1、建 表
建表语句:
create table 表名(字 段名1 字符类型1(字符长度),字段名2 字符类型2(字符长度),字段名3 字符类型3(字符长度))default charset=utf8 ;
default charset=utf8 ; 默认编码格式,当插入中文数值显示???解决
案例:create table hz10(id int(10),name char(20),age int(10))default charset=utf8 ;
desc 表名 查看结构 ;
建表约束:
表中的限制条件
作用:表在的时候加入约束的目的是为了保证表中的记录完整和有效。
(1)非空约束 (not null) 非空,用于保证字段的值不能为空,
(2)唯一约束 (unique) 保证字段值具有唯一性(不能重复),可以为空,唯一约束可以有多个
(3)主键约束(primary key) 保证字段值具有唯一性(不能重复),不能为空 ,一个表中只有一个主键
(4)外键约束 (forrign key) 两个表之间的主键建立关系
(5)默认 :(default) 当字段不填数据,默认数据
(6)自增长 :auto _increment 当添加一条数据,自动增长+1
(6.1) 与主键约束一起使用,一般针对id
(6.2)每插入一条数据,指定的字段值+1
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drop table 表名 删除表 ;
例如:DROP table aa ;
表字段:
添加字段:add
格式:alter table 表名 add 字段名 字符类型(字符长度);
alter table hz10 add fs int(10);
修改字段:change
格式:alter table 表名 change 原字段名 新字段名 字符类型(字符长度);
alter table hz10 change fs tz int(10);
删除字段:drop
格式:alter table 表名 drop 字段名 ;
alter table hz10 drop tz ;
重命名:rename
格式:alter table 原表名 rename 新表名;
alter table hz10 rename hzdcs10 ;
调换字段顺序:modify after 在......后面
格式:alter table 表名 MODIFY 移动的字段名 字段类型(字符长度) after 字段名 ;
alter table hzdcs10 MODIFY age int(10) after id ;
将字段添加到首行:
格式:alter table 表名 add 字段名 字符类型(字符长度) first ;
alter table hzdcs10 add no int(10) first ;
将字段添加到指定字段的后面:
格式:alter table 表名 add 新字段名 字符类型(字符长度 ) AFTER 原表名 ;
alter table hzdcs10 add tz int(10) AFTER age ;
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表中插入数据:
(1)表中插入数据:insert into 表名 values (插入值1,插入值2)
注意:插入的值,只有数值不需要加引号,字符要加上引号
如:INSERT into hz10 VALUES (1,"文琪",18);
(2)表中插入数据:insert into 表名 (id) values (插入值1)
INSERT into hz10(id) VALUES (2);
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2、单 表
查
(1)查询一个表中的所有数据
格式:select * from 表名;(——显示表格结果)
eg:select * from hz13;
注意:*号表示所有,每个语句后面加上分号
(2)查询表中表中具体的字段
格式:select 字段名1,字段名2,字段名3 FROM 表名;
eg:select name,age FROM hz13;
select name,age FROM hz13 ;
(3)查询表中的具体字段取别名(别名是为了在查看数据是方便)
格式:select 字段1 as “姓名” 字段2 “年龄” from 表,名;
select name as “姓名” age "年龄" from hz13;
备注:as 可以省略不写,as表示取别名
(4)查询指定内容,用where+条件
格式:select * from 表名 where 字段名=字段 ;
select * from emp where dept2=101 ;
where条件使用注意:
比较运算符
(1)where +条件(=(等于), !=(不等于) ,<>(不等于) ,<(小于),>(大于),>=(大于等于),<=(小于等于))
select * from emp where dept2=101 ;
select * from emp where dept2 !=101
select * from emp where dept2 <>101 ;
select * from emp where dept2 >102 ;
select * from emp where dept2 <102 ;
select * from emp where dept2 <=102 ;
select * from emp where dept2 >=102 ;
(2)and (与)
and 同时满足所有条件,比如:条件1和条件2都要满足
案例:select * from emp where dept2 >=102 and age=57 ;
(3)or(或)
or 当有多个条件时满足其中任意1个条件都显示,,条件1和条件同时存在,只要满足条件1就显示,只要满足条件2就显示。
select * from emp where dept2 =102 or age=55 ;
(4)between..... and.... 在什么范围之间
格式:select * from 表名 where 表字段 BETWEEN 值的范围 and 值的范围
select * from emp where age BETWEEN 24 and 55 ;
(5)in 在一组数据中匹配数据
格式:select * from 表名 where 字段名 in (数组值1,数组值2,数组值3);
select * from emp where dept2 in (101,102,110);
(6)is null 为空 或is not null 非空
6.1 is not null 非空
格式:select * from 表名 where 字段名 is not null ;
select * from emp where age is not null ;
6.2 is null 为空
格式:
select * from 表名 where 字段名 is null ;
多行注释:ctrl+/
