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问题描述
我的Hive表中有一个时间戳的字符串表示形式:
20130502081559999
$ c
$ b 我需要将它转换为如下所示的字符串:
2013-05-02 08:15:59
我试过以下{code} >>> {result}):
from_unixtime(unix_timestamp('20130502081559999','yyyyMMddHHmmss'))> >> 2013-05-03 00:54:59
from_unixtime(unix_timestamp('20130502081559999','yyyyMMddHHmmssMS'))>>> 2013-09-02 08:15:59
from_unixtime(unix_timestamp('20130502081559999','yyyyMMddHHmmssMS'))>>> 2013-05-02 08:10:39
转换为时间戳然后unixtime看起来很奇怪,是这样做的正确方法吗?
编辑
我想明白了。
from_unixtime(unix_timestamp(substr('20130502081559999',1,14),'yyyyMMddHHmmss'))>>> 2013-05-02 08:15:59
或
from_unixtime(unix_timestamp('20130502081559999','yyyyMMddHHmmssSSS'))>>> 2013-05-02 08:15:59
仍然...有更好的方法吗?
解决方案不确定您的意思是更好的方式,但您始终可以来处理日期转换。
I have the following string representation of a timestamp in my Hive table:
20130502081559999
I need to convert it to a string like so:
2013-05-02 08:15:59
I have tried following ({code} >>> {result}):
from_unixtime(unix_timestamp('20130502081559999', 'yyyyMMddHHmmss')) >>> 2013-05-03 00:54:59
from_unixtime(unix_timestamp('20130502081559999', 'yyyyMMddHHmmssMS')) >>> 2013-09-02 08:15:59
from_unixtime(unix_timestamp('20130502081559999', 'yyyyMMddHHmmssMS')) >>> 2013-05-02 08:10:39
Converting to a timestamp and then unixtime seems weird, what is the proper way to do this?
EDITI figured it out.
from_unixtime(unix_timestamp(substr('20130502081559999',1,14), 'yyyyMMddHHmmss')) >>> 2013-05-02 08:15:59
or
from_unixtime(unix_timestamp('20130502081559999', 'yyyyMMddHHmmssSSS')) >>> 2013-05-02 08:15:59
Still... Is there a better way?
解决方案 Not sure what you mean by "better way" but you can always write your own function to handle the date conversion.
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