含義：又稱多表查詢，當查詢的字段來自多個表時，就會用到連線查詢
笛卡爾乘積現象：表1有m行，表2有n行，結果就有m*n行
發生原因：沒有有效的連線條件
如何避免：新增有效的連線條件
分類：
按功能分類
①內連線（包括等值連線、非等值連線和自連線）
②外連線（包括左外連線、右外連線和全外連線）
③交叉連線
select name,boyname from boys,beauty;select name,boyname from boys,beauty
where beauty.boyfriend_id=boys.id;
②n表連線，至少需要n-1個連線條件；
③多表的順序沒有要求；
④一般需要為表起別名；
⑤可以搭配前面介紹的所有子句使用，如：排序、篩選、分組。
案例1：查詢女神名和對應的男神名
select name,boyname from boys,beauty
where beauty.boyfriend_id=boys.id;
select last_name,department_name
from employees,departments
where employees.department_id=departments.
`department_id`
;
注意：如果為表起別名，則查詢的字段就不能使用原來的表名去限定
案例：查詢員工名、工種號、工種名
select last_name,e.job_id,job_title
from employees as e,jobs j #生成虛擬檢視
where e.
`job_id`
=j.`job_id`
;
select e.last_name,e.job_id,j.job_title
from jobs j,employees e
where e.
`job_id`
=j.`job_id`
;
select last_name,department_name
from employees e,departments d
where e.
`department_id`
=d.`department_id`
and e.
`commission_pct`
isnot
null
;
select department_name,city
from departments d,locations l
where d.
`location_id`
=l.`location_id`
and city like
'_o%'
;
select
count(*
) 個數,city
from departments d,locations l
where d.
`location_id`
=l.`location_id`
group
by city;
select department_name,d.manager_id,
min(salary)
from departments d,employees e
where d.
`department_id`
=e.`department_id`
and commission_pct is
notnull
group
by department_name,d.manager_id;
select job_title,
count(*
)from employees e,jobs j
where e.
`job_id`
=j.`job_id`
group
by job_title
order
bycount(*
)desc
;
select last_name,department_name,city
from employees e,departments d,locations l
where e.
`department_id`
=d.`department_id`
and d.
`location_id`
=l.`location_id`
;
select salary,grade_level
from employees e,job_grades g
where e.
`salary`
between g.
`lowest_sal`
and g.
`highest_sal`
;
由於原資料庫中沒有工資級別的表，所以要進行建立，create table語句的語法結構如下：
create
table
表(列名1
>
資料類該列所需約束(可選)
>
,列名2
>
資料類該列所需約..
..);
create
table job_grades
(grade_level varchar(3
),lowest_sal int
, highest_sal int);
insert
into job_grades
values
('a'
,1000
,2999);
insert
into job_grades
values
('b'
,3000
,5999);
insert
into job_grades
values
('c'
,6000
,9999);
insert
into job_grades
values
('d'
,10000
,14999);
insert
into job_grades
values
('e'
,15000
,24999);
insert
into job_grades
values
('f'
,25000
,40000
);
select
*from job_grades;
select e.employee_id 員工,e.last_name,m.employee_id 上級,m.last_name
from employees e,employees m#e為員工表，m為領導表
where e.manager_id=m.employee_id;
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