取消多行注释:ctrl+shift+/
单行注释:直接#号
排序:
对表中的数据进行排序:order by asc 升序(可以省略) desc 降序
升序:
格式:select * from 表名 order by 字段名 asc ;
案例1:select * from emp order by age asc ;
案例2:select * from emp order by age asc ;(不填asc)
降序:
格式:select * from 表名 order by 字段名 desc ;
案例:select * from emp order by age desc ;
二次排序:
格式:select * from 表名 order by 字段1 desc , 字段2 asc ;
案例:select * from emp order by dept2 desc , incoming asc ;
模糊匹配查询 like
%:表会匹配0个字符或多个字符
_:表示一个字符
select * from emp where dept2 like "1%" #匹配1开头的部门编号数据
select * from emp where dept2 like "%2%" #匹配的部门包含2编号数据
select * from emp where dept2 like "%2" #匹配结尾包含2编号数据
select * from emp where dept2 like "__2" #匹配固定的字符2编号数据
限制查询(limit) 后面接连个值 ,第一个值 表示下标(索引), 第二值是步长
备注:一个表中索引是从0开始
格式:select * from 表名 limit 索引,行数 ;
案例1:select * from emp limit 2,3 ;
案例2:select * from emp limit 2 ; 直接查询两行
分组查询 group by
group by 一般和having 组合
案例:SELECT dept2,max(age) from emp group by dept2 ;
SELECT dept2,max(age) from emp group by dept2 ;
SELECT dept2,max(age) from emp group by dept2 ;
group by 一般不会单独使用个,通常都是和函数组合使用。
group by 后查询出来的结果,在需要的条件下可以接having
案例:SELECT dept2 ,max(age) as s from emp group by dept2 HAVING s>50 ;
having +条件 和where +条件 用法一样,但是场景不一样,一般group by 的后面接having
备注:函数求出以后,需要设置成别名,设置别名的字段,进行语句的操作。
错误案例:
(1)SELECT name,max(age) from emp
(2)SELECT name,max(age) from emp GROUP BY dept2
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sql语句函数:
max 最大值
案例:SELECT dept2,max(age) from emp
min 最小值
案例:SELECT min(age) from emp
avg 平均值
案例:SELECT avg(age) from emp
count 统计总数
案例:SELECT count(age) from emp
sum 求和
案例:SELECT sum(age) from emp
distinct 去重
案例:SELECT distinct(dept2) from emp
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改:
update ......set 改
格式:update 表名 set 修改的字段名=字段新值 where 条件 ;
案例:update emp set name="成全" where sid=1789 ;
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删:
delete 删
truncate 快速删除表内数据
drop 删除
删除数据速度 :drop > truncate >delete
NOTE:
1、drop 是删除表和数据
2、truncate 删除无法恢复
3、delete 删除是可以恢复
delete:
(1)删除表中所有数据
格式:delete from 表名:
案例:delete from hz13;
(2)删除表中指定条件的数据
格式: delete from 表名 where 条件
案例: delete from emp where dept2=103;
(3)快速删除表数据
格式:truncate 表名;
案例:truncate hz15 ;
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表取别名:
格式: 表名设置别名,
案例:SELECT s.dept2,s.name from emp as s ; (把emp表设置成临时表s)
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备份表结构:
格式:create table 新表名 like 备份原表名 ;
create table hz110 like emp ;
备份数据:
insert into 新表 有表结构 select * from 备份表的原表
案例:INSERT into hz110 select * from emp ;
备份部分数据:
格式:INSERT into 表名(s字段名1,字段名2) select sid ,dept2 from emp ;
INSERT into hz110110(sid,dept2) select sid ,dept2 from emp ;
备份表结构和数据:
格式:create table 表名 as(select * from 备份原表 )
案例:create table hzdcs110 as(select * from emp )
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linux备份数据库:
备份用:>
格式:mysqldump -u root -p 原数据库>数据库脚本.后缀名
案例:mysqldump -u root -p hz10>/home/hz10bf.sql
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还原数据库:
1、先建立一个新的空库
2、在linux中还原<
3、还原:mysql -u root -p 新建的空库<数据库备份好的数据脚本
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3、多表之两表
def——什么是多表关联查询?
