2041. 面试中被录取的候选人 - 力扣（LeetCode）

1. 2041. 面试中被录取的候选人 - 力扣（LeetCode）

2. 目标

   1. 输入

      表：Candidates
      candidate_id	name	years_of_exp	interview_id
      11	Atticus	1	101
      9	Ruben	6	104
      6	Aliza	10	109
      8	Alfredo	0	107

      表：Rounds
      interview_id	round_id	score
      109	3	4
      101	2	8
      109	4	1
      107	1	3
      104	3	6
      109	1	4
      104	4	7
      104	1	2
      109	2	1
      104	2	7
      107	2	3
      101	1	8

   2. 输出

      输出

      candidate_id

      9

3. 分析

   编写解决方案，找出 至少有两年 工作经验、且面试分数之和 严格大于 15 的候选人的 ID 。
   																					表：Candidates					表：Rounds				输出
   																					candidate_id	name	years_of_exp	interview_id		interview_id	round_id	score		candidate_id
   																					11	Atticus	1	101		109	3	4		9
   																					9	Ruben	6	104		101	2	8
   																					6	Aliza	10	109		109	4	1
   																					8	Alfredo	0	107		107	1	3
   					104	3	6
   					109	1	4
   					104	4	7
   					104	1	2
   					109	2	1
   					104	2	7
   					107	2	3
   					101	1	8
   筛选至少有两年经验的id,左连接ROUNDS表,分组聚合筛选面试分大于15的id		candidate_id
   		9

4. 实现

   DROP TABLE IF EXISTS Candidates;
   DROP TABLE IF EXISTS Rounds;
   Create table If Not Exists Candidates (candidate_id int, name varchar(30), years_of_exp int, interview_id int);
   Create table If Not Exists Rounds (interview_id int, round_id int, score int);
   Truncate table Candidates;
   insert into Candidates (candidate_id, name, years_of_exp, interview_id) values ('11', 'Atticus', '1', '101');
   insert into Candidates (candidate_id, name, years_of_exp, interview_id) values ('9', 'Ruben', '6', '104');
   insert into Candidates (candidate_id, name, years_of_exp, interview_id) values ('6', 'Aliza', '10', '109');
   insert into Candidates (candidate_id, name, years_of_exp, interview_id) values ('8', 'Alfredo', '0', '107');
   Truncate table Rounds;
   insert into Rounds (interview_id, round_id, score) values ('109', '3', '4');
   insert into Rounds (interview_id, round_id, score) values ('101', '2', '8');
   insert into Rounds (interview_id, round_id, score) values ('109', '4', '1');
   insert into Rounds (interview_id, round_id, score) values ('107', '1', '3');
   insert into Rounds (interview_id, round_id, score) values ('104', '3', '6');
   insert into Rounds (interview_id, round_id, score) values ('109', '1', '4');
   insert into Rounds (interview_id, round_id, score) values ('104', '4', '7');
   insert into Rounds (interview_id, round_id, score) values ('104', '1', '2');
   insert into Rounds (interview_id, round_id, score) values ('109', '2', '1');
   insert into Rounds (interview_id, round_id, score) values ('104', '2', '7');
   insert into Rounds (interview_id, round_id, score) values ('107', '2', '3');
   insert into Rounds (interview_id, round_id, score) values ('101', '1', '8');
   SELECT * FROM Candidates;
   SELECT * FROM Rounds;
   # 筛选至少有两年经验的id,左连接ROUNDS表,分组聚合筛选面试分大于15的id
   SELECT candidate_id FROM Candidates
   LEFT JOIN Rounds R ON Candidates.interview_id = R.interview_id
   WHERE years_of_exp>=2
   GROUP BY candidate_id
   HAVING SUM(score)>15

5. 小结

        左连接,分组聚合