sql 根据身份证号获取出生日期并转成对应格式
sql server
查询判断身份证号是18位的
select SUBSTRING(SUBSTRING(IDCard,7,8),1,4)+'-'+SUBSTRING(SUBSTRING(IDCard,7,8),5,2)+'-'+SUBSTRING(SUBSTRING(IDCard,7,8),7,2) from 表 where Birthday is null and LEN(IDCard)=18
修改
update 表set Birthday=SUBSTRING(SUBSTRING(IDCard,7,8),1,4)+'-'+SUBSTRING(SUBSTRING(IDCard,7,8),5,2)+'-'+SUBSTRING(SUBSTRING(IDCard,7,8),7,2) where Birthday is null and LEN(IDCard)=18
oracle
SELECT (SUBSTR(IdNumStr,7,8 )) FROM DUAL;
输出是年月日字符串。再转换为对应需要的时间格式，例YYYY-MM-DD
SELECT to_char(to_date((SUBSTR(IdNumStr,7,8 )),'yyyy-mm-dd'),'yyyy-mm-dd') FROM DUAL;
create function U_GET_BIRTH(IDNUM in varchar2) return varchar2 as
begin
--根据身份证号取得对应的出生年月日，并转换为对应的日期格式
return to_char(to_date((SUBSTR(IDNUM ,7,8 )),'yyyy-mm-dd'),'yyyy-mm-dd');
end;