dateToNumber: Difference between revisions
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Lou Montana (talk | contribs) m (Text replacement - "\|seealso= *\[\[([^ ]+)\]\], \[\[([^ ]+)\]\]" to "|seealso= $1 ") |
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[[hint]] [[str]] ([[date]] [[call]] fnc_daysFromEpoc);</code> | [[hint]] [[str]] ([[date]] [[call]] fnc_daysFromEpoc);</code> | ||
|seealso= [[date]] [[dayTime]] | |seealso= [[date]] [[dayTime]] [[time]], [[numberToDate]] | ||
}} | }} |
Revision as of 14:56, 20 January 2022
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 a normal year the command returns in range 0...1 in a leap year it will return in range 0...1.00274 - Groups:
- Environment
Syntax
- Syntax:
- dateToNumber date
- Parameters:
- date: Array - array in date format
- Return Value:
- Number
Examples
- Example 1:
_float = dateToNumber [2035,7,6,12,0]; //0.510959
- Example 2:
dateToNumber date; //will return float number for the current date.
- Example 3:
- Calculate days from 1/1/1970:
fnc_daysFromEpoc = { private _year = param [0]; private _days = 0; for "_i" from 1970 to _year - 1 do { _days = _days + round linearConversion [0, 1, dateToNumber [_i, 12, 31, 23, 59], 0, 365, false]; }; _days + linearConversion [0, 1, dateToNumber _this, 0, 365, false]; }; hint str (date call fnc_daysFromEpoc);
Additional Information
- See also:
- date dayTime timenumberToDate
Notes
-
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