在 Firebird 中创建一个整数时间戳

Question

我需要一个时间戳作为整数（请不要问我为什么，我完全被这个困扰）。像往常一样，定义是 01.01.1970 之后的秒数。

我的初始值是具有以下格式的字符串

YYYYMMDDHHMMSS（例如 20140108154821）

我已经能够建立一个正常的时间戳

CAST(
(SUBSTRING ( t.thetime FROM 5 FOR 2 )||'/'||SUBSTRING ( t.thetime FROM 7 FOR 2 )||'/'||SUBSTRING ( t.thetime FROM 1 FOR 4 )||' ' ||SUBSTRING ( t.thetime FROM 9 FOR 2 )||':'||SUBSTRING ( t.thetime FROM 11 FOR 2 )||':'||SUBSTRING ( t.thetime FROM 13 FOR 2 )||'.00'
) AS TIMESTAMP)

然后我尝试使用这种技术将其转换为 int

CAST(cast(
(SUBSTRING ( t.thetime FROM 5 FOR 2 )||'/'||SUBSTRING ( t.thetime FROM 7 FOR 2 )||'/'||SUBSTRING ( t.thetime FROM 1 FOR 4 )||' '|| SUBSTRING ( t.thetime FROM 9 FOR 2 )||':'||SUBSTRING ( t.thetime FROM 11 FOR 2 )||':'||SUBSTRING ( t.thetime FROM 13 FOR 2 )||'.00'
) AS TIMESTAMP)
-cast('01/01/1970 00:00:00' as timestamp) as BIGINT) as TIMESTAMP_INT

我确实得到了一个整数，但它看起来好像返回的值是天数或类似的东西。

例如 08.01.2015, 16:33:01.114 变成 16444

关于为什么会发生这种情况或是否有其他方法可以做到这一点的任何建议？

我使用的是 Firebird 2.5 版

提前致谢

PS：

SELECT 
CAST(cast(
    (SUBSTRING ( '20140108154821' FROM 5 FOR 2 )||'/'||SUBSTRING ( '20140108154821' FROM 7 FOR 2 )||'/'||SUBSTRING ( '20140108154821' FROM 1 FOR 4 )||' '
    ||
    SUBSTRING ( '20140108154821' FROM 9 FOR 2 )||':'||SUBSTRING ( '20140108154821' FROM 11 FOR 2 )||':'||SUBSTRING ( '20140108154821' FROM 13 FOR 2 )||'.00'
    ) AS TIMESTAMP)
    -cast('01/01/1970 00:00:00' as timestamp) as BIGINT) 
    from sometable

;

Answer 1

一种方法是使用DATEDIFF：

SELECT DATEDIFF(second, TIMESTAMP'1970-01-01 00:00', CURRENT_TIMESTAMP) 
FROM RDB$DATABASE

这假设时间是 UTC，否则您可能需要偏移时区偏移量来进行补偿（例如，对于 CET (UTC+1)，您可以TIMESTAMP'1970-01-01 01:00'改为使用）。