定义:查询数据来源于多张表。
主要讲:
1、内连接(基本内连接和隐藏内连接)
2、左连接(左外连接)
3、右连链(右外连接)
4、全链接(全外连接)
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1、讲解过程先建两个表 :dept表 和 emp 表
dept表:
dept1部门编号
dept_name 部门
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emp表:
sid 员工编号
name 员工姓名
age 员工年龄
worktime_start 入职时间
incoming 员工收入
dept2 部门编号
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关联关系(即有关联的字段):dept1 =dept2
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1、笛卡尔积查询(了解)
根据两张表所乘得到的结果:比如:左边有4条数据,右边有6条数据,查询出来就是4*6=24条数据,这种查询数据包含大量的错误结果,通常不会使用这种查询。
格式:select * from 表1,表2;
案例:select * from dept,emp;
2、内连接(普通内连接,隐藏内连接):查询两个表共有的关联数据。
格式:select * from 表1 inner join 表2 on 表1.关联字段1=表2.关联字段2
案例1:select * from dept inner join emp on dept.dept1=emp.dept2 ;
案例2:select name,dept_name from dept inner join emp on dept.dept1=emp.dept2 where sid=1789 ;
格式:select * from 表1,表2 where 表1.关联字段1=表2.关联字段2;
案例:select * from dept,emp where dept.dept1=emp.dept2 ;
3、左连接:以左表为主(显示完整的左表),右表关联的数据就显示,没有的数据就以null显示
格式:select * from 表1 left join 表2 on 表1.关联字段1=表2.关联字段2 ;
案例:select * from dept left join emp on dept.dept1=emp.dept2 ;
4、右连接:右表是以右表数据为主(显示整个右表)左表有关联的数据就显示,没有就以null值显示。
格式:select * from 表1 right join 表2 on 表1.关联字段1=表2.关联字段2 ;
案例:select* from dept right join emp on dept.dept1=emp.dept2 ;
5、左表独有数据:左表中独有的数据显示,(方法,左独有,以右表字段为空查询)
格式:select * from 表1 left join 表2 on 表1.关联字段1=表2.关联字段2 where 右表字段 is null
案例:select * from dept left join emp on dept.dept1=emp.dept2 where name is null;
6、右表独有数据:右表中独有的数据显示,(方法,右独有,以左表字段为空查询)
格式:select * from 表1 right join 表2 on 表1.关联字段1=表2.关联字段2 where 左表字段 is null
案例:select * from dept right join emp on dept.dept1=emp.dept2 where dept1 is null;
7、左表独有数据+右表独有数据:
union 拼接
左独有 union 右独有
select * from dept left join emp on dept.dept1=emp.dept2 where name is null
union
select * from dept right join emp on dept.dept1=emp.dept2 where dept1 is null;
8、全连接
方法一:左独有+右独有+内连接
select * from dept left join emp on dept.dept1=emp.dept2 where name is null
union
select * from dept right join emp on dept.dept1=emp.dept2 where dept1 is null
union
select * from dept inner join emp on dept.dept1=emp.dept2 ;
方法二:左连接+右独有
select * from dept left join emp on dept.dept1=emp.dept2 UNION
select * from dept right join emp on dept.dept1=emp.dept2 where dept1 is null ;
方法三:右连接+左独有
select * from dept right join emp on dept.dept1=emp.dept2 UNION
select * from dept left join emp on dept.dept1=emp.dept2 where name is null ;
总结:
普通内连接:
select * from 表1 inner join 表2 on 表1 的关联字段=表2的关联字段 ;
隐藏内连接:
select * from 表1 ,表2 where 表1 的关联字段=表2的关联字段 ;
左连接:
select * from 表1 left join 表2 on 表1 的关联字段=表2的关联字段 ;
右连接:
select * from 表1 right join 表2 on 表1 的关联字段=表2的关联字段 ;
左独有数据:
select * from 表1 left join 表2 on 表1 的关联字段=表2的关联字段 where 右表字段为空;
右独有数据:
select * from 表1 right join 表2 on 表1 的关联字段=表2的关联字段 where 左表字段为空;
全外连接:
1、左独有+右独有+内连接
2、左连接+右独有
3、右连接+左独有
mysql 之子查询
定义:一个查询嵌套另一个查询
请问:求出IT技术部门的员工姓名和员工编号?
