hive time_stamp 转换为 UTC,UTC 为 time_offset
hive time_stamp convert into UTC with time_offset in UTC
我有 2 列:time_stamp 和 time_offset。两者都是 STRING
数据类型。
我们如何借助 UTC 中的第二列将一列值转换为 UTC?他们是将 time_stamp 列转换为 UTC 的任何配置单元还是来自 unix 的解决方案?
hive> select time_stamp from table1 limit 2;
OK
20170717-22:31:57.348
20170719-21:10:15.393
[yyyymmdd-hh:mm:ss.msc] this column is in local time
hive> select time_offset from table1 limit 2;
OK
-05:00
+05:00
[‘+hh:mm’ or ‘-hh:mm’ ] this column is in UTC
您可以使用 Hive Date Functions unix_timestamp
和 from_unixtime
来执行转换。
代码
WITH table1 AS (
SELECT '20170717-22:31:57.348' AS time_stamp, '-05:00' AS time_offset UNION ALL
SELECT '20170719-21:10:15.393' AS time_stamp, '+05:00' AS time_offset
)
SELECT
time_stamp,
time_offset,
unix_timestamp(concat(time_stamp, ' ', time_offset), 'yyyyMMdd-HH:mm:ss.SSS X') AS unix_timestamp_with_offset,
from_unixtime(unix_timestamp(concat(time_stamp, ' ', time_offset), 'yyyyMMdd-HH:mm:ss.SSS X'), 'yyyyMMdd-HH:mm:ss.SSS') AS string_timestamp_with_offset
FROM table1
;
结果集
+------------------------+--------------+-----------------------------+-------------------------------+--+
| time_stamp | time_offset | unix_timestamp_with_offset | string_timestamp_with_offset |
+------------------------+--------------+-----------------------------+-------------------------------+--+
| 20170717-22:31:57.348 | -05:00 | 1500348717 | 20170717-20:31:57.000 |
| 20170719-21:10:15.393 | +05:00 | 1500480615 | 20170719-09:10:15.000 |
+------------------------+--------------+-----------------------------+-------------------------------+--+
说明
unix_timestamp
可以接受与 Java SimpleDateFormat
. I am guessing that your offsets are using the ISO 8601 syntax, so let's use the X
format specifier. Then, we can use the concat
String Operator 相同语法的可选格式字符串,以便在传递给 [=12 之前组合 time_stamp
和 time_offset
=].
unix_timestamp
函数生成指定为自纪元以来的秒数的数字时间戳。要将其转换回字符串表示形式,我们可以将从 unix_timestamp
获得的结果传递到 from_unixtime
,这次指定我们的原始格式说明符。
(请彻底测试以确保结果在您的环境中有意义。时区数学可能很棘手。)
我有 2 列:time_stamp 和 time_offset。两者都是 STRING
数据类型。
我们如何借助 UTC 中的第二列将一列值转换为 UTC?他们是将 time_stamp 列转换为 UTC 的任何配置单元还是来自 unix 的解决方案?
hive> select time_stamp from table1 limit 2;
OK
20170717-22:31:57.348
20170719-21:10:15.393
[yyyymmdd-hh:mm:ss.msc] this column is in local time
hive> select time_offset from table1 limit 2;
OK
-05:00
+05:00
[‘+hh:mm’ or ‘-hh:mm’ ] this column is in UTC
您可以使用 Hive Date Functions unix_timestamp
和 from_unixtime
来执行转换。
代码
WITH table1 AS (
SELECT '20170717-22:31:57.348' AS time_stamp, '-05:00' AS time_offset UNION ALL
SELECT '20170719-21:10:15.393' AS time_stamp, '+05:00' AS time_offset
)
SELECT
time_stamp,
time_offset,
unix_timestamp(concat(time_stamp, ' ', time_offset), 'yyyyMMdd-HH:mm:ss.SSS X') AS unix_timestamp_with_offset,
from_unixtime(unix_timestamp(concat(time_stamp, ' ', time_offset), 'yyyyMMdd-HH:mm:ss.SSS X'), 'yyyyMMdd-HH:mm:ss.SSS') AS string_timestamp_with_offset
FROM table1
;
结果集
+------------------------+--------------+-----------------------------+-------------------------------+--+
| time_stamp | time_offset | unix_timestamp_with_offset | string_timestamp_with_offset |
+------------------------+--------------+-----------------------------+-------------------------------+--+
| 20170717-22:31:57.348 | -05:00 | 1500348717 | 20170717-20:31:57.000 |
| 20170719-21:10:15.393 | +05:00 | 1500480615 | 20170719-09:10:15.000 |
+------------------------+--------------+-----------------------------+-------------------------------+--+
说明
unix_timestamp
可以接受与 Java SimpleDateFormat
. I am guessing that your offsets are using the ISO 8601 syntax, so let's use the X
format specifier. Then, we can use the concat
String Operator 相同语法的可选格式字符串,以便在传递给 [=12 之前组合 time_stamp
和 time_offset
=].
unix_timestamp
函数生成指定为自纪元以来的秒数的数字时间戳。要将其转换回字符串表示形式,我们可以将从 unix_timestamp
获得的结果传递到 from_unixtime
,这次指定我们的原始格式说明符。
(请彻底测试以确保结果在您的环境中有意义。时区数学可能很棘手。)