連表查詢
上面兩張表經過笛卡爾積獲得一個全量拼接的大表；html
笛卡爾積：python
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1
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select
*
from
employee,department;
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內鏈接(inner join)
雙方可以互相匹配的項纔會被顯示出來；ide
select * from 表1 inner join 表2 on 條件
例如：
select * from employee inner join department
on employee.dep_id = department.id;
表的重命名：post
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1
2
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select t1.name,t2.name
from
employee as t1 inner join department as t2
on t1.dep_id
=
t2.
id
；
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外鏈接
左外鏈接
左外鏈接(left join) 只完整的顯示左表中的全部內容,以及右表中與左表匹配的項；url
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select
*
from
表
1
left join 表
2
on 條件
例如：
select
*
from
employee left join department
on employee.dep_id
=
department.
id
；
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右外鏈接
右外鏈接(right join) 只完整的顯示右表中的全部內容,以及左表中與右表匹配的項；spa
|
1
2
3
4
5
|
select
*
from
表
1
right join 表
2
on 條件
例如：
select
*
from
employee right join department
on employee.dep_id
=
department.
id
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全外鏈接
全外鏈接 永遠顯示左表和右表中全部的項，關鍵字(union)code
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例如：
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
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子查詢
#1：子查詢是將一個查詢語句嵌套在另外一個查詢語句中。 #2：內層查詢語句的查詢結果，能夠爲外層查詢語句提供查詢條件。 #3：子查詢中能夠包含：IN、NOT IN、ANY、ALL、EXISTS 和 NOT EXISTS等關鍵字 #4：還能夠包含比較運算符：= 、 !=、> 、<等
# 查"技術"部的全部員工的名字
# select name from employee where dep_id = (select id from department where name = '技術');
# 查"技術"和"銷售"部的全部員工的名字
# select name from employee where dep_id in (select id from department where name in ('技術','銷售'));
練習：
1.找出年齡大於25歲的員工以及員工所在的部門
select e.name,d.name from employee e inner join department d
on e.dep_id = d.id where e.age>25
2.之內鏈接的方式查詢employee和department表，而且以age字段的升序方式顯示
select * from employee e inner join department d
on e.dep_id = d.id order by e.age
1. 查詢平均年齡在25歲以上的部門名 select dep_id from employee group by dep_id having avg(age)>25 select name from department where id in (select dep_id from employee group by dep_id having avg(age)>25); 2.查看技術部員工姓名 select name from employee where dep_id = (select id from department where name = '技術'); 3.查看不足1人的部門名(子查詢獲得的是有人的部門id) 在員工表中不存在的一個部門id,在department表裏 在department表裏的id字段中找到一個在員工表的dep_id中不存在的項 select name from department where id not in (select dep_id from employee group by dep_id); 把員工表裏全部的人所在的dep_id都查出來 子查詢
1.查詢大於全部人平均年齡的員工名與年齡 select avg(age) from employee; select name,age from employee where age > (select avg(age) from employee); 2.查詢大於部門內平均年齡的員工名、年齡 select dep_id,avg(age) from employee group by dep_id; select * from employee inner join (select dep_id,avg(age) as avg_age from employee group by dep_id) as t2 on employee.dep_id = t2.dep_id where employee.age > t2.avg_age; 帶比較的子查詢