这一篇属于加强版,问题和sql语句如下。
创建users表,设置id,name,gender,sal字段,其中id为主键
drop table if exists users; create table if not exists users( id int(5) primary key auto_increment, name varchar(10) unique not null, gender varchar(1) not null, sal int(5) not null ); insert into users(name,gender,sal) values('AA','男',1000); insert into users(name,gender,sal) values('BB','女',1200);
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一对一:AA的身份号是多少
drop table if exists users; create table if not exists users( id int(5) primary key auto_increment, name varchar(10) unique not null, gender varchar(1) not null, sal int(5) not null ); insert into users(name,gender,sal) values('AA','男',1000); insert into users(name,gender,sal) values('BB','女',1200); drop table if exists cards; create table if not exists cards( id int(5) primary key auto_increment, num int(3) not null unique, loc varchar(10) not null, uid int(5) not null unique, constraint uid_fk foreign key(uid) references users(id) ); insert into cards(num,loc,uid) values(111,'北京',1); insert into cards(num,loc,uid) values(222,'上海',2);
【注:inner join表示内连接】
select u.name "姓名",c.num "身份证号" from users u inner join cards c on u.id = c.uid where u.name = 'AA'; -- select u.name "姓名",c.num "身份证号" from users u inner join cards c on u.id = c.uid where name = 'AA';
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一对多:查询"开发部"有哪些员工
创建groups表
drop table if exists groups; create table if not exists groups( id int(5) primary key auto_increment, name varchar(10) not null ); insert into groups(name) values('开发部'); insert into groups(name) values('销售部');
创建emps表
drop table if exists emps; create table if not exists emps( id int(5) primary key auto_increment, name varchar(10) not null, gid int(5) not null, constraint gid_fk foreign key(gid) references groups(id) ); insert into emps(name,gid) values('哈哈',1); insert into emps(name,gid) values('呵呵',1); insert into emps(name,gid) values('嘻嘻',2); insert into emps(name,gid) values('笨笨',2);
查询"开发部"有哪些员工
select g.name "部门",e.name "员工" from groups g inner join emps e on g.id = e.gid where g.name = '开发部'; -- select g.name "部门",e.name "员工" from groups g inner join emps e on g.id = e.gid where g.name = '开发部';
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多对多:查询"赵"教过哪些学生
创建students表
drop table if exists students; create table if not exists students( id int(5) primary key auto_increment, name varchar(10) not null ); insert into students(name) values('哈哈'); insert into students(name) values('嘻嘻');
创建teachers表
drop table if exists teachers; create table if not exists teachers( id int(5) primary key auto_increment, name varchar(10) not null ); insert into teachers(name) values('赵'); insert into teachers(name) values('刘');
创建middles表 primary key(sid,tid) 表示联合主键,这两个字段的整体要唯一
drop table if exists middles; create table if not exists middles( sid int(5), constraint sid_fk foreign key(sid) references students(id), tid int(5), constraint tid_fk foreign key(tid) references teachers(id), primary key(sid,tid) ); insert into middles(sid,tid) values(1,1); insert into middles(sid,tid) values(1,2); insert into middles(sid,tid) values(2,1); insert into middles(sid,tid) values(2,2);
查询"赵"教过哪些学生
select t.name "老师",s.name "学生" from students s inner join middles m inner join teachers t on (s.id=m.sid) and (m.tid=t.id) where t.name = '赵'; -- select t.name "老师",s.name "学生" from students s inner join middles m inner join teachers t on (s.id=m.sid) and (t.id=m.tid) where t.name = "赵";
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将5000元(含)以上的员工标识为"高薪",否则标识为"起薪"
将薪水为NULL的员工标识为"无薪"
将5000元(含)以上的员工标识为"高薪",否则标识为"起薪"
将7000元的员工标识为"高薪",6000元的员工标识为"中薪",5000元则标识为"起薪",否则标识为"试用薪"
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内连接(等值连接):查询客户姓名,订单编号,订单价格
【注:customers c inner join orders o使用了别名,以后o就代表orders】
select c.name "客户姓名",o.isbn "订单编号",o.price "订单价格" from customers c inner join orders o on c.id = o.customers_id; -- select c.name "客户姓名",o.isbn "订单编号",o.price "订单价格" from customers c inner join orsers o on c.id = o.customers_id;
on+两张表连接的条件.一张表的主键,一张表的外键
内连接:只能查询出二张表中根据连接条件都存在的记录,有点类似于数学中交集
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外连接:按客户分组,查询每个客户的姓名和订单数
外连接:既可以根据连接条件查询出二张表中都存在的记录,也能根据一方,强行将另一方就算不满兄条件的记录也能查询出来
外连接可以细分为:
<左外连接 : 以左侧为参照,left outer join表示 select c.name,count(o.isbn) from customers c left outer join orders o on c.id = o.customers_id group by c.name; -- >右外连接 : 以右侧为参照,right outer join表示 select c.name,count(o.isbn) from orders o right outer join customers c on c.id = o.customers_id group by c.name;
left outer join表示左边的内容都会显现出来,例如customers c left out join 表示会把customers中的某列所有内容都找出来
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自连接:求出AA的老板是EE。把自己想象成两张表。左右各一张
select users.ename,bosss.ename from emps users inner join emps bosss on users.mgr = bosss.empno; select users.ename,bosss.ename from emps users left outer join emps bosss on users.mgr = bosss.empno;
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演示MySQL中的函数(查询手册)
日期时间函数:
select addtime('2016-8-7 23:23:23','1:1:1'); 时间相加 select current_date(); select current_time(); select now(); select year( now() ); s