【发布时间】:2018-06-18 02:59:53
【问题描述】:
问题:
Construct the SQL statement to find all of the people that have meetings only before Dec. 25, 2016 at noon using INNER JOINs. Display the following columns:
Person’s first name
Person’s last name
Meeting ID
Meeting start date and time
Meeting end date and time
表格: 这个数据库有5个表(person, building, room, meeting, person_meeting
+-----------+------------+------------+
| person_id | first_name | last_name |
+-----------+------------+------------+
| 1 | Tom | Hanks |
| 2 | Anne | Hathaway |
| 3 | Tom | Cruise |
| 4 | Meryl | Streep |
| 5 | Chris | Pratt |
| 6 | Halle | Berry |
| 7 | Robert | De Niro |
| 8 | Julia | Roberts |
| 9 | Denzel | Washington |
| 10 | Melissa | McCarthy |
+-----------+------------+------------+
+-------------+----------------------+
| building_id | building_name |
+-------------+----------------------+
| 1 | Headquarters |
| 2 | Main Street Buidling |
+-------------+----------------------+
+---------+-------------+-------------+----------+
| room_id | room_number | building_id | capacity |
+---------+-------------+-------------+----------+
| 1 | 100 | 1 | 5 |
| 2 | 200 | 1 | 4 |
| 3 | 300 | 1 | 10 |
| 4 | 10 | 2 | 4 |
| 5 | 20 | 2 | 4 |
+---------+-------------+-------------+----------+
+------------+---------+---------------------+---------------------+
| meeting_id | room_id | meeting_start | meeting_end |
+------------+---------+---------------------+---------------------+
| 1 | 1 | 2016-12-25 09:00:00 | 2016-12-25 10:00:00 |
| 2 | 1 | 2016-12-25 10:00:00 | 2016-12-25 12:00:00 |
| 3 | 1 | 2016-12-25 11:00:00 | 2016-12-25 12:00:00 |
| 4 | 2 | 2016-12-25 09:00:00 | 2016-12-25 10:00:00 |
| 5 | 4 | 2016-12-25 09:00:00 | 2016-12-25 10:00:00 |
| 6 | 5 | 2016-12-25 14:00:00 | 2016-12-25 16:00:00 |
+------------+---------+---------------------+---------------------+
+-----------+------------+
| person_id | meeting_id |
+-----------+------------+
| 1 | 1 |
| 10 | 1 |
| 1 | 2 |
| 2 | 2 |
| 3 | 2 |
| 4 | 2 |
| 5 | 2 |
| 6 | 2 |
| 7 | 2 |
| 8 | 2 |
| 9 | 3 |
| 10 | 3 |
| 1 | 4 |
| 2 | 4 |
| 8 | 5 |
| 9 | 5 |
| 1 | 6 |
| 2 | 6 |
| 3 | 6 |
+-----------+------------+
到目前为止我的解决方案:
SELECT first_name,last_name ,building_name,meeting_start,meeting_end 来自 P 人 INNER JOIN B楼 ON P.person_id=PM.person_id INNER JOIN person_meeting PM ON M.room_id
我在完成 SQL 语句时遇到问题,请尽可能提供帮助。
【问题讨论】: