MySQl數(shù)據(jù)庫必知必會的加強版sql語句

這一篇屬于加強版，問題和sql語句如下。

創(chuàng)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);

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一對一：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(nèi)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';

---------------------------------------------

一對多：查詢"開發(fā)部"有哪些員工

創(chuàng)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(fā)部'); 
insert into groups(name) values('銷售部');

創(chuàng)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(fā)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(fā)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(fā)部';

------------------------------------------------------

多對多：查詢"趙"教過哪些學生

創(chuàng)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(chuàng)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(chuàng)建middles表 primary key(sid,tid) 表示聯(lián)合主鍵，這兩個字段的整體要唯一

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(nèi)連接（等值連接）：查詢客戶姓名，訂單編號，訂單價格

【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 "訂單編號",o.price "訂單價格" 
from customers c inner join orders o 
on c.id = o.customers_id; 
-- 
select c.name "客戶姓名",o.isbn "訂單編號",o.price "訂單價格" 
from customers c inner join orsers o 
on c.id = o.customers_id;

on+兩張表連接的條件.一張表的主鍵，一張表的外鍵

內(nèi)連接：只能查詢出二張表中根據(jù)連接條件都存在的記錄，有點類似于數(shù)學中交集

----------------------------------------------------

外連接：按客戶分組，查詢每個客戶的姓名和訂單數(shù)

外連接：既可以根據(jù)連接條件查詢出二張表中都存在的記錄，也能根據(jù)一方，強行將另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 : 以左側為參照，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 : 以右側為參照，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表示左邊的內(nèi)容都會顯現(xiàn)出來，例如customers c left out join 表示會把customers中的某列所有內(nèi)容都找出來

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自連接：求出AA的老板是EE。把自己想象成兩張表。左右各一張

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;

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演示MySQL中的函數(shù)（查詢手冊）

日期時間函數(shù)：

select addtime('2016-8-7 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('2016-12-31',now());

字符串函數(shù)：

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(shù)學函數(shù)：

select bin(10); 
select floor(3.14);//比3.14小的最大整數(shù)---正3 
select floor(-3.14);//比-3.14小的最大整數(shù)---負4 
select ceiling(3.14);//比3.14大的最小整數(shù)---正4 
select ceiling(-3.14);//比-3.14大的最小整數(shù)---負3，一定是整數(shù)值 
select format(3.1415926,3);保留小數(shù)點后3位，四舍五入 
select mod(10,3);//取余數(shù) 
select rand();//

加密函數(shù)：

select md5('123456');

返回32位16進制數(shù) 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") "描述" 
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 "描述" 
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 "描述" 
from users;

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