Last Friday of each month
Write a program or a script that returns the last Fridays of each month of a given year. The year may be given through any simple input method in your language (command line, std in, etc.).
Example of an expected output:
./last_fridays 2012 2012-01-27 2012-02-24 2012-03-30 2012-04-27 2012-05-25 2012-06-29 2012-07-27 2012-08-31 2012-09-28 2012-10-26 2012-11-30 2012-12-28
- Cf.
Ada
<lang Ada>with Ada.Text_IO; use Ada.Text_IO; with Ada.Calendar; use Ada.Calendar; with Ada.Command_Line; use Ada.Command_Line; with Ada.Calendar.Formatting; use Ada.Calendar.Formatting; procedure Fridays is
T : Ada.Calendar.Time;
Year : Year_Number := Integer'Value (Argument (1));
MLength : constant array (1 .. 12) of Positive :=
(4 | 6 | 9 | 11 => 30, 2 => 28, others => 31);
begin
for month in 1 .. 12 loop
for day in reverse 1 .. MLength (month) loop
T := Ada.Calendar.Time_Of (Year, month, day);
if Day_Of_Week (T) = Friday then
Put_Line (Image (Date => T)(1 .. 10)); exit;
end if;
end loop;
end loop;
end Fridays;</lang>
- Output:
./fridays 2012 2012-01-27 2012-02-24 2012-03-30 2012-04-27 2012-05-25 2012-06-29 2012-07-27 2012-08-31 2012-09-28 2012-10-26 2012-11-30 2012-12-28
AutoHotkey
<lang AHK>if 1 = ; no parameter passed { InputBox, 1, Last Fridays of year, Enter a year:, , , , , , , , %A_YYYY% If ErrorLevel ExitApp }
YYYY = %1% ; retrieve command line parameter Stmp = %YYYY%0101000000 count= 0
While count < 12 { FormatTime, ddd, %stmp%, ddd FormatTime, M, %stmp%, M If (ddd = "Fri"){ if (M-1 = count){ t := stmp stmp += 7, days } else res .= SubStr(t, 1, 4) "-" SubStr(t, 5, 2) "-" SubStr(t, 7, 2) "`n" ,count++ ,stmp := YYYY . SubStr("0" M, -1) . "01" } else stmp += 1, days } MsgBox % res</lang> Output for 2012:
2012-01-27 2012-02-24 2012-03-30 2012-04-27 2012-05-25 2012-06-29 2012-07-27 2012-08-31 2012-09-28 2012-10-26 2012-11-30 2012-12-28
C
Doesn't work with Julian calendar (then again, you probably don't need to plan your weekends for middle ages).
<lang c>#include <stdio.h>
- include <stdlib.h>
int main(int c, char *v[]) { int days[] = {31,29,31,30,31,30,31,31,30,31,30,31}; int m, y, w;
if (c < 2 || (y = atoi(v[1])) <= 1700) return 1;
days[1] -= (y % 4) || (!(y % 100) && (y % 400));
w = y * 365 + (y - 1) / 4 - (y - 1) / 100 + (y - 1) / 400 + 6;
for(m = 0; m < 12; m++) { w = (w + days[m]) % 7; printf("%d-%02d-%d\n", y, m + 1, days[m] + (w < 5 ? -2 : 5) - w); }
return 0; }</lang>
C++
called with ./last_fridays 2012
<lang cpp>#include <boost/date_time/gregorian/gregorian.hpp>
- include <iostream>
- include <cstdlib>
int main( int argc , char* argv[ ] ) {
using namespace boost::gregorian ;
greg_month months[ ] = { Jan , Feb , Mar , Apr , May , Jun , Jul ,
Aug , Sep , Oct , Nov , Dec } ;
greg_year gy = atoi( argv[ 1 ] ) ;
for ( int i = 0 ; i < 12 ; i++ ) {
last_day_of_the_week_in_month lwdm ( Friday , months[ i ] ) ;
date d = lwdm.get_date( gy ) ;
std::cout << d << std::endl ;
}
return 0 ;
}</lang> Output:
2012-Jan-27 2012-Feb-24 2012-Mar-30 2012-Apr-27 2012-May-25 2012-Jun-29 2012-Jul-27 2012-Aug-31 2012-Sep-28 2012-Oct-26 2012-Nov-30 2012-Dec-28
CoffeeScript
<lang coffeescript> last_friday_of_month = (year, month) ->
# month is 1-based, JS API is 0-based, then we use
# non-positive indexes to work backward relative to the
# first day of the next month
i = 0
while true
last_day = new Date(year, month, i)
if last_day.getDay() == 5
return last_day.toDateString()
i -= 1
print_last_fridays_of_month = (year) ->
for month in [1..12] console.log last_friday_of_month year, month
do ->
year = parseInt process.argv[2] print_last_fridays_of_month year
</lang> output <lang> > coffee last_friday.coffee 2012 Fri Jan 27 2012 Fri Feb 24 2012 Fri Mar 30 2012 Fri Apr 27 2012 Fri May 25 2012 Fri Jun 29 2012 Fri Jul 27 2012 Fri Aug 31 2012 Fri Sep 28 2012 Fri Oct 26 2012 Fri Nov 30 2012 Fri Dec 28 2012 </lang>
D
<lang d>import std.stdio, std.datetime, std.traits;
void main() {
lastFridays(2012);
}
void lastFridays(in uint year) {
auto date = Date(year, 1, 1);
foreach (_; EnumMembers!Month) {
date.day(date.daysInMonth);
date.roll!"days"(- (date.dayOfWeek + 2) % 7);
writeln(date);
date.add!"months"(1, AllowDayOverflow.no);
}
}</lang>
2012-Jan-27 2012-Feb-24 2012-Mar-30 2012-Apr-27 2012-May-25 2012-Jun-29 2012-Jul-27 2012-Aug-31 2012-Sep-28 2012-Oct-26 2012-Nov-30 2012-Dec-28
Go
<lang go>package main
import (
"fmt" "os" "strconv" "time"
)
func main() {
y := time.Now().Year()
if len(os.Args) == 2 {
if i, err := strconv.Atoi(os.Args[1]); err == nil {
y = i
}
}
for m := time.January; m <= time.December; m++ {
d := time.Date(y, m+1, 1, 0, 0, 0, 0, time.UTC).Add(-24 * time.Hour)
d = d.Add(-time.Duration((d.Weekday() + 2) % 7) * 24 * time.Hour)
fmt.Println(d.Format("2006-01-02"))
}
}</lang> Output:
> ./fridays 2012 2012-01-27 2012-02-24 2012-03-30 2012-04-27 2012-05-25 2012-06-29 2012-07-27 2012-08-31 2012-09-28 2012-10-26 2012-11-30 2012-12-28
Icon and Unicon
This will write the last fridays for every year given as an argument. There is no error checking on the year.
<lang Icon>procedure main(A) every write(lastfridays(!A)) end
procedure lastfridays(year) every m := 1 to 12 do {
d := case m of {
2 : if IsLeapYear(year) then 29 else 28
4|6|9|11 : 30
default : 31
} # last day of month
z := 0
j := julian(m,d,year) + 1 # first day of next month
until (j-:=1)%7 = 4 do z -:=1 # backup to last friday=4
suspend sprintf("%d-%d-%d",year,m,d+z)
}
end
link datetime, printf</lang>
printf.icn provides formatting datetime.icn provides julian and IsLeapYear
Output:
last_fridays.exe 2012 2012-1-27 2012-2-24 2012-3-30 2012-4-27 2012-5-25 2012-6-29 2012-7-27 2012-8-31 2012-9-28 2012-10-26 2012-11-30 2012-12-28
J
<lang j>require'dates' last_fridays=: 12 {. [:({:/.~ }:"1)@(#~ 5 = weekday)@todate (i.366) + todayno@,&1 1</lang>
In other words, start from January 1 of the given year, and count forward for 366 days, keeping the fridays. Then pick the last friday within each represented month. Then pick the first 12 (since on a non-leap year which ends on a thursday we would get an extra friday).
