I am try to get the year and week of year off of a given date in my code:
$dueDate->format('W , Y');
In the code above, duedate is a datetime object with this date value:
December 31, 2018
When I output the format I specified above, I get this:
01 , 2018
Looking at each value separately the function is correct. However, together it is confusing.
It seems to be reading December 31st as the first week because it falls on a Monday, so technically it is right, it is the first week of 2019. In that case though, I would want the year to roll over and read 2019.
How can I resolve this to roll over the year in this case only? Any help is appreciated.
You need to use the ISO-8601 week numbering year which is o if you want the year for the ISO-8601 week. From the docs:
ISO-8601 week-numbering year. 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)
$dueDate->format('W , o');
I am trying to determine if a given date is the last day of that type (Sunday, Monday, Tuesday, etc.) in its month. For example, for March of 2016, the last Friday of the month is the 25th.
Please note I am not trying to figure out what the last day of the month is. Also, 'weekday' in the question refers to any of the 7 days, not just non-weekend days.
I am using excel and php as tags because I can work with either approach.
Just determine how close the given date is to the end of the month. For the given date in A1, use:
=IF(DATE(YEAR(A1),MONTH(A1)+1,0)-A1<7,TRUE,FALSE)
Can someone explain why this is so?
<?php
echo date("W", strtotime("2015-01-01"))."<br>"; //returns 1
echo date("W", strtotime("2014-01-01"))."<br>"; //returns 1
echo date("W", strtotime("2016-01-01"))."<br>"; //returns 53
?>
When outputing the week number of the day in the month , result is incorrect for some cases. Here the output, week number is 1 which is true for first two cases but not in the third case.
The reason it outputs 53 and not 1 for "2016-01-01" is that it counts the first week of the year from the first monday of the year which in the year 2016 lies on the 4th of Jan ("2016-01-04").
W ISO-8601 week number of year, weeks starting on Monday (added in PHP
4.1.0) Example: 42 (the 42nd week in the year)
See full documentation for php date function here.
As mentioned, W ISO-8601 week number of year is the number that you get.
For more information on W ISO-8601 week number of year see below,
An ISO week-numbering year (also called ISO year informally) has 52 or
53 full weeks. That is 364 or 371 days instead of the usual 365 or 366
days. The extra week is referred to here as a leap week, although ISO
8601 does not use this term. Weeks start with Monday. The first week
of a year is the week that contains the first Thursday of the year
(and, hence, always contains 4 January). ISO week year numbering
therefore slightly deviates from the Gregorian for some days close to
1 January.
Source: Wiki
2016-01-01 belongs to the last week of year 2015, as the week begins in that year, so the result is correct.
As the headline says, PHP's date("W") function gives back the calendar week (for the current day). Unfortunatly it gives back 52 or 53 for the first day(s) of most years. This is, in a simple thinking way, correct, but very annoying, as January 1st of 2012 is not calendar week 52, it's NOT a calendar week of the current year. Most calendars define this as week 0 or week 52 of the previous year.
This is very tricky when you group each day of the year by their calendar week: 1st of January 2012 and 31st of December 2012 are both put into the same calendar week group.
So my question is: Is there a (native) year-sensitive alternative to PHP's date("W") ?
EDIT: I think I wrote the first version of this question in a very unclear way, so this is my edit: I'm searching for a function that gives back the correct calendar week for the first day(s) of the year. PHP's date("W") gives back 52 for the 1st of January 2012, which is "wrong". It should be 0 or null. According to official sources, the first calendar week of a year starts on the first monday of the year. So, if the first day of a year is not a monday, it's not week 1 ! It's week 0. The wikipedia article says
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.
This becomes tricky as the last days of the year are also in week 52/53. date("W") does not divide into current year and previous year.
This solution converts the excess of december to week 53 and everything in january prior to week 1 to week 0.
$w=(int)date('W');
$m=(int)date('n');
$w=$w==1?($m==12?53:1):($w>=51?($m==1?0:$w):$w);
echo "week $w in ".date('Y');
2013-12-31 ==> week 53 in 2013
2014-01-01 ==> week 1 in 2014
2015-12-31 ==> week 52 in 2015
2016-01-01 ==> week 0 in 2016
And a small test run, so you can see for yourself ;-)
$id=array(25,26,27,28,29,30,31,1,2,3,4,5,6,7,8);
for($iy=2013;$iy<2067;++$iy){foreach($id as $k=>$v){if($k<7){$im=12;}else{$im=1;}
if($k==7){++$iy;echo '====<br>';}$tme=strtotime("$im/$v/$iy");
echo date('d-m-Y',$tme),' * * ';
//THE ACTUAL CODE =================
$w=(int)date('W',$tme);
$m=(int)date('n',$tme);
$w=$w==1?($m==12?53:1):($w>=51?($m==1?0:$w):$w);
//THE ACTUAL CODE =================
echo '<b>WEEK: ',$w,' --- ','YEAR: ',date('Y',$tme),'</b><br>';}--$iy;
echo '----------------------------------<br>';}
Is there a (native) year-sensitive alternative to PHP's date("W") ?
