dateToNumber: Difference between revisions
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In a leap year there are 366 days, so the range will increase by 1 day:<br> | In a leap year there are 366 days, so the range will increase by 1 day:<br> | ||
<tt>1 / 365 * 366 <nowiki>=</nowiki> 1.00274</tt><br> | <tt>1 / 365 * 366 <nowiki>=</nowiki> 1.00274</tt><br> | ||
In short, | In short, in normal year the command returns in range 0...1 in a leap year it will return in range 0...1.00274|= Description | ||
____________________________________________________________________________________________ | ____________________________________________________________________________________________ | ||
Revision as of 01:33, 27 February 2016
Description
- Description:
- Convert a date to a float number, based on Jan 1st 00:00:00 = 0 and Dec 31st 23:59:59 = 1. The same day and time in leap year will be different after 28th of February and 23:59 on 31st of December will be 1.00274
This is how this command works. The 365 days of the year are presented in range 0...1. So each day will be:
1 / 365 = 0.00273973
In a leap year there are 366 days, so the range will increase by 1 day:
1 / 365 * 366 = 1.00274
In short, in normal year the command returns in range 0...1 in a leap year it will return in range 0...1.00274 - Groups:
- Uncategorised
Syntax
Examples
- Example 1:
_float = dateToNumber [2035,7,6,12,0]; //0.510959
- Example 2:
dateToNumber date; //will return float number for the current date.
Additional Information
- See also:
- datedaytimetimenumberToDate
Notes
-
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