Example use:
<lang j> last_fridays 2012 2012 1 27 2012 2 24 2012 3 30 2012 4 27 2012 5 25 2012 6 29 2012 7 27 2012 8 31 2012 9 28 2012 10 26 2012 11 30 2012 12 28</lang>
Java
<lang java5>import java.text.*; import java.util.*;
public class LastFridays {
public static void main(String[] args) throws Exception {
int year = Integer.parseInt(args[0]);
GregorianCalendar c = new GregorianCalendar(year, 0, 1);
for (String m : new DateFormatSymbols(Locale.US).getShortMonths()) {
int totalDaysOfMonth = c.getActualMaximum(Calendar.DAY_OF_MONTH);
c.set(Calendar.DAY_OF_MONTH, totalDaysOfMonth);
int daysToRollBack = (c.get(Calendar.DAY_OF_WEEK) + 1) % 7;
c.roll(Calendar.DAY_OF_MONTH, -daysToRollBack);
int d = c.get(Calendar.DAY_OF_MONTH);
System.out.printf("%d %s %d\n", year, m, d);
c.add(Calendar.DAY_OF_YEAR, daysToRollBack + 1);
}
}
}</lang>
Output (for java LastFridays 2012):
2012 jan 27 2012 Jan 27 2012 Feb 24 2012 Mar 30 2012 Apr 27 2012 May 25 2012 Jun 29 2012 Jul 27 2012 Aug 31 2012 Sep 28 2012 Oct 26 2012 Nov 30 2012 Dec 28
Mathematica
<lang Mathematica>Needs["Calendar`"] FridaysOfTheYear[Y_] := Cases[Map[{#,DayOfWeek[#]}&,DaysPlus[{Y,1,2}, #]&/@Range[365],{1}],List[x_,Friday]->x]; Last[SortBy[Cases[FridaysOfTheYear[2011], {_,#,_}], #3 &]]& /@ Range[12] // Column</lang> Output:
{2012,1,27}
{2012,2,24}
{2012,3,30}
{2012,4,27}
{2012,5,25}
{2012,6,29}
{2012,7,27}
{2012,8,31}
{2012,9,28}
{2012,10,26}
{2012,11,30}
{2012,12,28}
MATLAB / Octave
<lang Matlab> function t = last_fridays_of_year(y)
t1 = datenum([y,1,1,0,0,0]);
t2 = datenum([y,12,31,0,0,0]);
t = datevec(t1:t2);
t = t(strmatch('Friday', datestr(t,'dddd')), :); % find all Fridays
t = t([find(diff(t(:,2)) > 0); end], :); % find Fridays before change of month
end;
datestr(last_fridays_of_year(2012),'yyyy-mm-dd')
</lang>
Output:
ans = 2012-01-27 2012-02-24 2012-03-30 2012-04-27 2012-05-25 2012-06-29 2012-07-27 2012-08-31 2012-09-28 2012-10-26 2012-11-30 2012-12-28
OCaml
Using the module Unix from the standard OCaml library:
<lang ocaml>#load "unix.cma" open Unix
let usage() =
Printf.eprintf "%s <year>\n" Sys.argv.(0); exit 1
let print_date t =
Printf.printf "%d-%02d-%02d\n" (t.tm_year + 1900) (t.tm_mon + 1) t.tm_mday
let is_date_ok tm t =
(tm.tm_year = t.tm_year && tm.tm_mon = t.tm_mon && tm.tm_mday = t.tm_mday)
let () =
let _year =
try int_of_string Sys.argv.(1)
with _ -> usage()
in
let year = _year - 1900 in
let fridays = Array.make 12 (Unix.gmtime 0.0) in
for month = 0 to 11 do
for day_of_month = 1 to 31 do
let tm = { (Unix.gmtime 0.0) with
tm_year = year;
tm_mon = month;
tm_mday = day_of_month;
} in
let _, t = Unix.mktime tm in
if is_date_ok tm t (* check for months that have less than 31 days *)
&& t.tm_wday = 5 (* is a friday *)
then fridays.(month) <- t
done;
done;
Array.iter print_date fridays</lang>
Output:
$ ocaml last_fridays.ml 2012 2012-01-27 2012-02-24 2012-03-30 2012-04-27 2012-05-25 2012-06-29 2012-07-27 2012-08-31 2012-09-28 2012-10-26 2012-11-30 2012-12-28
With a dedicated library
<lang ocaml>open CalendarLib
let usage() =
Printf.eprintf "%s <year>\n" Sys.argv.(0); exit 1
let print_date (year, month, day) =
Printf.printf "%d-%02d-%02d\n" year month day