No, there isn't.
According to official sources, the first calendar week of a year starts on the first monday of the year.
I'm not sure what official sources you're referring to.
PHP's date("W") returns the week number according to ISO 8601. As an international standard, ISO 8601 counts as one of possibly many "official sources". If its definition of week numbers doesn't fit your application, you're free to use anything else you like.
If you use a non-standard definition of "first week of the year", or if you use an official source that's not widely recognized, expect to have to write your own function to replace date("W"). (I'm pretty sure you'll need to write a function.)
The date 2012-01-01 was a Sunday. ISO 8601, Wikipedia, and php agree that the ISO week number for 2012-01-01 is 52.
ISO 8601 doesn't define a week 0.
So, if the first day of a year is not a monday, it's not week 1 !
Neither ISO nor Wikipedia say that. ISO 8601 defines week number 1 as the week that has the year's first Thursday in it. For 2012, the first Thursday was on Jan 5, so week number 1 was Jan 2 to Jan 8. 2012-01-01 was in the final week of the previous year, in terms of ISO weeks.
If you want something different, you can play with arithmetic, division, and so on. (Try dividing date("z") by 7, for example.) Or you can store that data in a database, and have your weeks any way you like.
If you're dealing with accounting periods, I'd almost certainly store that data in a table in a database. It's pretty easy to generate that kind of data with a spreadsheet.
The text of data in a table is much easier to audit than the text of a php function, no matter how simple that function is. And the data is certain to be the same for any program that accesses it, no matter what language it's written in. (So if your database someday has programs written in 5 different languages accessing it, you don't have to write, test, and maintain 5 different functions to get the week number.)
$d = new DateTime('first monday january '.date('Y'));
echo $d->format("W");
Google brought me here, and I wanted to post the following to help others like me...
I am in the US, and use DayPilot, and it works as follows:
Week starts on Sun, not Mon.
Jan 1st is always Week 1.
If Jan 1st is not a Sunday, Week 1 is less than 7 days.
This all makes a lot of since to me!
Here is my PHP function to copy that behavior:
function ProperWeekNum($inDate)
{$outNum = $inDate->format('W');
//Make week start on Sunday
if ($inDate->format('D') == 'Sun') {$outNum++;}
//Fix begining of year
if (($outNum >= 52) && ($inDate->format('M') == 'Jan')) {$outNum = 1;}
//Fix WEEK #1 is 1-day (Sat)
else //...without this 2022 was off by 1 all year
{$jan1st = new DateTime($inDate->format('Y').'-01-01');
if ($jan1st->format('D') == 'Sat') {$outNum++;}
}
//Return without leading zero
return ltrim($outNum, '0');
}
I use the function as follows, so when I click on DayPilot, my custom popup's Week # always matches DayPilot's Week #:
$weeknum = ProperWeekNum($startdate);
if ($weeknum != ProperWeekNum($enddate))
{$weeknum .= '-'.ProperWeekNum($enddate);}
Probably won't help the OP, but hopefully it helps someone.
echo date("W",strtotime('2010-01-01'));
This outputs 53. I would expect it to output 1. Can anyone else confirm this behavior, or maybe explain why? I couldn't find a bug report on it.
This isn't a bug at all, it's expected behaviour. From PHP's Date Page:
W: ISO-8601 week number of year, weeks starting on Monday
Jan 1, 2010, fell on a Friday, so its week number would belong to 2009, making it part of the 53rd week of 2009. Jan 4, 2010 would be week 1.
A week which begins in December and ends in January the following year belongs to the year where most of its days lie. Therefore a week with at least 4 days in December is the last week of that year and a week with at least 4 days in January is the first week in the new year.
So... the last week of a year always contains the 28th day of December. If you take date("W") on that day of a given year you always get the correct number of weeks for that year. The other end of that definition is that the 4th day of January always lies in the first week of a year.
It returns the ISO-8601 week number of year.
From Wikipedia:
There are mutually equivalent descriptions of week 01:
the week with the year's first Thursday in it (the formal ISO definition),
the week with 4 January in it,
the first week with the majority (four or more) of its days in the starting year, and
the week starting with the Monday in the period 29 December – 4 January.
Since 2010-01-01 was a Friday, non of the conditions is met.