MySQL怎么完成多表查詢？MySQL多表查詢的語句

創建表

# 創建表
create table department(id int,name varchar(20));
create table employee1(
id int primary key auto_increment,
name varchar(20),
sex enum('male','female') not null default 'male',
age int,
dep_id int
);
# 插入數據
insert into department values(200,'技術'),(201,'人力資源'),(202,'銷售'),(203,'運營');

insert into employee1(name,sex,age,dep_id) values('egon','male',18,200),('alex','female',48,201),('tom','male',38,201),('yuanhao','female',28,202),('lidawei','male',18,200),('jinkezhou','female',18,204);

# 查看表
mysql> select * from employee1;
+----+-----------+--------+------+--------+
  id   name        sex      age    dep_id  
+----+-----------+--------+------+--------+
   1   egon        male       18      200  
   2   alex        female     48      201  
   3   tom         male       38      201  
   4   yuanhao     female     28      202  
   5   lidawei     male       18      200  
   6   jinkezhou   female     18      204  
+----+-----------+--------+------+--------+
6 rows in set (0.00 sec)
mysql> select * from department;
+------+--------------+
  id     name          
+------+--------------+
   200   技術        
   201   人力資源    
   202   銷售        
   203   運營        
+------+--------------+
4 rows in set (0.00 sec)

多表連接查詢

交叉連接

交叉連接：不適用任何匹配條件。生成笛卡爾積

mysql> select * from employee1 ,department;

內連接

內連接：找兩張表共有的部分，相當于利用條件從笛卡爾積結果中篩選出了正確的結果。（只連接匹配的行）

# 找兩張表共有的部分，相當于利用條件從笛卡爾積結果中篩選出了正確的結果
#department沒有204這個部門，因而employee表中關于204這條員工信息沒有匹配出來
mysql> select * from employee1,department where employee1.dep_id=department.id;

#上面用where表示的可以用下面的內連接表示，建議使用下面的那種方法
mysql> select * from employee1 inner join department on employee1.dep_id=department.id;

# 也可以這樣表示哈
mysql> select employee1.id,employee1.name,employee1.age,employee1.sex,department.name from employee1,department where employee1.dep_id=department.id;

左連接left

優先顯示左表全部記錄。

#左鏈接：在按照on的條件取到兩張表共同部分的基礎上，保留左表的記錄
mysql> select * from employee1 left join department on department.id=employee1.dep_id;

mysql> select * from department left join  employee1 on department.id=employee1.dep_id;

右連接right

優先顯示右表全部記錄。

#右鏈接：在按照on的條件取到兩張表共同部分的基礎上，保留右表的記錄
mysql> select * from employee1 right join department on department.id=employee1.dep_id;
mysql> select * from department right join employee1 on department.id=employee1.dep_id;

全部連接join

mysql> select * from department full join employee1;

符合條件多表查詢

示例1：以內連接的方式查詢employee和department表，并且employee表中的age字段值必須大于25,
即找出公司所有部門中年齡大于25歲的員工

mysql> select * from employee1 inner join department on employee1.dep_id=department.id and age>25;

示例2：以內連接的方式查詢employee和department表，并且以age字段的升序方式顯示

mysql> select * from employee1 inner join department on employee1.dep_id=department.id and age>25 and age>25 order by age asc;

子查詢

#1：子查詢是將一個查詢語句嵌套在另一個查詢語句中。
#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符：= 、 !=、> 、<等

示例：

# 查詢平均年齡在25歲以上的部門名
mysql> select name from department where id in ( select dep_id from employee1 group by dep_id having avg(age) > 25 );

# 查看技術部員工姓名
mysql> select name from employee1 where dep_id = (select id from department where name='技術');

# 查看小于2人的部門名
mysql> select name from department where id in (select dep_id from employee1 group by dep_id having count(id) < 2) union select name from department where id not in (select distinct dep_id from employee1);
# 提取空部門                              
#有人的部門
mysql> select * from department where id not in (select distinct dep_id from employee1);

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