Question

我有一个形式的字符串：yyyy-MM-dd'T'HH:mm:ss.SSSXXX（即：2016-12-11T14：21：04.354-08：00），我想将其转换为时间戳。我试过了：unix_timestamp(时间戳, "yyyy-MM-dd'T'HH:mm:ss.SSSXXX") 但它抛出：WARNINGS: Bad date/time conversion format: yyyy-MM-ddHH:mm:ss.SSSXXX

    Query: select unix_timestamp('2016-12-11T14:21:04.354-08:00', "yyyy-MM-dd'T'HH:mm:ss.SSSXXX")
WARNINGS: Bad date/time conversion format: yyyy-MM-ddHH:mm:ss.SSSXXX

Bad date/time conversion format: yyyy-MM-ddHH:mm:ss.SSSXXX

Answer 1

对于Impala，您的时间格式字符串应如下所示：'yyyy-mm-ddthh:mm:ss.sss-hh:mm'

见这个例子：

[localhost:21000] > select unix_timestamp('2016-12-11T14:21:04.354-05:00', "yyyy-MM-ddTHH:mm:ss.SSS-hh:mm");
Query: select unix_timestamp('2016-12-11T14:21:04.354-05:00', "yyyy-MM-ddTHH:mm:ss.SSS-hh:mm")
Query submitted at: 2016-12-21 00:49:48 (Coordinator: http://lv-desktop:25000)
Query progress can be monitored at: http://lv-desktop:25000/query_plan?query_id=564166231bc9fd4d:36e038bd00000000
+----------------------------------------------------------------------------------+
| unix_timestamp('2016-12-11t14:21:04.354-05:00', 'yyyy-mm-ddthh:mm:ss.sss-hh:mm') |
+----------------------------------------------------------------------------------+
| 1481484064                                                                       |
+----------------------------------------------------------------------------------+
Fetched 1 row(s) in 0.01s
[localhost:21000] > select unix_timestamp('2016-12-11T14:21:04.354-08:00', "yyyy-MM-ddTHH:mm:ss.SSS-hh:mm");
Query: select unix_timestamp('2016-12-11T14:21:04.354-08:00', "yyyy-MM-ddTHH:mm:ss.SSS-hh:mm")
Query submitted at: 2016-12-21 00:49:56 (Coordinator: http://lv-desktop:25000)
Query progress can be monitored at: http://lv-desktop:25000/query_plan?query_id=484ba21cc5d78ff8:e8d8353a00000000
+----------------------------------------------------------------------------------+
| unix_timestamp('2016-12-11t14:21:04.354-08:00', 'yyyy-mm-ddthh:mm:ss.sss-hh:mm') |
+----------------------------------------------------------------------------------+
| 1481494864                                                                       |
+----------------------------------------------------------------------------------+
Fetched 1 row(s) in 0.01s
[localhost:21000] >

Answer 2

select from_unixtime( unix_timestamp( 
  concat(substr('2016-12-11T14:21:04.354-05:00', 1, 26), '00'), 'yyyy-MM-dd\'T\'HH:mm:ss.SSSZ'), 'yyyy-MM-dd HH:mm:ss Z');

0: jdbc:hive2://localhost:10000/default> select from_unixtime( unix_timestamp( concat(substr('2016-12-11T14:21:04.354-05:00', 1, 26), '00'), 'yyyy-MM-dd\'T\'HH:mm:ss.SSSZ'), 'yyyy-MM-dd HH:mm:ss Z');
INFO  : Compiling command(queryId=hive_20170307142929_85757745-69f9-46b8-bbaf-9a2b9d0d8904): select from_unixtime( unix_timestamp( concat(substr('2016-12-11T14:21:04.354-05:00', 1, 26), '00'), 'yyyy-MM-dd\'T\'HH:mm:ss.SSSZ'), 'yyyy-MM-dd HH:mm:ss Z')
INFO  : OK
+----------------------------+--+
|            _c0             |
+----------------------------+--+
| 2016-12-12 03:21:04 +0800  |
+----------------------------+--+
1 row selected (1.416 seconds)

在Impala中将ISO8601日期时间格式的字符串转换为时间戳

2 个答案: