Time/Date conversion year/week -> day/month/year

[Jezternz]

Okay. Simple as it sounds, but possibly complex.

I have $week (which is the week in a year 1-52) and I have $year.
I want to get the day of the saturday in the given $week and $year, so i want returned:
$day (being day of month, will be the saturday of he origonal given week), $month (the month 0-12) and $year.

Bit of a brain tease for me atleast. Any Idea's?

[Jezternz]

Okay found the problem after massive amounts of debugging, it was daylight savings, gah, drives me crazy. anyway, to solve this I just added 12 hours of seconds onto the end (12*60*60)

cheerz

[ronverdonk]

After some searching I found a VERY old (20 years?) C routine that I once wrote in the stone age. I managed to recode it to somehow acceptable PHP and the result is here.

P.S. It works for any week > 2 and < 54 and any year within the strtotime range. It also uses monday as the start of the week. But with some alteration it will do.[php]<?php
$monthtab=array (31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334, 365);

function showDayAndMonth ($dayno) {
global $monthtab;
$date_arr=array ();
for ($j=0; $j<12; $j++) {
if ($dayno <= $monthtab[$j]) {
$date_arr['day']=$dayno-$monthtab[$j-1];
$date_arr['mth']=++$j;
return $date_arr;
}
}
}

$year=2008;
$week=18;

$monday=0-(date("N", strtotime("1 January $year"))-2);
if ($year % 4 == 0)
for ($i=1; $i<12; $i++)
$monthtab[$i]++;
if ($week > 1 AND $week < 54) {
$week--; /* -1 voor offset */
$day_fr=($week * 7) + $monday; /* Dagnummer van weekstart */
$day_to=$day_fr +6; /* Dagnummer van weekeind */
$date_fr=showDa yAndMonth($day_ fr);
$date_to=showDa yAndMonth($day_ to);
}
else {
die ("Invalid week number");
}
echo "REQUESTED: week $week, year $year";
echo "<br>STARTD ATE: ".date("l F j, Y",strtotime($d ate_fr['day'].'-'.$date_fr['mth'].'-'.$year));
echo "<br>ENDDAT E: ".date("l F j, Y",strtotime($d ate_to['day'].'-'.$date_to['mth'].'-'.$year));
?>[/php]I can explain the code if anyone is interested.

Ronald

[TheServant]

Wow, that is really clever. Good job Ron!

[Jezternz]

Yeh that is impressive Ron, thanks heaps.
One thing I do wonder about though is wht would happen when february has 29 days not 28 :S.
$monthtab=array (31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334, 365);
But regardless, this is far further then I got, thanks heaps.

[TheServant]

Maybe just include an if statement with something like:

[PHP]if ($year == "is not a leap year") {
$monthtab=array (31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334, 365);
} elseif ($year == "is a leap year") {
$monthtab=array (31, 60, 91, 121, 152, 182, 213, 244, 274, 305, 336, 366);
}[/PHP]

You just nee to get a statement to define "is not a leap year". So probably something like divide the year by 4, make it a string, and search the string for a "." and if it is found, it is not a leap year. Then you also need to check if it is the 400th year, so if it can be divided by 400, if so it is also not a leap year.

[ronverdonk]

NO sir! Leap year handling is in the code already, otherwise it would not have worked for 2008! The leap year handling is done in the following statements in the code[php]if ($year % 4 == 0)
for ($i=1; $i<12; $i++)
$monthtab[$i]++;[/php]
and you don't want to worry about 400-year cycle because the previous one was 1600 and the next one is 2400. And the strtotime routine does not go that far (just until 2032(?).

Ronald

[TheServant]

Good point. I forgot the dates don't go back that far! And I didn't realize that it only goes upto 2032?! That's no good. Means that we will have to recode with a new date function :P

Ron, what does the line: if ($year % 4 == 0) mean?

[ronverdonk]

Ron, what does the line: if ($year % 4 == 0) mean?

This uses the modulo operator. The modulo operation finds the remainder of division of one number by another.

Given two numbers, $year (the dividend) and 4 (the divisor), a modulo 4 is the remainder, on division of $year by 4. For instance, the expression "2008 mod 4" would evaluate to 0, while "2007 mod 4" would evaluate to 3.

So when the modulo 4 of year is zero, it is a leap year.

Ronald

[TheServant]

This uses the modulo operator. The modulo operation finds the remainder of division of one number by another.

Given two numbers, $year (the dividend) and 4 (the divisor), a modulo 4 is the remainder, on division of $year by 4. For instance, the expression "2008 mod 4" would evaluate to 0, while "2007 mod 4" would evaluate to 3.

So when the modulo 4 of year is zero, it is a leap year.

Ronald

So:
[PHP]5 %4 = 0.25
4 %4 = 0
26 %6 = 0.333
30 %6 = 0[/PHP]Just making sure I understand, because this is useful!

[ronverdonk]

No sir, the modulus is the remainder of the division, e.g. 4 mod 3 = 1 because when you divide 4 by 3 the remainder is 1. So:

5 % 4 = 1
4 % 4 = 0
26 % 6 = 2
30 % 6 = 0

E.g. n mod 2 is often used to 'flip' between 2-entry indices or determine if a number is even or uneven, i.e. 1%2 = 1, 8%2 = 0, 133%2=1, 8490%2=0, etc.

Ronald