SQL计算年龄

[chuxuestar]

worker(workerno,workername,workersex,workerborndate)

显示所有职工的年龄，并按职工号递增排序。

表中只有员工的出生时间，例如1952-01-03


此sql语言如何写？如何利用系统时间计算年龄？
请高手帮忙1

[永生天地]

select workerno,workername,workersex,datediff(year,workerborndate,getdate())
from worker
order by workerno

[feixianxxx]

select workerno,workername,workersex,datediff(year,workerborndate,getdate())+1 as age
from worker
order by workerno

[zhengduan964532]

select workerno,
       workername,
       workersex,
       age=datediff(year,workerborndate,getdate())
from worker
order by workerno asc 

[yuyujulin]

if OBJECT_ID('worker') is not null
drop table worker
go
--worker(workerno,workername,workersex,workerborndate)
create table worker(
	workerno int,
	workername varchar(20),
	workersex int,
	workerborndate datetime
)
insert into worker 
select  3,'pjl',1,'1987-11-02' union all
select  2,'kzd',1,'1988-11-11' union all
select  1,'dyx',1,'1986-01-11'

select * from worker

select workerno,workername,workersex,DATEDIFF(MONTH,workerborndate,GETDATE())/12 as age
from worker
order by workerno
/*

[yuyujulin]

就这样结贴啦？

[lapheer]

----------------------------------------------------------------------
--计算年龄的函数
--算法原则：超过生日，年龄就得加1
--
--SELECT dbo.FUN_GetAge('1988-7-29', getdate()) as age
----------------------------------------------------------------------
CREATE    function [dbo].[FUN_GetAge]
(
   @birthday datetime,  --出生日期
   @today datetime      --截至日期
)
returns int
as
begin
	if @birthday > @today
	begin
		return 0;
	end	
	
	declare @age int
	
	select @age = datediff(year, @birthday, @today)--年份差值

	if datepart(month, @today) > datepart(month, @birthday)--月份超过
	begin
		select  @age = @age + 1
	end

	if datepart(month, @today) = datepart(month, @birthday)--月份一样
	begin
		if datepart(day, @today) >= datepart(day, @birthday)--日超过
		begin
			select  @age = @age + 1
		end
	end

	return @age ;
End