oracle - 在Oracle中将本地日期时间(带时区)转换为Unix时间戳

Question

我当前有一个 SQL 查询，它从数据库中的 Unix TIMESTAMP 列返回正确的本地 DATETIME。

以下是使用1539961967000的特定TIMESTAMP的示例:

SELECT FROM_TZ(CAST(DATE '1970-01-01' + 1539961967000 * (1/24/60/60/1000) AS TIMESTAMP), 'UTC') AT TIME ZONE 'America/Denver' DATETIME
FROM dual;

返回:

DATETIME
19-OCT-18 09.12.47.000000000 AM AMERICA/DENVER

我很难反转此查询以返回以本地 DATETIME 开头的 Unix TIMESTAMP。

有人遇到过这种情况吗？

Answer 1

您可以将带有时区的时间戳转换为 UTC，然后从中减去纪元:

select timestamp '2018-10-19 09:12:47.0 AMERICA/DENVER'
  - timestamp '1970-01-01 00:00:00.0 UTC' as diff
from dual;

这为您提供了间隔数据类型:

DIFF                  
----------------------
+17823 15:12:47.000000

然后您可以从中提取元素，并将每个元素乘以适当的因子以将其转换为毫秒(即天数为 60*60*24*1000)；然后将它们加在一起:

select extract(day from diff) * 86400000
  + extract(hour from diff) * 3600000
  + extract(minute from diff) * 60000
  + extract(second from diff) * 1000 as unixtime
from (
  select timestamp '2018-10-19 09:12:47.0 AMERICA/DENVER'
    - timestamp '1970-01-01 00:00:00.0 UTC' as diff
  from dual
);

            UNIXTIME
--------------------
       1539961967000

db<>fiddle

如果起始时间戳有毫秒，这也会保留毫秒(这会在保留它们的同时从“Unix”时间转换):

select (timestamp '1970-01-01 00:00:00.0 UTC' + (1539961967567 * interval '0.001' second))
  at time zone 'America/Denver' as denver_time
from dual;

DENVER_TIME                                 
--------------------------------------------
2018-10-19 09:12:47.567000000 AMERICA/DENVER

然后转换回来:

select extract(day from diff) * 86400000
  + extract(hour from diff) * 3600000
  + extract(minute from diff) * 60000
  + extract(second from diff) * 1000 as unixtime
from (
  select timestamp '2018-10-19 09:12:47.567 AMERICA/DENVER'
    - timestamp '1970-01-01 00:00:00.0 UTC' as diff
  from dual
);

            UNIXTIME
--------------------
       1539961967567

db<>fiddle

如果您的起始时间戳的精度高于此精度，那么您需要截断(或舍入/下限/上限/强制转换)以避免出现非整数结果；这个版本只是截断提取的毫秒部分:

select diff,
  extract(day from diff) * 86400000
  + extract(hour from diff) * 3600000
  + extract(minute from diff) * 60000
  + trunc(extract(second from diff) * 1000) as unixtime
from (
  select timestamp '2018-10-19 09:12:47.123456789 AMERICA/DENVER'
    - timestamp '1970-01-01 00:00:00.0 UTC' as diff
  from dual
);

DIFF                                  UNIXTIME
------------------------- --------------------
+17823 15:12:47.123456789        1539961967123

如果没有该截断(或同等截断)，您最终会得到 1539961967123.456789。

---

我忘记了闰秒的差异；如果您需要/想要处理这个问题，see this answer。