答:IT部门 dept 中的dept_name
name sid emp表
内连接(先合表在根据条件查询):select name ,sid from dept inner join emp on dept1=dept2 where dept_name="IT技术";
子查询案例:select name,sid from emp where dept2=(select dept1 from dept where dept_name="IT技术" )
子查询分类:
1、标量子查询
2、列子查询
3、行子查询
4、表子查询
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1、标量子查询(返回一个值):
标量子查询 把一个sql执行返回的一个值,作为另一sql执行的一个条件,得到的结果是一行一列,一般出现在where后面
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注意点:标量子查询允许使用符号:=、!=、>、<、>=、<=、<>(不等于)
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2、列子查询(返回一个列)
定义:返回的是一列值
注意:通常在where 条件后使用,in或not in ,不能使用=,<,>等比较运算符。
案例: select dept1 from dept where dept_name="财务" or dept_name="IT技术" ;
select name,sid from emp where dept2 in (select dept1 from dept where dept_name="财务" or dept_name="IT技术")
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3、行子查询:(返回一行多列)
行子查询返回的结果集是一行多列,一般出现在where后面
案例:select age,incoming from emp where name="马十二" ;
select * from emp where (age,incoming) =any (select age,incoming from emp where name="马十二" )
或
select * from emp where (age,incoming) in (select age,incoming from emp where name="马十二" )
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4、表子查询(重点)(返回一个表)
一般都是接from后面
select * from dept,emp where dept1=dept2 ;
select name,sid from ( select * from dept,emp where dept1=dept2)s where age=64;
临时表:给表取一个别名 聚合函数 最好取别名
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案例:求财务部门的收入总和;
解法一:
子查询方法:select sum(incoming) from emp where dept2=(select dept1 from dept where dept_name="财务") ;
解法二:
普通内连接:SELECT SUM(incoming) FROM dept inner join emp on dept.dept1=emp.dept2 WHERE dept_name="财务";
解法三:
表子查询:select sum(s.incoming) from (select *from emp ,dept where dept1=dept2)s where dept_name="财务" ;
练习题:
create table dept (
dept1 int(10),
dept_name char(20)
) default charset=utf8;
insert into dept values (101,'财务');
insert into dept values (102,'销售');
insert into dept values (103,'IT技术');
insert into dept values (104,'行政');
SELECT * from dept;
create table emp(
sid int primary key, -- 员工编号
name varchar(10), -- 员工姓名
age int, -- 员工年龄
worktime_start date, -- 入职时间
incoming double, -- 收入
dept2 int, -- 部门号
foreign key (dept2) references dept(dept2)) default charset=utf8; -- 关联dept表
-- alter table emp add foreign key (deptno) references dept(deptno);
insert into emp values(1789,'张三',35,"1980-01-01",4000,101);
insert into emp values(1674,'李四',32,'1983-04-01',3500,101);
insert into emp values(1776,'王五',24,'1990-07-01',2000,101);
insert into emp values(1568,'赵六',57,'1970-10-11',7500,102);
insert into emp values(1564,'荣七',64,'1963-10-11',8500,102);
insert into emp values(1879,'牛八',55,'1971-10-20',7300,103);
select * from emp;
Q1.列出每个部门的平均收入及部门名称;
核心词汇:emp 、 incoming 、 dept dept_name 、avg 、group by、
select avg(incoming ),dept_name from dept inner join emp on dept.dept1=emp.dept2.
方法一:select avg(incoming ),dept_name from (select * from dept left join emp on dept.dept1=emp.dept2)c group by dept_name;
方法二:select avg(incoming),dept_name from dept left JOIN emp on dept.dept1=emp.dept2 GROUP BY dept_name ;
Q2.财务部门的收入总和;
dept "财务"
emp incoming
sum
select sum(incoming ),dept_name from dept inner join emp on dept.dept1=emp.dept2 where dept_name="财务" ;
方法一:select sum(incoming) from (select * from dept,emp where dept.dept1=emp.dept2)c where dept_name='财务';
方法二:select sum(incoming ) from dept inner join emp on dept.dept1=emp.dept2 where dept_name="财务" ;
方法三:select sum(incoming) from emp where dept2=(select dept1 from dept where dept_name="财务");
Q3.It技术部入职员工的员工号
emp sid
dept it技术部门
name ,sid
方法一:select emp.sid from dept inner join emp on dept.dept1=emp.dept2 where dept_name="IT技术 ";
方法二:select sid from (select * from dept,emp where dept.dept1=emp.dept2)c where dept_name='IT技术';
方法三:select sid from emp where dept2=(select dept1 from dept where dept_name="IT技术") ;
方法四:select sid from dept,emp where dept1=dept2 and dept_name="IT技术" ;
Q4.财务部门收入超过2000元的员工姓名
emp incoming>2000
dept 财务
方法一:SELECT name,incoming FROM Dept INNER JOIN Emp on Dept1=DEPT2 WHERE incoming>2000 AND Dept_name="财务";
方法三:select name from (select * from dept,emp where dept.dept1=emp.dept2)c where incoming>2000 and dept_name= '财务';
方法四:select name from (select*from dept inner join emp on dept1=dept2 where incoming>2000)s where dept_name='财务'
Q5.找出销售部收入最低的员工的入职时间;
emp
dept
max
方法一:select name,woektime_start from (select *,min(incoming) from dept inner join emp on dept1=dept2 group by dept_name)s where dept_name='销售'
方法二:select woektime_start FROM emp INNER JOIN dept on dept.dept1=emp.dept2 where incoming =(select min(incoming) from emp INNER JOIN dept on emp.dept2=dept.dept1 where dept.dept_name="销售") and dept_name="销售";
方法三:(这种方法可以,弊端在与有重复的数据,只显示一条)
select a.dept_name,b.sid,b.name,b.woektime_start from dept a join emp b on a.dept1 =b.dept2 where a.dept_name = "销售" ORDER BY b.incoming asc LIMIT 1;
Q6.找出年龄小于平均年龄的员工的姓名,ID和部门名称
思路:条件 n:ame ,sid ,dept_name
方法一:SELECT name,sid,dept_name FROM emp INNER JOIN dept ON dept.dept1 = emp.dept2 where age< (SELECT AVG(age) FROM emp);
方法二:
select sid,name,dept_name from (select * from emp,dept where dept1=dept2)s where age<(select avg(age)from emp);
Q7.列出每个部门收入总和高于9000的部门名称
条件: group by dept_name
sum (incoming) >9000
结果:dept_name
方法一:
select s.dept_name from (select a.dept_name,sum(b.incoming) as k from dept a join emp b on a.dept1 = b.dept2 GROUP BY a.dept_name HAVING k >9000)s;
方法二:
SELECT dept_name FROM dept INNER JOIN emp on dept.dept1=emp.dept2 GROUP BY dept_name having sum(incoming)>90000;
方法三:
SELECT dept_name FROM ( SELECT sum( incoming ) a, dept_name FROM emp INNER JOIN dept ON dept.dept1 = emp.dept2 GROUP BY dept_name ) b
WHERE a > 9000;
方法四:SELECT dept_name from (SELECT sum(incoming) as s,dept2 from emp GROUP BY dept2 having s>9000)t1 on dept.dept1=t1.dept2;
Q8.查出财务部门工资少于3800元的员工姓名
dept_name =财务 dept
incoming<3800 emp
name
方法一:
SELECT name FROM dept INNER JOIN emp on dept.dept1=emp.dept2 where incoming<3800 and dept_name="财务";
方法二:
select name from emp where incoming<3800 and dept2=(SELECT dept1 from dept WHERE dept_name='财务');
方法三
select name from (select*from dept,emp where dept1=dept2)s where incoming<3800 and dept_name="财务"
Q9.求财务部门最低工资的员工姓名;
条件
min
dept表 dept_name=财务
emp 表 min( incoming )
结果:员工姓名 name
方法一:
select name from dept inner join emp on dept.dept1=emp.dept2 where
dept2=(select dept1 from dept where dept_name="财务" ) and incoming=(select min(incoming) from emp ,dept where emp.dept2=dept.dept1 and dept_name="财务");
方法二:
select name from emp ,dept where dept1=dept2 and dept_name="财务" ORDER BY incoming asc LIMIT 1
Q10.找出销售部门中年纪最大的员工的姓名
条件 :dept 表 dept_name="销售"
emp max(age)
结果:name
方法一:select name "姓名", age "年龄" from dept,emp where dept.dept1=emp.dept2 and
age=(select max(age) from dept,emp where dept.dept1=emp.dept2 and dept_name="销售")and dept_name="销售";
方法二:(缺陷是多个同样数据就显示一个)
SELECT name,age from emp INNER JOIN dept on emp.dept2=dept.dept1 WHERE dept_name="销售" ORDER BY age desc LIMIT 0,1;
Q11.求收入最低的员工姓名及所属部门名称:
条件:emp min(incoming )
结果:name,dept_name
方法一:
SELECT NAME,dept_name FROM dept inner join emp on dept.dept1=emp.dept2 WHERE incoming=(SELECT MIN(incoming) FROM emp );
方法二:
SELECT name ,dept_name FROM(SELECT * FROM emp INNER JOIN dept ON dept.dept1 = emp.dept2)a WHERE incoming =( SELECT MIN(incoming) FROM emp);
方法三:SELECT name,dept_name from emp left join dept on emp.dept2=dept.dept1 where incoming=(SELECT min(incoming) from emp left join dept on emp.dept2=dept.dept1);
Q12.求李四的收入及部门名称
条件:emp name=“李四”
结果:incoming dept_name
方法一:
SELECT incoming,dept_name FROM dept INNER JOIN emp on dept.dept1=emp.dept2 where name="李四"
方法二:
select name,dept_name from (select * from dept,emp where dept.dept1=emp.dept2)c where name='李四';
方法三:select incoming,dept_name from (select*from dept,emp where dept1=dept2)s where name="李四";
Q13.求员工收入小于4000元的员工部门编号及其部门名称
条件:
emp incoming<4000
结果:dept1 ,deptname
方法一:select dept1,dept_name from (select * from dept,emp where dept.dept1=emp.dept2)c where incoming<4000;
方法二:select dept1,dept_name from emp left join dept on emp.dept2=dept.dept1 where incoming<4000;
