MySQL实训4
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表名：[boys]
| Boy_id | Boy | Toy_id |
|---|---|---|
| 1 | Tony | 3 |
| 2 | Andy | 2 |
| 3 | Frank | 1 |
| 4 | Only | 2 |
| 4 | Only | 3 |
| 5 | Terrance | 4 |
| 5 | Terrance | 6 |
#创建表及表字段
mysql> create table boys(
-> boy_id int not null,
-> boy varchar(10),
-> toy_id int);
#查看表结构
mysql> desc boys;
+--------+-------------+------+-----+---------+-------+
| Field | Type | Null | Key | Default | Extra |
+--------+-------------+------+-----+---------+-------+
| boy_id | int(11) | NO | | NULL | |
| boy | varchar(10) | YES | | NULL | |
| toy_id | int(11) | YES | | NULL | |
+--------+-------------+------+-----+---------+-------+
#插入数据
mysql> 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);
#查询表的数据
mysql> select * from boys;
+--------+----------+--------+
| boy_id | boy | toy_id |
+--------+----------+--------+
| 1 | Tony | 3 |
| 2 | Andy | 2 |
| 3 | Frank | 1 |
| 4 | Only | 2 |
| 4 | Only | 3 |
| 5 | Terrance | 4 |
| 5 | Terrance | 6 |
+--------+----------+--------+
表：[toys]
| Toy_id | Toy |
|---|---|
| 1 | ToyA |
| 2 | ToyB |
| 3 | ToyC |
| 4 | ToyD |
| 5 | ToyE |
#创建表及表字段
mysql> create table toys(
-> toy_id int primary key not null,
-> toy varchar(10));
#查看表字段
mysql> desc toys;
+--------+-------------+------+-----+---------+-------+
| Field | Type | Null | Key | Default | Extra |
+--------+-------------+------+-----+---------+-------+
| toy_id | int(11) | NO | PRI | NULL | |
| toy | varchar(10) | YES | | NULL | |
+--------+-------------+------+-----+---------+-------+
#插入数据
mysql> insert into toys values
-> (1,"ToyA"),
-> (2,"ToyB"),
-> (3,"ToyC"),
-> (4,"ToyD"),
-> (5,"ToyE");
#查询表的数据
mysql> select * from toys;
+--------+------+
| toy_id | toy |
+--------+------+
| 1 | ToyA |
| 2 | ToyB |
| 3 | ToyC |
| 4 | ToyD |
| 5 | ToyE |
+--------+------+
1.请用left join写出查询代码，找出每个男孩买了哪个玩具，并写出输出结果集。
在表boys中，并没有玩具的名称，只有玩具的id，故需要将两个表进行联合，题目已经要求用left join联合，由表可知用toy_id进行连接。
由于题目要求知道每个男孩买的玩具，故需要对男孩进行分类(group by)，为了更好的显示玩具名，可使用group_concat函数将每个男孩买的玩具(toy)拼接到一起
mysql> select boy_id,boy,group_concat(toy separator ',') as toys
-> from boys left join toys on boys.toy_id=toys.toy_id
-> group by boy_id;
+--------+----------+-----------+
| boy_id | boy | toys |
+--------+----------+-----------+
| 1 | Tony | ToyC |
| 2 | Andy | ToyB |
| 3 | Frank | ToyA |
| 4 | Only | ToyB,ToyC |
| 5 | Terrance | ToyD |
+--------+----------+-----------+
2.找出既买过“ToyB”也买过”ToyC”的男孩
分析：
- 首先会想到使用in，看哪些人买过“ToyB”和”ToyC”，但是使用in只能查询买过，不能知道是否同时都买了。
- 随后想到能否对买了“ToyB”和”ToyC”的boy_id进行分组，看男孩买过的玩具是否不止一个。如果不止一个的话，则既买过“ToyB”也买过”ToyC”。
步骤：
- 首先使用(select Toy_id from toys where toy in(“ToyB”,“ToyC”))语句得到ToyB”和”ToyC”的toy_id。g使用in得到一个买的玩具只有“ToyB”和”ToyC”的表。
- 再对上一步的表进行分析，对boy_id进行分组，就能得到每个男孩买的玩具情况，再使用count统计boy_id看其是否不止出现一个
mysql> 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;
+------+
| boy |
+------+
| Only |
+------+
表名：[饮料信息]
| 饮料名称 | 价格 | 碳水化合物 | 颜色 | 加冰 | 卡路里 |
|---|---|---|---|---|---|
| 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 |
#创建表及表字段
mysql> 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);
#查看表结构
mysql> desc drink;
+------------+--------------+------+-----+---------+-------+
| Field | Type | Null | Key | Default | Extra |
+------------+--------------+------+-----+---------+-------+
| 名称 | varchar(5) | YES | | NULL | |
| 价格 | decimal(8,2) | YES | | NULL | |
| 碳水化合物 | decimal(8,2) | YES | | NULL | |
| 颜色 | varchar(20) | YES | | NULL | |
| 加冰 | varchar(10) | YES | | NULL | |
| 卡路里 | int(11) | YES | | NULL | |
+------------+--------------+------+-----+---------+-------+
#插入数据
mysql> 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);
#查询表的数据
mysql> select * from drink;
+------+------+------------+--------+------+--------+
| 名称 | 价格 | 碳水化合物 | 颜色 | 加冰 | 卡路里 |
+------+------+------------+--------+------+--------+
| A | 1.00 | 8.40 | Yellow | N | 33 |
| B | 2.50 | 3.20 | Blue | N | 12 |
| C | 3.50 | 8.80 | Orange | Y | 35 |
| D | 2.50 | 5.40 | Green | Y | 24 |
| E | 5.50 | 42.50 | Purple | Y | 171 |
+------+------+------------+--------+------+--------+
1.列出不加冰，且颜色为yellow，且卡路里大于30的饮料名称和价格
- 需要显示的数据是“名称”，“价格”
- 条件是不加冰(加冰=“N”)，颜色为yellow(颜色=“yellow”)，卡路里大于30(卡路里>30)
mysql> select 名称记一次Mysql魔鬼实训