(1)你可以使用以下语句创建示例表:
mysql> CREATE TABLE shop (
-> article INT(4) UNSIGNED ZEROFILL DEFAULT '0000' NOT NULL,
-> dealer CHAR(20) DEFAULT '' NOT NULL,
-> price DOUBLE(16,2) DEFAULT '0.00' NOT NULL,
-> PRIMARY KEY(article, dealer));
mysql> INSERT INTO shop VALUES
-> (1,'A',3.45),(1,'B',3.99),(2,'A',10.99),(3,'B',1.45),
-> (3,'C',1.69),(3,'D',1.25),(4,'D',19.95);
(2)执行语句后,表应包含以下内容: SELECT * FROM shop;
(3)列的最大值
选取价格最贵:select max(price) as article from shop;
(4)拥有某个列的最大值的行
任务:找出最贵物品的编号、销售商和价格。
方法一:
mysql> select article,dealer,price
-> from shop
-> where price=(select max(price) from shop);
方法二:按价格降序排序所有行并用MySQL特定LIMIT子句只得到第一行,这里用到desc前面讲过,是降序的意思。
mysql> select article ,dealer,price
-> from shop
-> order by price desc
-> limit 1;
注:如果有多项最贵的物品( 例如每个的价格为19.95),LIMIT解决方案仅仅显示其中一个!
(5)列的最大值:按组
任务:每项物品的的最高价格是多少?
mysql> SELECT article, MAX(price) AS price
-> FROM shop
-> GROUP BY article
-> ;
(6)拥有某个字段的组间最大值的行
任务:对每项物品,找出最贵价格的物品的经销商。
SELECT article, dealer, price
FROM shop s1
WHERE price=(SELECT MAX(s2.price)
FROM shop s2
WHERE s1.article = s2.article);
(7)使用用户变量
任务:要找出价格最高或最低的物品的
SELECT @min_price:=MIN(price),@max_price:=MAX(price) FROM shop;
SELECT * FROM shop WHERE price=@min_price OR price=@max_price;
(8)使用外键
CREATE TABLE person (
id SMALLINT UNSIGNED NOT NULL AUTO_INCREMENT,
name CHAR(60) NOT NULL,
PRIMARY KEY (id)
);
CREATE TABLE shirt (
id SMALLINT UNSIGNED NOT NULL AUTO_INCREMENT,
style ENUM('t-shirt', 'polo', 'dress') NOT NULL,
color ENUM('red', 'blue', 'orange', 'white', 'black') NOT NULL,
owner SMALLINT UNSIGNED NOT NULL REFERENCES person(id),
PRIMARY KEY (id)
);
INSERT INTO person VALUES (NULL, 'Antonio Paz');
SELECT @last := LAST_INSERT_ID();
INSERT INTO shirt VALUES
(NULL, 'polo', 'blue', @last),
(NULL, 'dress', 'white', @last),
(NULL, 't-shirt', 'blue', @last);
INSERT INTO person VALUES (NULL, 'Lilliana Angelovska');
SELECT @last := LAST_INSERT_ID();
INSERT INTO shirt VALUES
(NULL, 'dress', 'orange', @last),
(NULL, 'polo', 'red', @last),
(NULL, 'dress', 'blue', @last),
(NULL, 't-shirt', 'white', @last);
SELECT * FROM person;
SELECT * FROM shirt;
SELECT s.* FROM person p, shirt s
WHERE p.name LIKE 'Lilliana%'
AND s.owner = p.id
AND s.color <> 'white';
(8)根据天计算访问量
CREATE TABLE t1 (year YEAR(4), month INT(2) UNSIGNED ZEROFILL,
day INT(2) UNSIGNED ZEROFILL);
INSERT INTO t1 VALUES(2000,1,1),(2000,1,20),(2000,1,30),(2000,2,2),
(2000,2,23),(2000,2,23);
SELECT year,month,BIT_COUNT(BIT_OR(1<<day)) AS days FROM t1
GROUP BY year,month;
(9)使用AUTO_INCREMENT
CREATE TABLE animals (
id MEDIUMINT NOT NULL AUTO_INCREMENT,
name CHAR(30) NOT NULL,
PRIMARY KEY (id)
);
INSERT INTO animals (name) VALUES
('dog'),('cat'),('penguin'),
('lax'),('whale'),('ostrich');
SELECT * FROM animals;