let () =
let year =
try int_of_string Sys.argv.(1)
with _ -> usage()
in
let fridays = ref [] in
for month = 1 to 12 do
let num_days = Date.days_in_month (Date.make_year_month year month) in
let rec aux day =
if Date.day_of_week (Date.make year month day) = Date.Fri
then fridays := (year, month, day) :: !fridays
else aux (pred day)
in
aux num_days
done;
List.iter print_date (List.rev !fridays)</lang>
Run this script with the command:
ocaml unix.cma str.cma -I +calendar calendarLib.cma last_fridays.ml 2012
Perl
<lang Perl>#!/usr/bin/perl -w use strict ; use DateTime ; use feature qw( say ) ;
foreach my $month ( 1..12 ) {
my $dt = DateTime->last_day_of_month( year => $ARGV[ 0 ] , month => $month ) ;
while ( $dt->day_of_week != 5 ) {
$dt->subtract( days => 1 ) ;
}
say $dt->ymd ;
}</lang> Output:
2012-01-27 2012-02-24 2012-03-30 2012-04-27 2012-05-25 2012-06-29 2012-07-27 2012-08-31 2012-09-28 2012-10-26 2012-11-30 2012-12-28
Perl 6
<lang perl6>sub MAIN (Int $year = Date.today.year) {
my @fri;
for Date.new("$year-01-01") .. Date.new("$year-12-31") {
@fri[.month] = .Str if .day-of-week == 5;
}
.say for @fri[1..12];
}</lang>
Example:
$ ./lastfri 2038 2038-01-29 2038-02-26 2038-03-26 2038-04-30 2038-05-28 2038-06-25 2038-07-30 2038-08-27 2038-09-24 2038-10-29 2038-11-26 2038-12-31
A solution without a result array to store things in:
<lang perl6>sub MAIN (Int $year = Date.today.year) {
say ~.value.reverse.first: *.day-of-week == 5
for classify *.month, Date.new("$year-01-01") .. Date.new("$year-12-31");
}</lang>
Here, classify sorts the dates into one bin per month (but preserves the order in each bin). We then take the list inside each bin (.value) and find the last (.reverse.first) date which is a Friday.
PHP
PHP is generally used for web apps, so I am not implementing the command-line component of this task.
<lang PHP><?php function last_friday_of_month($year, $month) {
$day = 0;
while(True) {
$last_day = mktime(0, 0, 0, $month+1, $day, $year);
if (date("w", $last_day) == 5) {
return date("Y-m-d", $last_day);
}
$day -= 1;
}
}
function print_last_fridays_of_month($year) {
foreach(range(1, 12) as $month) {
echo last_friday_of_month($year, $month), "
";
}
}
date_default_timezone_set("GMT"); $year = 2012; print_last_fridays_of_month($year); ?></lang>
Output in browser:
2012-01-27 2012-02-24 2012-03-30 2012-04-27 2012-05-25 2012-06-29 2012-07-27 2012-08-31 2012-09-28 2012-10-26 2012-11-30 2012-12-28
PicoLisp
<lang PicoLisp>(de lastFridays (Y)
(for M `(range 1 12)
(prinl
(dat$
(find '((D) (= "Friday" (day D)))
(mapcar '((D) (date Y M D)) `(range 31 22)) )
"-" ) ) ) )</lang>
Test: <lang PicoLisp>: (lastFridays 2012) 2012-01-27 2012-02-24 2012-03-30 2012-04-27 2012-05-25 2012-06-29 2012-07-27 2012-08-31 2012-09-28 2012-10-26 2012-11-30 2012-12-28</lang>
Pike
<lang Pike>int(0..1) last_friday(object day) {
return day->week_day() == 5 &&
day->month_day() > day->month()->number_of_days()-7;
}
int main(int argc, array argv) {
array days = filter(Calendar.Year((int)argv[1])->months()->days()[*], last_friday);
write("%{%s\n%}", days->format_ymd());
return 0;
}</lang>
Python
<lang python>import calendar c=calendar.Calendar() fridays={} year=raw_input("year") for item in c.yeardatescalendar(int(year)):
