PostgreSQL的select语句

Question

考虑下表：

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 

和一个额外的表：座位包含的编号一样，row_no在预订与区域名称的座位名单。

已经分组使用此语句所预订的座位：

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; 

主要生产：

row_no | area_name 
    --------+-------------- 
     1 | rear stalls 
     2 | front stalls 

我怎么能现在使用的计数行和AREA_NAME编写一个SQL语句来产生一个列表显示节目名称，演出日期和时间以及每个区域的预订座位数量？

我已经试过这样：

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) 

与解决任何帮助，将不胜感激。

Answer 1

在每个表演条目上，查找所有预订。在这些预订中，链接到座位表以确定区域名称。然后，简单的COUNT（*）按照每个节目/区域的标准分组。

SELECT 
     p.perf_date, 
     p.perf_time, 
     p.title, 
     s.area_name, 
     COUNT(*) as SeatsSold 
    from 
     performance p 

     JOIN booking b 
      ON p.perf_date = b.perf_date 
      AND p.perf_time = b.perf_time 

      JOIN seat s 
       ON b.row_no = s.row_no 
    group by 
     p.perf_date, 
     p.perf_time, 
     p.title, 
     s.area_name 

Answer 2

您的行不重复。请注意结果中date和area_name的差异。这一切对我来说都很好。

Answer 3

你应该考虑合并列perf_date date和perf_time time成一个单一的timestamp柱：

perf_timestamp timestamp 

如果你需要一个date或time从一个时间戳，可简单地把它像这样：

SELECT perf_timestamp::time; 
SELECT perf_timestamp::date; 

我通常会建议使用代理主键。 “表演”（电影标题）的名称不是自然的关键 - 它不是唯一的。或者，正如@user_unknown已经提到的那样：开始时间对于演出来说并不是实际的主要关键。您可以使用serial列。整个安装可能看起来是这样的：

-- 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); 

然后将查询看起来是这样的：

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