相当于 to_date(b.birthday,'yyyymmddhh24miss') 是

from_unixtime(unix_timestamp(b.birthday,'yyyyMMddHHmmss'))

然而，我不明白这一点 extract(year from age(iDATE, to_date(b.birthday,'yyyymmddhh24miss'))) . 你需要不同的出生日期和出生日期吗？
如果是，您可以使用以年为单位获取日期差异。

datediff(iDate,from_unixtime(unix_timestamp(b.birthday,'yyyyMMddHHmmss')) )/365

Hive也有 extract (year from datetimestamp) 功能。

提取时间戳的另一种方法是使用regexp\u replace：

regexp_replace(b.birthday,'^(\\d{4})(\\d{2})(\\d{2})(\\d{2})(\\d{2})(\\d{2})$','$1-$2-$3 $4:$5:$6')

年龄单位为：

cast(datediff(current_date,
             regexp_replace(b.birthday,'^(\\d{4})(\\d{2})(\\d{2})(\\d{2})(\\d{2})(\\d{2})$','$1-$2-$3 $4:$5:$6')
             )/365.25 as int)