Question

我目前有一个SQL查询，该查询从数据库中的Unix DATETIME列返回正确的本地TIMESTAMP。

以下是使用TIMESTAMP中的特定1539961967000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

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

以前有人遇到过吗？

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;

为您提供间隔数据类型：

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

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

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

如果开始时间戳包含毫秒，它也会保留毫秒（这是从“ Unix”时间转换而保留的）：

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

然后转换回去：

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

如果起始时间戳的精度比后者高，那么您将需要截断（或舍入/舍入/上限/下标/投射）以避免产生非整数的结果；这个版本只是截断了提取的毫秒部分：

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

如果没有这种截断（或等效截断），您最终会得到1539961967123.456789。

我忘记了the秒的差异；如果您需要/想要处理此问题，请see this answer。

Answer 2

主要问题是Oracle有两种方法（至少）将秒数转换为秒间隔-使用函数或对间隔文字进行简单的算术运算-但没有直接方法反之。

在下面的两个查询中，我首先展示如何在不损失毫秒的情况下将UNIX时间戳（自大纪元以来的毫秒数）转换为Oracle时间戳。（请参阅我在“问题”下的评论，其中指出您的方法将丢失毫秒。）然后，我将展示如何逆转该过程。

像您一样，我忽略了“ Unix时间戳”导致的“ UTC时间戳”和“ Unix时间戳”之间的差异，忽略了leap秒。您的企业必须确定这是否重要。

Unix时间戳到带有时区（保留毫秒）的Oracle时间戳：

with
  inputs (unix_timestamp) as (
    select 1539961967186 from dual
  )
select from_tz(timestamp '1970-01-01 00:00:00' 
               + interval '1' second * (unix_timestamp/1000), 'UTC')
                   at time zone 'America/Denver' as oracle_ts_with_timezone
from   inputs
;

ORACLE_TS_WITH_TIMEZONE               
--------------------------------------
2018-10-19 09:12:47.186 America/Denver

Oracle时间戳，时区为Unix时间戳（保留毫秒）：

with
  sample_data (oracle_ts_with_timezone) as (
    select to_timestamp_tz('2018-10-19 09:12:47.186 America/Denver', 
                           'yyyy-mm-dd hh24:mi:ss.ff tzr')           from dual
  )
select ( extract(second from ts) 
         + (trunc(ts, 'mi') - date '1970-01-01') * (24 * 60 * 60)
       ) * 1000 as unix_timestamp
from   ( select cast(oracle_ts_with_timezone at time zone 'UTC' 
                     as timestamp) as ts
         from   sample_data
       )
;

  UNIX_TIMESTAMP
----------------
   1539961967186

将本地日期时间（带有时区）转换为Oracle中的Unix时间戳

2 个答案: