[lp_date_monthLastDay] at tagSwap.net: Open Source Lasso Code

[lp_date_monthLastDay]

Description

Link: [lp_date_monthLastDay]

Author: Bil Corry

Category: Date

Version: 8.x

License: Public Domain

Posted: Dec. 02, 2005

Updated: Jan. 01, 0001

More by this author...

Returns the last day of a given month. May specify the month and year either as a [date] or as -month and -year. Omitting the month chooses current month. Omitting the year choose current year.

Note: Requires [lp_date_leapyear]

Parameters

-date	date, optional	Find the last day of the month of the given date.
-month	integer, optional	Month number (1 - 12) to use, default is current month.
-year	integer, optional	Year (4-digit) to use, default is current year.

Sample Usage

lp_date_monthlastday: -month=2, -year=2000;

returns:

29

Source Code

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[

define_tag:'lp_date_monthLastDay',
	-description='Returns the last day of a given month.',
	-priority='replace',
	-optional='date',
	-optional='month',
	-optional='year';

	local:'month_selected'=(date)->month;
	local:'year_selected'=(date)->year;

	// get date
	if: local_defined:'date' && #date->type == 'date';
		#month_selected = #date->month;
		#year_selected = #date->year;
	/if;
	if: local_defined:'month' && (integer:#month) >= 1 && (integer:#month) <= 12;
		#month_selected = integer:#month;
	/if;
	if: local_defined:'year' && (integer:#year) >= 1000 && (integer:#year) <= 9999;
		#year_selected = integer:#year;
	/if;
	
	select: #month_selected;
		case: 1;
			return: 31;
		case: 2;
			if: (lp_date_leapyear: #year_selected);
				return: 29;
			else;
				return: 28;
			/if;
		case: 3;
			return: 31;
		case: 4;
			return: 30;
		case: 5;
			return: 31;
		case: 6;
			return: 30;
		case: 7;
			return: 31;
		case: 8;
			return: 31;
		case: 9;
			return: 30;
		case: 10;
			return: 31;
		case: 11;
			return: 30;
		case: 12;
			return: 31;
	/select;
	return: 0; // error
/define_tag;

]

Related Tags

[calendar_grid]

Comments

02/06/2006, Bil Corry
Possibly
In order to subtract one second, you would have to extract the month and year from the date passed, add one month (and possibly one year if December) to the date, make sure the time is set to midnight, then subtract some delta (I would recommend something safer than one second, like 12 hours), then extract the ->day and return it.

02/05/2006, deco rior
Another option
I believe a faster way is typically to get the first of the next month and subtract a second.