MySQl數據庫必會sql語句加強版
來源：程序員人生 發布時間：2016-07-11 15:26:27 閱讀次數：3609次
這篇承接上1篇《mysql必會sql語句》：http://blog.csdn.net/qq_32059827/article/details/51763950
這1篇屬于加強版，問題和sql語句以下。
創建users表，設置id,name,gender,sal字段，其中id為主鍵
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);
--------------------------------------------------------------------------------------
1對1：AA的身份號是多少
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表示內連接】
select u.name "姓名",c.num "身份證號"
from users u inner join cards c
on u.id = c.uid
where u.name = 'AA';
--
select u.name "姓名",c.num "身份證號"
from users u inner join cards c
on u.id = c.uid
where name = 'AA';
---------------------------------------------
1對多：查詢"開發部"有哪些員工
創建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('開發部');
insert into groups(name) values('銷售部');
創建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);
查詢"開發部"有哪些員工
select g.name "部門",e.name "員工"
from groups g inner join emps e
on g.id = e.gid
where g.name = '開發部';
--
select g.name "部門",e.name "員工"
from groups g inner join emps e
on g.id = e.gid
where g.name = '開發部';
------------------------------------------------------
多對多：查詢"趙"教過哪些學生
創建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');
創建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('趙');
insert into teachers(name) values('劉');
創建middles表 primary key(sid,tid) 表示聯合主鍵，這兩個字段的整體要唯1
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);
查詢"趙"教過哪些學生
select t.name "老師",s.name "學生"
from students s inner join middles m inner join teachers t
on (s.id=m.sid) and (m.tid=t.id)
where t.name = '趙';
--
select t.name "老師",s.name "學生"
from students s inner join middles m inner join teachers t
on (s.id=m.sid) and (t.id=m.tid)
where t.name = "趙";
--------------------------------------------------------------------------------------------------------
將5000元（含）以上的員工標識為"高薪"，否則標識為"起薪"
將薪水為NULL的員工標識為"無薪"
將5000元（含）以上的員工標識為"高薪"，否則標識為"起薪"
將7000元的員工標識為"高薪",6000元的員工標識為"中薪",5000元則標識為"起薪",否則標識為"試用薪"
---------------------------------------------------------------------------------------------------------
內連接（等值連接）：查詢客戶姓名，定單編號，定單價格
【注：customers c inner join orders o使用了別名，以后o就代表orders】
select c.name "客戶姓名",o.isbn "定單編號",o.price "定單價格"
from customers c inner join orders o
on c.id = o.customers_id;
--
select c.name "客戶姓名",o.isbn "定單編號",o.price "定單價格"
from customers c inner join orsers o
on c.id = o.customers_id;
on+兩張表連接的條件.1張表的主鍵，1張表的外鍵
內連接：只能查詢出2張表中根據連接條件都存在的記錄，有點類似于數學中交集
----------------------------------------------------
外連接：按客戶分組，查詢每一個客戶的姓名和定單數
外連接：既可以根據連接條件查詢出2張表中都存在的記錄，也能根據1方，強即將另外一方就算不滿兄條件的記錄也能查詢出來
外連接可以細分為：
<左外連接 : 以左邊為參照，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;
--
>右外連接 : 以右邊為參照，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表示左側的內容都會顯現出來，例如customers c left out join 表示會把customers中的某列所有內容都找出來
------------------------------------------------------
自連接：求出AA的老板是EE。把自己想象成兩張表。左右各1張
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;
-----------------------------------------------------------------------------------------------
演示MySQL中的函數（查詢手冊）
日期時間函數：
select addtime('2016⑻⑺ 23:23:23','1:1:1'); 時間相加
select current_date();
select current_time();
select now();
select year( now() );
select month( now() );
select day( now() );
select datediff('2016⑴2⑶1',now());
字符串函數：
select charset('哈哈');
select concat('你好','哈哈','嗎');
select instr('www.baidu.com','baidu');
select substring('www.baidu.com',5,3);
數學函數：
select bin(10);
select floor(3.14);//比3.14小的最大整數---正3
select floor(⑶.14);//比⑶.14小的最大整數---負4
select ceiling(3.14);//比3.14大的最小整數---正4
select ceiling(⑶.14);//比⑶.14大的最小整數---負3，1定是整數值
select format(3.1415926,3);保存小數點后3位，4舍5入
select mod(10,3);//取余數
select rand();//
加密函數：
select md5('123456');
返回32位16進制數 e10adc3949ba59abbe56e057f20f883e
演示MySQL中流程控制語句
use json;
drop table if exists users;
create table if not exists users(
id int(5) primary key auto_increment,
name varchar(10) not null unique,
sal int(5)
);
insert into users(name,sal) values('哈哈',3000);
insert into users(name,sal) values('呵呵',4000);
insert into users(name,sal) values('嘻嘻',5000);
insert into users(name,sal) values('笨笨',6000);
insert into users(name,sal) values('明明',7000);
insert into users(name,sal) values('絲絲',8000);
insert into users(name,sal) values('君君',9000);
insert into users(name,sal) values('趙趙',10000);
insert into users(name,sal) values('無名',NULL);
將5000元（含）以上的員工標識為"高薪"，否則標識為"起薪"
select name "姓名",sal "薪水",
if(sal>=5000,"高薪","起薪") "描寫"
from users;
將薪水為NULL的員工標識為"無薪"
select name "姓名",ifnull(sal,"無薪") "薪水"
from users;
將5000元（含）以上的員工標識為"高薪"，否則標識為"起薪"
select name "姓名",sal "薪水",
case when sal>=5000 then "高薪"
else "起薪" end "描寫"
from users;
將7000元的員工標識為"高薪"，6000元的員工標識為"中薪",5000元則標識為"起薪"，否則標識為"試用薪"
select name "姓名",sal "薪水",
case sal
when 3000 then "低薪"
when 4000 then "起薪"
when 5000 then "試用薪"
when 6000 then "中薪"
when 7000 then "較好薪"
when 8000 then "不錯薪"
when 9000 then "高薪"
else "重薪"
end "描寫"
from users;
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