如何将'％Y％m％d％H％i％s'列拆分为两列：一列包含％Y％m％d％，其中包含H％i％s in hive？

Question

我有一列显示'％Y％m％d％H％i％s'（例如20150125145900）如何将其转换为两列，一列为“ymd”，另一列为“his”（例如2015/01 / 25和14:59:00）？

Answer 1

select datetime[0] as年,datetime[1] as {小时{1}}

from (select split(from_unixtime(unix_timestamp('20150125145900','yyyyMMddhhmmss'),'yyyy-MM-dd HH:mm:ss'),' ') as datetime)e

Answer 2

<强>蜂房

select  ts[0] as year
       ,ts[1] as hour

from   (select  split(from_unixtime(unix_timestamp('20150125145900','yyyyMMddHHmmss')
                    ,'yyyy-MM-dd HH:mm:ss'),' ') as ts
        ) t

+------------+----------+
|    year    |   hour   |
+------------+----------+
| 2015-01-25 | 14:59:00 |
+------------+----------+

<强>帕拉

select  split_part(ts,' ',1) as year
       ,split_part(ts,' ',2) as hour

from   (select  from_unixtime(unix_timestamp('20150125145900','yyyyMMddHHmmss')
                    ,'yyyy-MM-dd HH:mm:ss') as ts
        ) t
;

+------------+----------+
| year       | hour     |
+------------+----------+
| 2015-01-25 | 14:59:00 |
+------------+----------+