【发布时间】:2015-11-15 05:23:00
【问题描述】:
这是我需要从函数重新创建的结果。
+-----------------+------------+------------+-----------+
| exhibition_name | start_date | end_date | Paintings |
+-----------------+------------+------------+-----------+
| E1 | 12-18-2015 | 12-18-2015 | 7 |
| E2 | 11-10-2015 | 11-13-2015 | 2 |
| Scupture show | 10-20-2015 | 10-25-2015 | 4 |
+-----------------+------------+------------+-----------+
下面的这些表格是我正在使用的,我已经在其中插入了值
CREATE TABLE Exhibition(
name VARCHAR(15) NOT NULL,
start_date VARCHAR(11),
end_date VARCHAR(11),
PRIMARY KEY (name)
);
CREATE TABLE Shown_at(
art INT NOT NULL,
exhibition_name VARCHAR(15) NOT NULL,
PRIMARY KEY (art, exhibition_name),
FOREIGN KEY (art) REFERENCES Art_object(id_no),
FOREIGN KEY (exhibition_name) REFERENCES Exhibition(name)
);
这是我尝试完成的功能。
CREATE FUNCTION No_of_painting_exhibited(exhb_name VARCHAR(15), startd VARCHAR(11), enddVARCHAR(11), num_paint INT)
RETURNS VARCHAR(15), VARCHAR(11), VARCHAR(11), INT
DETERMINISTIC
SELECT exhibition_name, start_date, end_date, COUNT(*) as Paintings
FROM Shown_at, Exhibition
WHERE name = exhibition_name
Group by exhibition_name;
END
【问题讨论】:
-
嗨,天使,请提供您在尝试中遇到的具体问题,这样您的问题将在 SO 中得到更好的关注。
标签: mysql