方法三:select dept1,dept_name from dept where dept1 in (select dept2 from emp where incoming<4000) ;
Q14.列出每个部门中收入最高的员工姓名,部门名称,收入,并按照收入降序;
方法一:(缺陷:重复就显示一个)
SELECT name "姓名",incoming "收入",dept_name"部门" from (select * from dept,emp where dept.dept1=emp.dept2 order by incoming desc ) b group by dept_name ORDER BY incoming desc ;
方法二:
SELECT name "姓名",incoming "收入",dept_name"部门" from (select * from dept,emp where dept.dept1=emp.dept2)a where
方法三:
select b.name,a.dept_name, b.incoming from (select max(incoming) as c,dept_name from dept, emp where dept1=dept2 group by dept_name)a LEFT JOIN ( select * from emp ,dept where dept1=dept2)b on a.dept_name=b.dept_name and a.c =b.incoming order by b.incoming desc;
Q15.求出财务部门收益最高的俩位员工的姓名,工号,收益
条件: dept dept_name="财务" max(incoming) limit 2
结果 : emp name sid incoming
方法一:select name,sid,incoming from emp inner join dept on dept1=dept2 where dept_name='财务' order by incoming desc limit 2;
方法二:select * from (select name,sid,incoming from dept left join emp on dept.dept1=emp.dept2 where dept_name="财务" order by incoming desc)as s where sid limit 0,2
方法三: select a.dept1,a.dept_name,b.name,b.incoming from dept a , emp b where a.dept1 = b.dept2 and a.dept_name ='财务' ORDER BY b.incoming desc LIMIT 2;
Q16.查询财务部低于平均收入的员工号与员工姓名:(按所有的平均薪资)
条件: dept dept_name=“财务” avg(incoming) >
结果:sid name
方法一:
select sid,name from emp where incoming<(select avg(incoming)from emp) and dept2=(select dept1 from dept where dept_name='财务');
方法二:
SELECT sid,name FROM dept INNER JOIN emp on dept.dept1=emp.dept2 WHERE incoming<(SELECT avg(incoming) FROM emp) and dept_name="财务";
备注(在多表中求得平均值)SELECT sid,name from emp left join dept on emp.dept2=dept.dept1 where dept_name="财务" and incoming<(SELECT avg(incoming) from emp left join dept on emp.dept2=dept.dept1)
方法三:
select sid,name from (select * from dept,emp where dept.dept1=emp.dept2)c where incoming<(select avg(incoming) from emp) and dept_name='财务';
Q17.列出部门员工数大于1个的部门名称;
条件:dept count(name)
结果:dept_name
方法一:select dept_name from dept where dept1 in(select dept2 from emp group by dept2 having count(dept2)>1)
方法二:SELECT DISTINCT(dept_name) FROM emp INNER JOIN dept ON emp.dept2 = dept.dept1 WHERE (SELECT count(name) from emp)>1;
方法三:SELECT dept_name from (SELECT dept_name ,count(name)s from dept INNER JOIN emp on dept1=dept2 GROUP BY dept_name having s>1)d;
Q18.列出部门员工收入不超过7500,且大于3000的员工年纪及部门编号;
条件:emp incoming<=750 and incoming>3000
结果:age ,dept_name
方法一:select age, dept2 from emp where incoming >3000 and incoming<=7500;
方法二:
select age,dept2 from (select * from dept,emp where dept.dept1=emp.dept2)c where 7500>=incoming and incoming>3000;
方法三:SELECT name,age,dept1 FROM dept inner JOIN emp on dept.dept1=emp.dept2 WHERE incoming BETWEEN 3000 and 7500 and incoming!=3000;
错误写法:
SELECT name,age,dept1 FROM dept inner JOIN emp on dept.dept1=emp.dept2 WHERE incoming BETWEEN 3000 and 7500; (包含3000了)
Q19.求入职于20世纪70年代的员工所属部门名称;
条件:20世纪 表示1900-1999内 , 70年代 :大于等于1970~小于1980 woerktime =“197%”
结果: dept dept_name
方法一:select dept_name from emp left join dept on emp.dept2=dept.dept1 where woektime_start >="19700101" and woektime_start<"19800101";
方法二:SELECT name,dept_name FROM dept RIGHT JOIN emp on dept.dept1=emp.dept2 WHERE worktime BETWEEN "1970-01-01" and "1979-12-31";
方法三:select dept_name "部门" from dept,emp where dept.dept1=emp.dept2 and woektime_start like "197%" ;
方法四:select dept_name from emp inner join dept on dept1=dept2 where woektime_start >="19700101" and woektime_start<="19791231";
Q20.查找张三所在的部门名称;
条件: emp name=“张三”
结果:dept dept_name
方法一:
select dept_name from emp inner join dept on dept1=dept2 where name='张三';
方法二:
select dept_name from dept where dept1=(select dept2 from emp where name='张三');
方法三:
select dept_name "部门" from dept,emp where dept2=dept1 and name="张三";
Q21.列出每一个部门中年纪最大的员工姓名,部门名称;
条件: group by dept_name max(age)
结果:name ,dept_name
方法一:sel ect name "姓名",dept_name "部门名" from dept,emp where dept1=dept2 and
