#建表 create table department( id int, name varchar(20) ); create table employee( 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 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) ; #查看錶結構和數據 mysql> desc department; +-------+-------------+------+-----+---------+-------+ | Field | Type | Null | Key | Default | Extra | +-------+-------------+------+-----+---------+-------+ | id | int(11) | YES | | NULL | | | name | varchar(20) | YES | | NULL | | +-------+-------------+------+-----+---------+-------+ mysql> desc employee; +--------+-----------------------+------+-----+---------+----------------+ | Field | Type | Null | Key | Default | Extra | +--------+-----------------------+------+-----+---------+----------------+ | id | int(11) | NO | PRI | NULL | auto_increment | | name | varchar(20) | YES | | NULL | | | sex | enum('male','female') | NO | | male | | | age | int(11) | YES | | NULL | | | dep_id | int(11) | YES | | NULL | | +--------+-----------------------+------+-----+---------+----------------+ mysql> select * from department; +------+--------------+ | id | name | +------+--------------+ | 200 | 技術 | | 201 | 人力資源 | | 202 | 銷售 | | 203 | 運營 | +------+--------------+ mysql> select * from employee; +----+------------+--------+------+--------+ | 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 | +----+------------+--------+------+--------+ 表department與employee sql示例
SELECT 字段列表 FROM 表1 INNER|LEFT|RIGHT JOIN 表2 ON 表1.字段 = 表2.字段;mysql
mysql> select * from employee,department; +----+------------+--------+------+--------+------+--------------+ | id | name | sex | age | dep_id | id | name | +----+------------+--------+------+--------+------+--------------+ | 1 | egon | male | 18 | 200 | 200 | 技術 | | 1 | egon | male | 18 | 200 | 201 | 人力資源 | | 1 | egon | male | 18 | 200 | 202 | 銷售 | | 1 | egon | male | 18 | 200 | 203 | 運營 | | 2 | alex | female | 48 | 201 | 200 | 技術 | | 2 | alex | female | 48 | 201 | 201 | 人力資源 | | 2 | alex | female | 48 | 201 | 202 | 銷售 | | 2 | alex | female | 48 | 201 | 203 | 運營 | | 3 | wupeiqi | male | 38 | 201 | 200 | 技術 | | 3 | wupeiqi | male | 38 | 201 | 201 | 人力資源 | | 3 | wupeiqi | male | 38 | 201 | 202 | 銷售 | | 3 | wupeiqi | male | 38 | 201 | 203 | 運營 | | 4 | yuanhao | female | 28 | 202 | 200 | 技術 | | 4 | yuanhao | female | 28 | 202 | 201 | 人力資源 | | 4 | yuanhao | female | 28 | 202 | 202 | 銷售 | | 4 | yuanhao | female | 28 | 202 | 203 | 運營 | | 5 | liwenzhou | male | 18 | 200 | 200 | 技術 | | 5 | liwenzhou | male | 18 | 200 | 201 | 人力資源 | | 5 | liwenzhou | male | 18 | 200 | 202 | 銷售 | | 5 | liwenzhou | male | 18 | 200 | 203 | 運營 | | 6 | jingliyang | female | 18 | 204 | 200 | 技術 | | 6 | jingliyang | female | 18 | 204 | 201 | 人力資源 | | 6 | jingliyang | female | 18 | 204 | 202 | 銷售 | | 6 | jingliyang | female | 18 | 204 | 203 | 運營 | +----+------------+--------+------+--------+------+--------------+
#找兩張表共有的部分,至關於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,於是employee表中關於204這條員工信息沒有匹配出來 mysql> select employee.id,employee.name,employee.age,employee.sex,department.name from employee inner join department on employee.dep_id=department.id; +----+-----------+------+--------+--------------+ | id | name | age | sex | name | +----+-----------+------+--------+--------------+ | 1 | egon | 18 | male | 技術 | | 2 | alex | 48 | female | 人力資源 | | 3 | wupeiqi | 38 | male | 人力資源 | | 4 | yuanhao | 28 | female | 銷售 | | 5 | liwenzhou | 18 | male | 技術 | +----+-----------+------+--------+--------------+ #上述sql等同於 mysql> select employee.id,employee.name,employee.age,employee.sex,department.name from employee,department where employee.dep_id=department.id;
#以左表爲準,即找出全部員工信息,固然包括沒有部門的員工 #本質就是:在內鏈接的基礎上增長左邊有右邊沒有的結果 mysql> select employee.id,employee.name,department.name as depart_name from employee left join department on employee.dep_id=department.id; +----+------------+--------------+ | id | name | depart_name | +----+------------+--------------+ | 1 | egon | 技術 | | 5 | liwenzhou | 技術 | | 2 | alex | 人力資源 | | 3 | wupeiqi | 人力資源 | | 4 | yuanhao | 銷售 | | 6 | jingliyang | NULL | +----+------------+--------------+
#以右表爲準,即找出全部部門信息,包括沒有員工的部門 #本質就是:在內鏈接的基礎上增長右邊有左邊沒有的結果 mysql> select employee.id,employee.name,department.name as depart_name from employee right join department on employee.dep_id=department.id; +------+-----------+--------------+ | id | name | depart_name | +------+-----------+--------------+ | 1 | egon | 技術 | | 2 | alex | 人力資源 | | 3 | wupeiqi | 人力資源 | | 4 | yuanhao | 銷售 | | 5 | liwenzhou | 技術 | | NULL | NULL | 運營 | +------+-----------+--------------+
全外鏈接:在內鏈接的基礎上增長左邊有右邊沒有的和右邊有左邊沒有的結果 #注意:mysql不支持全外鏈接 full JOIN #強調:mysql能夠使用此種方式間接實現全外鏈接 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 ; #查看結果 +------+------------+--------+------+--------+------+--------------+ | id | name | sex | age | dep_id | id | name | +------+------------+--------+------+--------+------+--------------+ | 1 | egon | male | 18 | 200 | 200 | 技術 | | 5 | liwenzhou | male | 18 | 200 | 200 | 技術 | | 2 | alex | female | 48 | 201 | 201 | 人力資源 | | 3 | wupeiqi | male | 38 | 201 | 201 | 人力資源 | | 4 | yuanhao | female | 28 | 202 | 202 | 銷售 | | 6 | jingliyang | female | 18 | 204 | NULL | NULL | | NULL | NULL | NULL | NULL | NULL | 203 | 運營 | +------+------------+--------+------+--------+------+--------------+ #注意 union與union all的區別:union會去掉相同的紀錄
#示例1:之內鏈接的方式查詢employee和department表,而且employee表中的age字段值必須大於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: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 employee.id,employee.name,employee.age,department.name from employee,department where employee.dep_id = department.id and age > 25 order by age asc;
#1:子查詢是將一個查詢語句嵌套在另外一個查詢語句中。sql
#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。spa
#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code
#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); #查看技術部員工姓名 select name from employee where dep_id in (select id from department where name='技術'); #查看不足1人的部門名(子查詢獲得的是有人的部門id) select name from department where id not in (select distinct dep_id from employee);
#比較運算符:=、!=、>、>=、<、<=、<> #查詢大於全部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 | +---------+------+ 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;
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;當返回值爲False時,外層查詢語句不進行查詢blog
#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)