bob@leetcode.com', 'Bob');
insert into Users (user_id, mail, name) values ('3', 'alice@leetcode.com', 'Alice');
insert into Users (user_id, mail, name) values ('4', 'hercy@leetcode.com', 'Hercy');
insert into Users (user_id, mail, name) values ('5', 'quarz@leetcode.com', 'Quarz');

输入表

contests表

users表

分析数据

分析一：为了容易连接，将金牌，银牌，铜牌设置成一列。

select contest_id,'gold' type,gold_medal user_id from contests
 union all
 select contest_id,'silver' type,silver_medal user_id from contests
 union all
 select contest_id,'bronze' type,bronze_medal user_id from contests

分析二：该用户连续三场及更多比赛中赢得任意奖牌


with t as (
    select contest_id,'gold' type,gold_medal user_id from contests
    union all
    select contest_id,'silver' type,silver_medal user_id from contests
    union all
    select contest_id,'bronze' type,bronze_medal user_id from contests
)
   , t1 as (
    select t1.user_id from t t1,t t2,t t3
    where t1.user_id = t2.user_id and t2.user_id = t3.user_id
      and t1.contest_id + 1 = t2.contest_id and t2.contest_id + 1 = t3.contest_id)
select * from t1;

分析三：该用户在三场及更多不同的比赛中赢得金牌（可不连续）

with t as (
    select contest_id,'gold' type,gold_medal user_id from contests
    union all
    select contest_id,'silver' type,silver_medal user_id from contests
    union all
    select contest_id,'bronze' type,bronze_medal user_id from contests
)select user_id from t where type = 'gold' group by user_id having count(*) >= 3;

实现代码

with t as (
    select contest_id,'gold' type,gold_medal user_id from contests
    union all
    select contest_id,'silver' type,silver_medal user_id from contests
    union all
    select contest_id,'bronze' type,bronze_medal user_id from contests
)
   , t1 as (
    select t1.user_id from t t1,t t2,t t3
    where t1.user_id = t2.user_id and t2.user_id = t3.user_id
      and t1.contest_id + 1 = t2.contest_id and t2.contest_id + 1 = t3.contest_id)
   , t2 as (
    select user_id from t where type = 'gold' group by user_id having count(*) >= 3
)
   , t3 as (
    select user_id from t1 union select user_id from t2
)
select u.name,u.mail from t3 , users u where t3.user_id = u.user_id;