给出下表:
shows:
title | basic_ticket_price
----------------+--------------------
Inception | $3.50
Romeo & Juliet | $2.00
performance:
perf_date | perf_time | title
------------+-----------+----------------
2012-08-14 | 00:08:00 | Inception
2012-08-12 | 00:12:00 | Romeo & Juliet
booking:
ticket_no | perf_date | perf_time | row_no | person_id
-----------+------------+-----------+--------+-----------
1 | 2012-08-14 | 00:08:00 | P01 | 1
2 | 2012-08-12 | 00:12:00 | O05 | 4
3 | 2012-08-12 | 00:12:00 | A01 | 2
还有一个附加表:seat,其中包含编号与预订中的 row_no 相同的座位列表,并带有区域名称。
使用此语句对预订的座位进行分组:
select count(row_no) AS row_no,
area_name
from seat
where exists (select row_no
from booking
where booking.row_no = seat.row_no)
group by area_name;
产生:
row_no | area_name
--------+--------------
1 | rear stalls
2 | front stalls
我现在如何使用计数的行和area_name编写单个SQL语句来生成一个列表,显示演出名称、演出日期和时间以及每个区域的预订座位数?
我已经尝试过这个:
select s.title,
perf_date,
perf_time,
count(row_no) AS row_no,
area_name
from shows s,
performance,
seat
where exists (select row_no
from booking
where booking.row_no = seat.row_no)
group by area_name,s.title,performance.perf_date,performance.perf_time;
但它显示重复的行:
title | perf_date | perf_time | row_no | area_name
----------------+------------+-----------+--------+--------------
Romeo & Juliet | 2012-08-12 | 00:12:00 | 1 | rear stalls
Romeo & Juliet | 2012-08-14 | 00:08:00 | 2 | front stalls
Inception | 2012-08-12 | 00:12:00 | 1 | rear stalls
Inception | 2012-08-14 | 00:08:00 | 2 | front stalls
Inception | 2012-08-14 | 00:08:00 | 1 | rear stalls
Inception | 2012-08-12 | 00:12:00 | 2 | front stalls
Romeo & Juliet | 2012-08-14 | 00:08:00 | 1 | rear stalls
Romeo & Juliet | 2012-08-12 | 00:12:00 | 2 | front stalls
(8 rows)
任何解决此问题的帮助将不胜感激。
最佳答案
您应该考虑将 perf_date date
和 perf_time time
列合并到单个 timestamp
中。栏目:
perf_timestamp timestamp
如果您需要时间戳中的日期
或时间
,只需按如下所示进行转换:
SELECT perf_timestamp::time;
SELECT perf_timestamp::date;
我通常建议使用代理主键。 “节目”(电影标题)的名称不是自然键 - 它不是唯一的。或者,正如 @user_unknown 已经提到的:开始时间并不是性能的实际主键。您可以使用 serial为此列。整个设置可能如下所示:
-- show:
CREATE TEMP TABLE show (
show_id serial PRIMARY KEY
,title text
,basic_ticket_price money -- or numeric
);
INSERT INTO show (title, basic_ticket_price) VALUES
('Inception', 3.50)
,('Romeo & Juliet', 2.00);
-- performance:
CREATE TEMP TABLE performance (
performance_id serial PRIMARY KEY
,show_id int REFERENCES show(show_id) ON UPDATE CASCADE
,perf_start timestamp
);
INSERT INTO performance (show_id, perf_start) VALUES
(1, '2012-08-14 00:08')
,(2, '2012-08-12 00:12');
-- seat:
CREATE TEMP TABLE seat (
row_no text PRIMARY KEY
,area_name text
);
INSERT INTO seat (row_no, area_name) VALUES
('P01', 'rear stalls')
,('O05', 'front stalls')
,('A01', 'front stalls');
-- booking:
CREATE TEMP TABLE booking (
ticket_id serial PRIMARY KEY
,performance_id int REFERENCES performance(performance_id) ON UPDATE CASCADE
,row_no text REFERENCES seat(row_no) ON UPDATE CASCADE
,person_id int -- REFERENCES ?
);
INSERT INTO booking (performance_id, row_no, person_id) VALUES
(1, 'P01', 1)
,(2, 'O05', 4)
,(2, 'A01', 2);
那么您的查询可能如下所示:
SELECT p.perf_start
,sh.title
,s.area_name
,count(*) booked
FROM booking b
JOIN seat s USING (row_no)
JOIN performance p USING (performance_id)
JOIN show sh USING (show_id)
GROUP BY 1,2,3
ORDER BY 1,2,3;
结果:
perf_start | title | area_name | booked
---------------------+----------------+--------------+-------
2012-08-12 00:12:00 | Romeo & Juliet | front stalls | 2
2012-08-14 00:08:00 | Inception | rear stalls | 1
关于sql - PostgreSQL 选择语句,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/9859821/