(dept_name,age) in (select dept_name,max(age) from dept,emp where dept1=dept2 group by dept_name );
方法二:
select b.name,a.dept_name, b.incoming from (select max(age) as c,dept_name from dept, emp where dept1=dept2 group by dept_name)a LEFT JOIN ( select * from emp ,dept where dept1=dept2)b on a.dept_name=b.dept_name and a.c =b.age
方法三:(是一种方法,当出现多条相同数据,就显示一条)
select name,dept_name from (select * from emp inner join dept on dept.dept1=emp.dept2 order by age desc) a group by dept_name ;
Q22.列出每一个部门的员工总收入及部门名称;
条件: group by dept_name
结果; sum(incoming) dept_name
方法一:SELECT sum(incoming),dept_name FROM dept left JOIN emp on dept.dept1=emp.dept2 GROUP BY dept_name
方法二:select sum(incoming),dept_name from dept left join emp on dept.dept1=emp.dept2 group by dept1;
方法三:(不合理的写法,答案正确)
SELECT sum(incoming),dept_name FROM dept iNNER JOIN emp on dept.dept1=emp.dept2 WHERE (SELECT sum(incoming) FROM emp ) GROUP by dept_name;
方法四:select dept_name,sum(incoming) from (select * from dept,emp where dept.dept1=emp.dept2)c group by dept_name;
Q23.列出部门员工收入大于7000的员工号,部门名称;
条件: dept dept_name
emp incoming>7000 sid
结果: sid dept_name
方法一:SELECT sid,dept_name FROM dept left JOIN emp on dept1=dept2 where incoming>7000
方法二:select a.sid,a.dept_name from(select * from emp inner join dept on emp.dept2=dept.dept1 where incoming>7000) a;
方法三:select sid,dept_name from (select * from dept,emp where dept.dept1=emp.dept2)c where incoming>7000;
Q24.找出哪个部门还没有员工入职;
条件: dept_name name或sid 判断
结果:dept_name
方法一:
select dept_name from dept left join emp on dept.dept1=emp.dept2 group by dept1 having (count(sid)=0)
方法二:select * from dept where dept1 not in ( select dept2 from emp);
方法三:select dept_name from emp right join dept on dept1=dept2 where name is null;
方法四:select dept_name from dept left join emp on dept.dept1=emp.dept2 where name is NULL;
方法五:SELECT dept_name FROM (SELECT * FROM emp RIGHT JOIN dept ON emp.dept2 = dept.dept1 )a WHERE sid is NULL;
Q25.先按部门号大小排序,再依据入职时间由早到晚排序员工信息表 ;
条件:二次排序 order by desc 从大到小 早到晚 asc
结果:所有信息
方法一:
SELECT * from dept left join emp on dept.dept1=emp.dept2 ORDER BY dept1 desc,woektime_start asc ;
方法二:(部门编号104不显示,内连接方法)
select * from dept a ,emp b where a.dept1 = b.dept2 ORDER BY a.dept1 desc,b.woektime_start asc;
方法三:SELECT * from emp ORDER BY dept2 DESC ,woektime_start asc;
Q26.求出财务部门工资最高员工的姓名和员工号
条件: dept_name ="财务" max(incoming)
结果:name ,sid
方法一:SELECT name,sid from dept left join emp on dept.dept1=emp.dept2 where incoming=(SELECT max(incoming) from dept left join emp on dept.dept1=emp.dept2 where dept_name="财务")and dept_name="财务" ;
方法 二:select name, sid from emp inner join dept on dept1= dept2 where dept_name='财务' order by incoming desc limit 1
方法三:SELECT name,sid FROM dept,emp WHERE dept.dept1=emp.dept2 AND (SELECT max(incoming) from (SELECT incoming from dept a ,emp b where a.dept1=b.dept2 and dept_name="财务")s )=incoming and dept_name="财务";
方法四:(缺陷有重复数据)
select name,sid FROM dept d INNER JOIN emp e on d.dept1=e.dept2 WHERE dept_name="财务" AND incoming in(SELECT MAX(incoming) FROM emp GROUP BY dept2);
方法五:SELECT name,dept1 FROM emp INNER JOIN dept ON emp.dept2 = dept.dept1 WHERE dept_name='财务' ORDER BY incoming DESC LIMIT 0,1;
Q27.求出工资在7500到8500之间,年龄最大的员工的姓名和部门名称。
条件: emp incming max ( age)
结果:name dept_name
方法一:
select name,dept_name from dept left join emp on dept.dept1=emp.dept2 where age=(SELECT max(age) from dept left join emp on dept.dept1=emp.dept2 WHERE incoming between 7500 and 8500 ) and incoming between 7500 and 8500 ;
方法二:
SELECT name,dept_name from (SELECT * from emp INNER JOIN dept on emp.dept2=dept.dept1 ORDER BY age desc) a where incoming>=7500 and incoming<=8500 LIMIT 1;
方法三:
SELECT name,dept_name FROM (SELECT * FROM dept inne JOIN emp on dept1=dept2 where incoming between 7500 and 8500)b WHERE age=(SELECT max(age) FROM (SELECT * FROM dept inne JOIN emp on dept1=dept2 where incoming between 7500 and 8500)b)
方法四:
SELECT name,dept_name FROM (SELECT * FROM dept inne JOIN emp on dept1=dept2 where incoming between 7500 and 8500)b ORDER BY age desc LIMIT 1