for i1 in item:
for i2 in i1:
for i3 in i2:
if "Fri" in i3.ctime() and year in i3.ctime():
month,day=str(i3).rsplit("-",1)
fridays[month]=day
for item in sorted((month+"-"+day for month,day in fridays.items()),
key=lambda x:int(x.split("-")[1])):
print item</lang>
Using reduce
<lang python>import calendar c=calendar.Calendar() fridays={} year=raw_input("year") add=list.__add__ for day in reduce(add,reduce(add,reduce(add,c.yeardatescalendar(int(year))))):
if "Fri" in day.ctime() and year in day.ctime():
month,day=str(day).rsplit("-",1)
fridays[month]=day
for item in sorted((month+"-"+day for month,day in fridays.items()),
key=lambda x:int(x.split("-")[1])):
print item</lang>
using itertools
<lang python>import calendar from itertools import chain f=chain.from_iterable c=calendar.Calendar() fridays={} year=raw_input("year") add=list.__add__
for day in f(f(f(c.yeardatescalendar(int(year))))):
if "Fri" in day.ctime() and year in day.ctime():
month,day=str(day).rsplit("-",1)
fridays[month]=day
for item in sorted((month+"-"+day for month,day in fridays.items()),
key=lambda x:int(x.split("-")[1])):
print item</lang>
REXX
This REXX program will find the last day-of-week (for any day) of all the months for any year. <lang rexx>/*REXX program displays dates of last Fridays of each month for any year*/
parse arg yyyy
do j=1 for 12
say lastDOW('Friday',j,yyyy)
end /*j*/
exit /*stick a fork in it, we're done.*/
/*┌────────────────────────────────────────────────---─────────────────┐
│ lastDOW: procedure to return the date of the last day-of-week of │ │ any particular month of any particular year. │ │ │ │ The day-of-week must be specified (it can be in any case, │ │ (lower-/mixed-/upper-case) as an English name of the spelled day │ │ of the week, with a minimum length that causes no ambiguity. │ │ I.E.: W for Wednesday, Sa for Saturday, Su for Sunday ... │ │ │ │ The month can be specified as an integer 1 ──► 12 │ │ 1=January 2=February 3=March ... 12=December │ │ or the English name of the month, with a minimum length that │ │ causes no ambiguity. I.E.: Jun for June, D for December. │ │ If omitted [or an asterisk(*)], the current month is used. │ │ │ │ The year is specified as an integer or just the last two digits │ │ (two digit years are assumed to be in the current century, and │ │ there is no windowing for a two-digit year). │ │ If omitted [or an asterisk(*)], the current year is used. │ │ Years < 100 must be specified with (at least 2) leading zeroes.│ │ │ │ Method used: find the "day number" of the 1st of the next month, │ │ then subtract one (this gives the "day number" of the last day of │ │ the month, bypassing the leapday mess). The last day-of-week is │ │ then obtained straightforwardly, or via subtraction. │ └───────────────────────────────────────────────────---──────────────┘*/
lastdow: procedure; arg dow .,mm .,yy . /*DOW = day of week*/ parse arg a.1,a.2,a.3 /*orig args, errmsg*/ if mm== | mm=='*' then mm=left(date('U'),2) /*use default month*/ if yy== | yy=='*' then yy=left(date('S'),4) /*use default year */ if length(yy)==2 then yy=left(date('S'),2)yy /*append century. */
