SQL 总结

       sql 总结一下，以备后面用   

---create database jqtest
create database jctest   --创建数据库    
drop database jctest     --删除数据库
create table users (id int not null primary key,sex int)--创建创建表
create table users_news (id int not null primary key,sex int)--创建创建表
alter table users add username varchar(50)
alter table users add passwords varchar(100)
alter table users add primary key(id)--添加主键
alter table users add updatetime datetime
--经常查询的字段需要设置聚集索引，
create  unique index username_index on users(username)
--创建视图
create view NewView as select n.id as nid,u.id as [uid] from users u,users_news n
select * from  NewView--查询
insert into users(id,username,passwords,sex) values(001,'姜强','1234567',1)--增
update users set updatetime=getdate() where id=001--改
delete users where id=001--删
select sum(id)from users--和
select min(id)from users--最小值
select max(id)from users--最大值
select avg(id)from users--平均值
----------------------------高级查询-----------------------------
--union all 全部数据，union 消除重复行
--union 可以将两个完全相同的表数据完全查出来
select * from users union all select * from users_news
--except 可以查出table1存在table2不存在的数据
select * from users except  select * from users_news
--intersect(相交);intersect 可以查出table1与table2都有的数据
select * from users intersect select * from users_news
----------------------------连接查询------------------------------
--左连接
select * from users u left join users_news n on u.id=n.id
--右连接
select *from users u right join users_news n on u.id=n.id
--全连接
select * from users u full join users_news n on u.id=n.id
---------------------------特殊函数用法--------------------------
--between
select * from users u where id between 1 and 2
--datediff(参数，开始时间，结束时间) 返回的是结束时间-开始时间（参数为单位）
select *,datediff(second,updatetime,getdate())as [second] from users where datediff(second,updatetime,getdate())>200
--newid()获取随即ID为87605956-8004-483A-B902-1A22813FE735
select top 2 *,newid() as [newid] from users order by newid()
--删除重复数据 
--得到不重复的列的ID，得到重复列的最大ID delete from table id not in (不重复ID)and id not in (重复列最大ID)
delete from users where id not in 
----得到重复列最大ID 
(select max(id) from users group by passwords having count(passwords)>1)
 and id not in
-----得到不重复ID 
(select max(id) from users group by passwords having count(passwords)=1)

-----------------------------------------存储过程------------------------------------------------
---分页存储
create proc Pager
  @pagesize int,
  @pageindex int
as
begin
 declare @start int,@end int
 set @start=((@pageindex-1) * @pagesize)+1;
 set @end=@pageindex * @pagesize;
--row_number() over(order by id asc) 这是sql2005之后加的函数，将数据按照id 排序并生成新的排序号
 select * from (select *,row_number() over(order by id asc) as number from users) t where t.number between @start and @end
end
exec Pager 2,3

select * from users
select distinct(passwords) from users
select * from users where username in('姜强北京','姜强拉萨')

-------------------------事物-------------------------------------------
alter proc shiwu
@pagesize int,
@pageindex int
as
   begin tran
   insert into users(id,username,passwords,sex)values(006,'姜强北京','姜强北京',1);
   insert into users(id,username,passwords,sex) values(008,'321564','2313',1)--增
   declare @start int,@end int
   set @start=((@pageindex-1)*@pagesize)+1;
   set @end=@pageindex*@pagesize;
   select * from (select *,row_number()over(order by id asc) as number from users) t where t.number between @start and @end
   print @@error
   if @@error<>0
     begin rollback tran
   end
   else
     begin commit tran
   end
exec shiwu 1,1