=================================================================================================================================
4、多表之三表
以练习题为例讲解:
表结构:
表一:Student学生表(学号、姓名、性别、年龄、编辑)
表二:Course课程表(编号、课程名称)
表三:sc选课表(选课编号、学号、课程编号、成绩)
Question:
(1)写一个SQL语句,查询选修了“计算机原理”的学生学号和姓名
(2)写一个SQL语句,查询“小明”同学选修的课程名称
(3)写一个SQL语句,查询选修了5门课程的学生学号和姓名
create table student(
stu_no int,
stu_name varchar(10),
sex char(1),
age int(3),
edit varchar(20) )
DEFAULT charset=utf8;
insert into student values
(1,'wang','男',21,'hello'),
(2,'小明','女',22,'haha2'),
(3,'hu','女',23,'haha3'),
(4,'li','男',25,'haha4');
create table course(
c_no int,
c_name varchar(10))
DEFAULT charset=utf8;
insert into course values
(1,'计算机原理'),
(2,'java'),
(3,'c'),
(4,'php'),
(5,'py');
#rop table sc;
create table sc(
sc_no int,
stu_no int,
c_no int,
score int(3))
DEFAULT charset=utf8;
insert into sc values
(1,1,1,80),
(2,2,2,90),
(3,2,1,85),
(4,2,3,70),
(5,2,4,95),
(6,2,5,89);
=================================================================================================================================
select * from student ;
select * from course ;
select * from sc ;
=================================================================================================================================
file:///E:/noteyoudao/qqC00975CCA04FE5F25136875459BFDA42/dc2bfb5be6fc4842bf7e6c11813be1e3/clipboard.png
file:///E:/noteyoudao/qqC00975CCA04FE5F25136875459BFDA42/9aacc30c07fd4de59946839bce27f2c8/clipboard.png
file:///E:/noteyoudao/qqC00975CCA04FE5F25136875459BFDA42/acae6802bc05490fbcef63e6e1802127/clipboard.png
Student 学生表(学号、姓名、性别、年龄、编辑) as a
Course 课程表(课程编号、课程名称) as b
sc 选课表(选课编号、学号、课程编号、成绩) as c
=================================================================================================================================
三表连接
方法一:三表隐藏内连接
格式:select * from 表1 , 表2 ,表3 where 表1.关联的字段=表3.关联字段 and 表2.关联字段=表3.表3字段
select * from student as a , course as b , sc as c where a.stu_no=c.stu_no and b.c_no=c.c_no
方法二:三表普通内连接
格式:select * from 表1 INNER JOIN 表3 on 表1.关联的字段=表3.关联字段 inner join 表2 on 表2.关联字段=表3.表3字段
select * from student as a INNER JOIN sc as c on a.stu_no=c.stu_no inner join course as b on b.c_no=c.c_no
方法三:三表普通左连接
格式:select * from 表1 left JOIN 表3 on 表1.关联的字段=表3.关联字段 left join 表2 on 表2.关联字段=表3.表3字段
select * from student as a left JOIN sc as c on a.stu_no=c.stu_no left join course as b on b.c_no=c.c_no
方法四:三表普通右连接
格式:select * from 表1 right JOIN 表3 on 表1.关联的字段=表3.关联字段 right join 表2 on 表2.关联字段=表3.表3字段
select * from student as a right JOIN sc as c on a.stu_no=c.stu_no right join course as b on b.c_no=c.c_no
方法五:先合两表,在两表合并成一表与第三个表合并
select s.stu_no ,s.stu_name from ( select a.stu_no ,stu_name,c_no from student a ,sc c where a.stu_no=c.stu_no )s ,course as b where s.c_no=b.c_no
=================================================================================================================================
Question:(1)写一个SQL语句,查询选修了“计算机原理”的学生学号和姓名
条件: c_name =“计算机原理” course表
结果:stu_no , name student表
方法一: 普通内连接
select a.stu_no, a.stu_name from student a join sc c on a.stu_no = c.stu_no join course b on c.c_no = b.c_no where b.c_name = "计算机原理"
Question:(2)写一个SQL语句,查询“小明”同学选修的课程名称
方法一:
select b.c_name from student1 as a , course as b , sc as c where a.stu_no=c.stu_no and b.c_no=c.c_no and a.stu_name='小明';
方法二:
select b.c_name from(select c_no,stu_name from student a inner join sc as c on a.stu_no=c.stu_no)s inner join course as b on s.c_no=b.c_no where s.stu_name=‘小明’;
方法三:
SELECT c_name from Student as a, course as b, sc as c where a.stu_no=c.stu_no and b.c_no=c.c_no and stu_name="小明";
Question:(3)写一个SQL语句,查询选修了5门课程的学生学号和姓名
方法一:
select student.stu_no,stu_name from student,course,sc where student.stu_no=sc.stu_no and sc.c_no=course.c_no group by student.stu_no having count(*)=5;
方法二:select a.stu_no,a.stu_name from student1 as a,course as b, sc as c where a.stu_no= c.stu_no and b.c_no=c.c_no group by a.stu_no having count(c_name)=5;
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