sql 18位身份证常用操作，sql18位身份证

sql 18位身份证常用操作，sql18位身份证

1.年龄

select datediff(yy,cast(substring('300082198206028878',7,8) as datetime),getdate()) as 年龄

2.出生年月

select substring('300082198206028878',7,4) + '-' + substring('300082198206028878',11,2) + '-' + substring('300082198206028878',13,2) as 出生年月

3.性别

select      性别  =  case when (left(right('300082198206028878',2),1) % 2 = 0) then '女' else '男' end  where len('300082198206028878')=18 and left(right('300082198206028878',2),1)<>'X'