/*Note mandatory leading blank in strings below.*/
$=" Monday TUesday Wednesday THursday Friday SAturday SUnday" !=" JAnuary February MARch APril MAY JUNe JULy AUgust September",
" October November December"
upper $ ! /*uppercase strings*/ if dow== then call .er "wasn't specified",1 if arg()>3 then call .er 'arguments specified',4
do j=1 for 3 /*any plural args ?*/ if words(arg(j))>1 then call .er 'is illegal:',j end
dw=pos(' 'dow,$) /*find day-of-week*/ if dw==0 then call .er 'is invalid:',1 if dw\==lastpos(' 'dow,$) then call .er 'is ambigious:',1
if datatype(mm,'month') then /*if MM is alpha...*/
do
m=pos(' 'mm,!) /*maybe its good...*/
if m==0 then call .er 'is invalid:',1
if m\==lastpos(' 'mm,!) then call .er 'is ambigious:',2
mm=wordpos(word(substr(!,m),1),!)-1 /*now, use true Mon*/
end
if \datatype(mm,'W') then call .er "isn't an integer:",2 if \datatype(yy,'W') then call .er "isn't an integer:",3 if mm<1 | mm>12 then call .er "isn't in range 1──►12:",2 if yy=0 then call .er "can't be 0 (zero):",3 if yy<0 then call .er "can't be negative:",3 if yy>9999 then call .er "can't be > 9999:",3
tdow=wordpos(word(substr($,dw),1),$)-1 /*target DOW, 0──►6*/
/*day# of last dom.*/
_=date('B',right(yy+(mm=12),4)right(mm//12+1,2,0)"01",'S')-1 ?=_//7 /*calc. DOW, 0──►6*/ if ?\==tdow then _=_-?-7+tdow+7*(?>tdow) /*not DOW? Adjust.*/ return date('weekday',_,"B") date(,_,'B') /*return the answer*/
.er: arg ,_;say; say '***error!*** (in LASTDOW)';say /*tell error, and */
say word('day-of-week month year excess',arg(2)) arg(1) a._
say; exit 13 /*... then exit. */</lang>
output when using the following input: 2012 or 12
Friday 27 Jan 2012 Friday 24 Feb 2012 Friday 30 Mar 2012 Friday 27 Apr 2012 Friday 25 May 2012 Friday 29 Jun 2012 Friday 27 Jul 2012 Friday 31 Aug 2012 Friday 28 Sep 2012 Friday 26 Oct 2012 Friday 30 Nov 2012 Friday 28 Dec 2012
Ruby
<lang ruby>require 'date'
def last_friday(year, month)
# Find end of month = beginning of month + 1 month - 1 day. d = Date.new(year, month, 1).>>(1) - 1 d -= (d.wday - 5) % 7 # Subtract days after Friday.
end
year = Integer(ARGV.shift) (1..12).each {|month| puts last_friday(year, month)}</lang>
Friday is d.wday == 5; the expression (d.wday - 5) % 7 counts days after Friday.
Using the ActiveSupport library for some convenience methods
<lang ruby>require 'rubygems' require 'activesupport'
def last_friday(year, month)
d = Date.new(year, month, 1).end_of_month until d.wday == 5 d = d.yesterday end d
end</lang>
Scala
<lang scala>import java.util.Calendar import java.text.SimpleDateFormat
object Fridays {
def lastFridayOfMonth(year:Int, month:Int)={
val cal=Calendar.getInstance
cal.set(Calendar.YEAR, year)
cal.set(Calendar.MONTH, month)
cal.set(Calendar.DAY_OF_WEEK, Calendar.FRIDAY)
cal.set(Calendar.DAY_OF_WEEK_IN_MONTH, -1)
cal.getTime
}
def fridaysOfYear(year:Int)=for(month <- 0 to 11) yield lastFridayOfMonth(year, month)
def main(args:Array[String]){
val year=args(0).toInt
val formatter=new SimpleDateFormat("yyyy-MMM-dd")
fridaysOfYear(year).foreach{date=>
println(formatter.format(date))
}
}
}</lang> Output:
2012-Jan-27 2012-Feb-24 2012-Mrz-30 2012-Apr-27 2012-Mai-25 2012-Jun-29 2012-Jul-27 2012-Aug-31 2012-Sep-28 2012-Okt-26 2012-Nov-30 2012-Dez-28
