手擼Mysql原生語句--多表

AndreasZhou發表於2020-09-26
MySql

在開始之前，我們需要建立表，做建表和資料的準備的工作。

1.建表
create table department(
	id int,
    name varchar(20)
);
create table employee(
	id int primary key auto_increment,
    name varchar(20),
    sex enmu('male','female') not null default = 'male',
    age int,
    dep_id int,
    constraint fk_id foreign key(dep_id) references department(id),
);

2.插入資料
insert into department values 
(200,'技術'),
(201,'人力資源'),
(202,'銷售'),
(203,'運營');

insert into employee （name,sex,age,dep_id）values
('egon','male',18,200),
('alex','female',48,201),
('wupeiqi','male',38,201),
('yuanhao','female',28,202),
('liwenzhou','male',18,200),
('jingliyang','female',18,204);

# 檢視錶結構
desc department;
desc employee;

# 檢視我們插入的資料
select * from department;
select * from employee;

多表的查詢，我們主要是分為兩種查詢，這兩種查詢分別是多表連線查詢和子查詢兩種情況。

多表連線查詢

1.交叉連線：不適用任何匹配條件。生成笛卡爾積。

select * from employee,department;

2.內連線

# 找兩張表共有部分，相當於利用笛卡兒積結果中篩選出正確的結果。
# department沒有204這個部門，因而employee表中關於204這條員工資訊沒有匹配出來
select employee.id,employee.name,employee.age,employee.sex,department.name from employee inner join department on employee.dep_id=department.id;

上面的sql語句等價於下面的sql語句
select employee.id,employee.name,employee.age,employee.sex,department.name from employee,department where employee.dep_id=department.id;

3.外連線之左連線：優先顯示左表全部記錄

# 以左表為標準，即找出所有員工資訊，當然包括沒有部門的員工
# 本質就是：在內連線的基礎上增加左邊有而右邊沒有的結果
select employee.id,employee.name,department.name as depart_name from employee letf join department on employee.dep_id=department.id;

4.外連線之右連線：優先顯示右表全部記錄

# 以右表為標準，即找出所有部門資訊，包括沒有員工的部門
# 本質就是：在內連線的基礎上增加右邊有而左邊沒有的結果
select employee.id,employee.name,department.name as depart_name from employee right join department on empolyee.dep_id=department.id;

5.全外連線：顯示左右兩個表全部的記錄

select * from employee left join department on employee.dep_id = department.id
union
select * from employee right join department on employee.dep_id = department.id;
#注意 union與union all的區別：union會去掉相同的紀錄

6.符合條件連線查詢

# 例項1：以內連線的方式查詢employee和department表，並且employee表中的age欄位值必須大於25，即找出年齡大於25歲的員工以及員工所在的部門
select employee.name,department.name from employee inner join department on employee.dep_id=department.id where age > 25;
#示例2：以內連線的方式查詢employee和department表，並且以age欄位的升序方式顯示
select employee.id,employee.name,employee.age,department.name from employee inner join department on employee.dep_id=department.id order by age asc;

子查詢

# 1.子查詢是將一個查詢語句巢狀在另外一個查詢語句中。
# 2.內層查詢語句的查詢結果，可以為外層查詢語句提供查詢條件。
#3：子查詢中可以包含：IN、NOT IN、ANY、ALL、EXISTS 和 NOT EXISTS等關鍵字
#4：還可以包含比較運算子：= 、 !=、> 、<等

1.帶IN關鍵字的子查詢

#查詢平均年齡在25歲以上的部門名
select id,name from department
    where id in 
        (select dep_id from employee group by dep_id having avg(age) > 25);
這兩個sql語句是等價的。
 select department.name from employee inner join department on employee.dep_id=department.id where employee.age > (select avg(age) from employee group by dep_id);
#檢視技術部員工姓名
select name from employee
    where dep_id in 
        (select id from department where name='技術');

#檢視不足1人的部門名(子查詢得到的是有人的部門id)
select name from department where id not in (select distinct dep_id from employee);

2.帶比較運算子的子查詢

#比較運算子：=、!=、>、>=、<、<=、<>
#查詢大於所有人平均年齡的員工名與年齡
mysql> select name,age from emp where age > (select avg(age) from emp);
+---------+------+
| name    | age  |
+---------+------+
| alex    | 48   |
| wupeiqi | 38   |
+---------+------+
2 rows in set (0.00 sec)


#查詢大於部門內平均年齡的員工名、年齡
select t1.name,t1.age from emp t1
inner join 
(select dep_id,avg(age) avg_age from emp group by dep_id) t2
on t1.dep_id = t2.dep_id
where t1.age > t2.avg_age;

3 帶EXISTS關鍵字的子查詢

EXISTS關字鍵字表示存在。在使用EXISTS關鍵字時，內層查詢語句不返回查詢的記錄。
而是返回一個真假值。True或False
當返回True時，外層查詢語句將進行查詢；當返回值為False時，外層查詢語句不進行查詢

#department表中存在dept_id=203，Ture
mysql> select * from employee
    ->     where exists
    ->         (select id from department where id=200);
+----+------------+--------+------+--------+
| id | name       | sex    | age  | dep_id |
+----+------------+--------+------+--------+
|  1 | egon       | male   |   18 |    200 |
|  2 | alex       | female |   48 |    201 |
|  3 | wupeiqi    | male   |   38 |    201 |
|  4 | yuanhao    | female |   28 |    202 |
|  5 | liwenzhou  | male   |   18 |    200 |
|  6 | jingliyang | female |   18 |    204 |
+----+------------+--------+------+--------+
#department表中存在dept_id=205，False
mysql> select * from employee
    ->     where exists
    ->         (select id from department where id=204);
Empty set (0.00 sec)

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