如何写一个返回值是int数据类型的newid()函数

[magnetmoon]

如何写一个返回值是int数据类型的newid()函数
能否写一个函数，就象限newid()一样，只是返回的数据类型是int型的，

[viptiger]

select cast(cast(replace(newid(),'-','')as binary)as int)

[dawugui]

select id = right('0000000000' + cast(cast(rand(checksum(newid()))*1000000000 as int) as varchar),10) from sysobjects

[dawugui]

--出来的是字符型.
select id = right('0000000000' + cast(cast(rand(checksum(newid()))*1000000000 as int) as varchar),10) from sysobjects

--如果要INT(估计得bigint才行)
select id = cast(left(cast(cast(rand(checksum(newid()))*1000000000 as int) as varchar) + '0000000000',10) as bigint) from sysobjects

[viptiger]

select cast(cast(replace(newid(),'-','')as binary)as bigint)

--int不行，要用bigint

[-狙击手-]

create function f_newid()
returns bigint
as
begin
return abs(cast(checksum(newid() )as bigint))
end
go

[wzy_love_sly]

create table tb(id int,name varchar(50)) 
insert into tb select 1,'a' 
insert into tb select 2,'b' 

alter table tb add randnum int 

update tb set randnum= abs(checksum(newid()))%100--控制在100以内的随机数 

select * from tb

[arrow_gx]

试试这个

create function f_newid() 
returns bigint 
as 
begin 
    return abs(cast(checksum(newid() )as bigint)) 
end 
go

[magnetmoon]

上面所有的我都试了，还是不行，
用一个20多万笔数据的表测试，还是有重复值，