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根据身份证得到生日函数
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--创建函数(函数来自csdn,作者不详)
create function [dbo].[Get_birthday]
(
@idcardno nvarchar(50)
)
returns varchar(10)
as
begin
declare @birthday datetime
if (len(@idcardno)=15 or len(@idcardno)=16) and substring(@idcardno,9,2) between 1 and 12 andsubstring(@idcardno,11,2) between 1 and 31
set @birthday= convert(varchar(10),‘19‘+substring(@idcardno,7,2)+‘-‘+substring(@idcardno,9,2)+‘-‘+substring(@idcardno,11,2),120)
else if len(@idcardno)=18 and substring(@idcardno,7,2)>=19 and substring(@idcardno,11,2) between 1 and 12 andsubstring(@idcardno,13,2) between 1 and 31
set @birthday= convert(varchar(10),substring(@idcardno,7,4)+‘-‘+substring(@idcardno,11,2)+‘-‘+substring(@idcardno,13,2),120)
else
set @birthday=null
return(convert(varchar(10),@birthday,120))
end
--测试示例
select dbo.[Get_birthday](‘222222198306043213‘)
--运行结果
/*
1983-06-04
*/
根据身份证得到生日函数