Timestamp to numeric conversion

[xtrim]

Hi,

How do I convert a standard timestamp to its numeric representation?

[Xenobiologist]

Any example? From xto y?

[jchd]

From which "standard" to which "numeric representation"?

[xtrim]

Hi,

for example

timestamp 13/04/2011 14:35:56 is represented as 40646.608287037

timestamp 30/04/2011 14:37:36 is represented as 40663.609444444

[andrei0x309]

Hi,

for example

timestamp 13/04/2011 14:35:56 is represented as 40646.608287037

timestamp 30/04/2011 14:37:36 is represented as 40663.609444444

I haven't seen yet this kind of date representation , you could try this :

it gives you a represenation like this : Tue, 12 Apr 2011 11:43:33 GMT is represented as 1302608613 just like mktime() form php check this .

[jchd]

@xtrim,

None of your examples is anywhere close to "standard" and your terminology is backward.

A standard (ISO 8601) date-time (string) is something like 2011-04-12T13:56:23+01:00 (my own current time).

The equivalent Unix timestamp is 1302609878 (in fact, it's a few seconds later!).

So the standard Unix timestamp for 2011-04-13 14:35:56 is 1302705356 and 2011-04-30 14:37:36 gives 1304174256.

There is not even a constant proportion from the timestamp integers to the values you provide, so I question what they actually represent.

[MvGulik]

Look like Ms-Excel Date-Time value's. or [DayCount(date).DayFraction(Time)] ... start point: "January 1, 1900"

[xtrim]

tnx

[jchd]

Ah, you mean that! You should have said so before.

#include <Date.au3>
Local $msTime = _DateDiff('s', '1899/12/30 00:00:00', '2011/04/13 14:35:56') / 86400
ConsoleWrite($msTime & @LF)

If you wonder why 30/12/1899, it's because Excel mimics a bug introduced by Lotus 1-2-3 and Multiplan which wrongly assumed the year 1900 to be leap.