create database mianshi5;
use mianshi5;
create table boys(
boy_id int not null,
boy varchar(10),
toy_id int);

create table toys(
toy_id int primary key not null,
toy varchar(10));

create table drink(
名称 varchar(5),
价格 decimal(8,2),
碳水化合物 decimal(8,2),
颜色 varchar(20),
加冰 varchar(10),
卡路里 int);

insert into boys values
(1,"Tony",3),
(2,"Andy",2), 
(3,"Frank",1), 
(4,"Only",2), 
(4,"Only",3), 
(5,"Terrance",4), 
(5,"Terrance",6);

insert into toys values
(1,"ToyA"),
(2,"ToyB"),
(3,"ToyC"),
(4,"ToyD"),
(5,"ToyE");


insert into drink values
("A",1,8.4,"Yellow","N",33),
("B",2.5,3.2,"Blue","N",12),
("C",3.5,8.8,"Orange","Y",35),
("D",2.5,5.4,"Green","Y",24),
("E",5.5,42.5,"Purple","Y",171);

#请用left join写出查询语句，找出每个男孩买了哪个玩具，并写出输出结果集

select * from boys left join toys on boys.toy_id=toys.toy_id order by boy_id;

#找出既买过“ToyB”也买过”ToyC”的男孩

select boy from boys where toy_id in
(select Toy_id from toys where toy in("ToyB","ToyC")) group by boy_id having count(boy_id)>1;

#列出加冰，且颜色为yellow，且卡路里大于30的饮料名称和价格

select 名称,价格 from drink where 颜色="yellow" and 加冰="Y" and 卡路里>30;

#列出碳水化合物小于4，或者加冰的饮料名称和颜色

select 名称,颜色 from drink where 碳水化合物<4 or 加冰="Y";

#所有卡路里小于100的饮料各一杯，需要多少钱

select sum(价格) from drink where 卡路里<100;