Date returns wrong Week and Year combination?

jimblos

Hi all,

I've got a date: 1136123520, which is "January 1, 2006, 1:52 pm".

Now i do:

date("W", $lastdate)."_".date("Y", $lastdate)

The result is reallllly strange:

52_2006

The exepected result should be:

52_2005

Is this a bug in php or am i'm missing something here?

I'm currently using php 4.x and can't upgrade to php 5.

Thanks you 🙂

Installer

I can't figure out what's strange about it. It's the behavior I would expect and hope for. The date is part of the last week in 2005, but the day itself is in 2006.

ClarkF1

I think it's a bug as the 2nd January 2006 is a Monday and shows up as 01_2006

Weeks start on a Monday in php

Weedpacket

jimblos wrote:
The exepected result should be:

Why would you expect the year that 1136123520 occurs in to be 2005 when you yourself say that it's 2006? 😉

Weedpacket

ClarkF1 wrote:
Weeks start on a Monday in php

Which means that the previous day (January 1) belongs to the previous week - the 52nd week of 2005, to be precise. So, what "bug" are you talking about?

And it's not just "in php". It's standard ISO-8601.

ClarkF1

The ISO standard definition:

Mutually equivalent definitions for week 01 are:

* the week with the year's first Thursday in it
* the week with the year's first working day in it (if Saturdays, Sundays, and 1 January are no working days)
* the week with January 4 in it
* the first week with the majority (four or more) of its days in the starting year
* the week starting with the Monday in the period 29 December - 4 January
* the week with the Thursday in the period 1 - 7 January
* If 1 January is on a Monday, Tuesday, Wednesday or Thursday, it is in week 01. If 1 January is on a Friday, Saturday or Sunday, it is in week 52 or 53 of the previous year.

The last line of that definition indicates that it should be 52_2005 as the OP suggested. It was a Sunday this year.

Installer

The last line of that definition indicates that it should be 52_2005 as the OP suggested.

The last line says "week 52...of the previous year". So extracting the current year should give you......the current year.

Weedpacket

Actually, ISO8601 only uses the first definition, and mentions the third in a note. The rest was compiled on Wikipedia....

jimblos

Thanks for the replies people.

Expecting 52_2005 if the date is "January 1, 2006, 1:52 pm" is indeed a little strange, but 52_2006 means week 52 in year 2006 and January 1 2006 isn't week 52 in the year 2006 🙂

I read the docs and saw that php5 has the likely solution to my problem:

date() function
o
ISO-8601 year number.
This has the same value as Y, except that if the ISO week number (W) belongs to the previous or next year, that year is used instead. (added in PHP 5.1.0)

But i'm not using php5 :bemused: Is there anyway i can get this thing to work in php4?

Installer

$yr = date('Y', $ts);
if ($yr != date('Y', strtotime('-2 weeks', $ts))) {
    $yr--;
}

Weedpacket

You could look at the month; since January isn't itself 52 weeks long....

if($week>51 && $month==1)
	$year--;
else if ($week==1 && $month==12)
	$year++;

Installer

Much better.