sql通过日期判断年龄函数的示例代码

定义函数：

create function [dbo].[getage]  
(  
@birthday nvarchar(20) --生日  
)  
returns varchar(20)  
as  
begin  
if(@birthday is null or @birthday='')
return '';
 -- declare the return variable here  
 declare @age varchar(20)  
 declare @years int  
 declare @months int  
 declare @days int  
 -- add the t-sql statements to compute the return value here  
 set @age = ''  
  
 set @years = year(getdate()) - year(@birthday)  
 set @months = month(getdate()) - month(@birthday)  
 if day(@birthday)<=day(getdate())  
   set @days = day(getdate()) - day(@birthday)  
 else  
   begin  
     set @months = @months - 1  
     if month(@birthday) in (1,3,5,7,8,10,12)  
       set @days = 31-day(@birthday)+day(getdate())  
     else if month(@birthday) in (4,6,9,11)  
       set @days = 30-day(@birthday)+day(getdate())  
     else if month(@birthday) = 2  
       if (year(@birthday)%4 = 0 and year(@birthday)%100 <> 0) or year(@birthday)%400 = 0  
         set @days = 29-day(@birthday)+day(getdate())  
       else  
         set @days = 28-day(@birthday)+day(getdate())  
   end  
 if @months < 0  
   begin  
     set @years = @years - 1  
     set @months = @months + 12  
   end  
 if @years = 0 and @months = 0  
 begin  
     return convert(varchar,@days+1) + '天'  
  end  
 if @years > 0  
   set @age = cast(@years as varchar(5)) + '岁'  
 if @years < 3 and @months > 0 and @years>-1  
 begin  
   set @age = @age + cast(@months as varchar(5)) + '月'  
 end  
 if @years<0  
 set @age=''  
 return @age  
end

使用函数：

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