Tcl
<lang tcl>package require Tcl 8.5 set year [lindex $argv 0] foreach dm {02/1 03/1 04/1 05/1 06/1 07/1 08/1 09/1 10/1 11/1 12/1 12/32} {
# The [clock scan] code is unhealthily clever; use it for our own evil purposes set t [clock scan "last friday" -base [clock scan $dm/$year -gmt 1] -gmt 1] # Print the interesting part puts [clock format $t -format "%Y-%m-%d" -gmt 1]
}</lang> Sample execution:
$ tclsh8.5 lastfri.tcl 2012 2012-01-27 2012-02-24 2012-03-30 2012-04-27 2012-05-25 2012-06-29 2012-07-27 2012-08-31 2012-09-28 2012-10-26 2012-11-30 2012-12-28
TUSCRIPT
<lang tuscript> $$ MODE TUSCRIPT year=2012 LOOP month=1,12
LOOP day=31,22,-1 dayofweek=DATE (number,day,month,year,nummer) IF (dayofweek==5) THEN PRINT year,"-",month,"-",day EXIT ENDIF ENDLOOP
ENDLOOP </lang> Output:
2012-1-27 2012-2-24 2012-3-30 2012-4-27 2012-5-25 2012-6-29 2012-7-27 2012-8-31 2012-9-28 2012-10-26 2012-11-30 2012-12-28
UNIX Shell
Using ncal. Will switch to Julian calender as ncal sees fit, and will not calculate past year 9999 (chances are you'll be too dead by then to worry about weekends anyway).
<lang bash>#!/bin/sh
if [ -z $1 ]; then exit 1; fi
- weed out multiple erros due to bad year
ncal 1 $1 > /dev/null && \ for m in 01 02 03 04 05 06 07 08 09 10 11 12; do echo $1-$m-`ncal $m $1 | grep Fr | sed 's/.* \([0-9]\)/\1/'` done</lang>
Using date --date from GNU date??? This code is not portable.
<lang bash>#!/bin/sh
- Free code, no limit work
- $Id: lastfridays,v 1.1 2011/11/10 00:48:16 gilles Exp gilles $
- usage :
- lastfridays 2012 # prints last fridays of months of year 2012
debug=${debug:-false}
- debug=true
epoch_year_day() { #set -x x_epoch=`expr ${2:-0} '*' 86400 + 43200` date --date="${1:-1970}-01-01 UTC $x_epoch seconds" +%s }
year_of_epoch() { date --date="1970-01-01 UTC ${1:-0} seconds" +%Y } day_of_epoch() { LC_ALL=C date --date="1970-01-01 UTC ${1:-0} seconds" +%A } date_of_epoch() { date --date="1970-01-01 UTC ${1:-0} seconds" "+%Y-%m-%d" } month_of_epoch() { date --date="1970-01-01 UTC ${1:-0} seconds" "+%m" }
last_fridays() { year=${1:-2012}
next_year=`expr $year + 1`
$debug && echo "next_year $next_year"
current_year=$year
day=0
previous_month=01
while test $current_year != $next_year; do
$debug && echo "day $day"
current_epoch=`epoch_year_day $year $day`
$debug && echo "current_epoch $current_epoch"
current_year=`year_of_epoch $current_epoch`
current_day=`day_of_epoch $current_epoch`
$debug && echo "current_day $current_day"
test $current_day = 'Friday' && current_friday=`date_of_epoch $current_epoch`
$debug && echo "current_friday $current_friday"
current_month=`month_of_epoch $current_epoch`
$debug && echo "current_month $current_month"
# Change of month => previous friday is the last of month
test "$previous_month" != "$current_month" \
&& echo $previous_friday
previous_month=$current_month
previous_friday=$current_friday
day=`expr $day + 1`
done
}
- main
last_fridays ${1:-2012}</lang>
Sample execution:
lastfridays 2012 2012-01-27 2012-02-24 2012-03-30 2012-04-27 2012-05-25 2012-06-29 2012-07-27 2012-08-31 2012-09-28 2012-10-26 2012-11-30 2012-12-28