Get week number

Anon

I need the week number of a year. Can anyone help me? I haven't got a clue how to do this...

Fredrik

Anon

You'd probably have to do some math stuff to work it out.

If date("d"); returns the day of the month like 03.....and date("m"); returns the month as a number then you could do some functions like:

day = date("d");
month = date("m");

if(($month == "1") && ($day <= "07"))
{
$week = "1";
}

Perhaps this would take a while to code though...since there would be 52 to cope with?

rgds,
scott d~

Anon

okay.
first you need to settle on your definition of 'week'; sunday to sat or mon to sun.
then you need to decide whether, if the new year is midweek, that week is week one, or the first full week is week one.

once you've got that, take the day of the year ('z'), subtract an offset equal to the day of the year of the start of week 1 (which, depending on your choices, may be negative), and divide the result by 7 and ceil() it.

if treated with care this should give you the answer you're looking for 🙂

--
rad

Anon

Found this:

<?php
$month = 9;
$day = 8;
$year = 2000;

print strftime("%W",mktime(0,0,0,$month,$day,$year));
?>

Anon

To get a weak number you need to tire out a strong number until it is exhausted and then it's sure to be weak! That was easy!

Just some friendly humor to mix up the soup, LOL all in fun!

Anon

Well it's the easy way, but there is a problem with it:

It works fine for 2001, because the first day of the first week of that year happens to be a MONDAY.

But when we get to 2002 the first day of the first week is a TUESDAY.

Now strftime("%W",mktime(0,0,0,$01,$01,$2002));

returns weeknumber '0' ..not '1'

only the first FULL week of a year (beginning on a monday) will start the weekcounter at 1.

So if you find it wrong to call the first week of the year 2002 'week 0' - and my calender calls it week 1 - then you have a problem. And it offsets all the weeks of that year by one ofcourse... I guess you have to compute the weeknumber in your own routine... I looked for the solution for a while, but havent found one, other than to do the math. (but the computer will do that VERY fast 😉

here's an quick idea I just got (but its not thought through or tested):

For every year it goes:
Either the first day of the year is a monday, and then we can use strftime(%W)

Else the first day of the year is NOT a monday: in that case $weeknumber = strftime(%W)+1

It will never be offset more than 1

.. i guess?

Anon

// getting the weeknumber

//getting current year
$year = date("Y");
// test what weeknumber is suggested for the firstweek this year
$testweek = strftimr("%W",mktime(0,0,0,01,01,$year));

// adjusting for offset
if ($testweek == 0) {
$weeknumber = strftimr("%W")+1
}else{
$weeknumber = strftimr("%W");
}

// now $weeknumber is ready for use
// I havent tested this.

Anon

Now tested: IT WORKS.

I made a typo above in 'strftimr' should have beed 'strftime'

$testweek = strftime("%W",mktime(0,0,0,01,01,$testyear));

Anon

stupid

Anon

Sorry people, it is NOT the way!!

Some years have 53 weeks!!!!
No, I am not kidding!

There is a standard for calculating weeknumbers, but I don't know them exatly ;(

I have a problem with it to.. so, does anybody know the right way??

Tnx allready! 😉

Grtz DePaasHaas

Anon

OH! Here I am again!

I found a JavaScript that calculates the week number correctly.. wait a sec. I'll rewrite it for php :_)

Oef.. harder than i thought..

OH.. it isn't a good function..

This one works perfect!!
100% bug free!

(monday is the start of new week!)

function is_leap_year($year)
{
if ((($year % 4) == 0 and ($year % 100)!=0) or ($year % 400) == 0)
return 1;
else
return 0;
}

function iso_week_days($yday, $wday) {
return $yday - (($yday - $wday + 382) % 7) + 3;
}

function get_week_number($timestamp) {
$d = getdate($timestamp);
$days = iso_week_days($d["yday"], $d["wday"]);

if ($days < 0) {
$d[ "yday"] += 365 + is_leap_year(--$d["year"]);
$days = iso_week_days($d["yday"], $d["wday"]);
}
else {
$d["yday"] -= 365 + is_leap_year($d["year"]);
$d2 = iso_week_days($d["yday"], $d["wday"]);
if (0 <= $d2)
$days = $d2;
}

return (int)($days / 7) + 1;
}

aeon_

How cool! This one really works.
Thank you!