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 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');
This question already has an answer here:
date_create_from_format() returns wrong value. [duplicate]
(1 answer)
Closed 6 years ago.
I am trying to create data from format using Datetime Class as following
$date = DateTime::createFromFormat('m-Y', '02-2016');
echo $date->format('Y-m-d');
Out put 2016-03-01, i was expecting to get 2016-02-01
is it a bug? or i am understating this function in a wrong way?
If you provide month and year, then PHP will provide a default day, using the current day of the current month (ie 30) today
That gives an effective 2016-02-30, which isn't a valid date..... but PHP allows these types of values where days is higher/lower than days in month, months are higher/lower than months in the year, and just increments/decrements to a valid date..... in this case, 1 additional day after the last day in month 2 or 2016 (29th February 2016) to give 1st March 2016
EDIT
Reference for day/month overflow/underflow behaviour
Note:
It is possible to over- and underflow the dd and DD format. Day 0 means the last day of previous month, whereas overflows count into the next month. This makes "2008-08-00" equivalent to "2008-07-31" and "2008-06-31" equivalent to "2008-07-01" (June only has 30 days).
It is also possible to underflow the mm and MM formats with the value 0. A month value of 0 means December of the previous year. As example "2008-00-22" is equivalent to "2007-12-22".
If you combine the previous two facts and underflow both the day and the month, the following happens: "2008-00-00" first gets converted to "2007-12-00" which then gets converted to "2007-11-30". This also happens with the string "0000-00-00", which gets transformed into "-0001-11-30" (the year -1 in the ISO 8601 calendar, which is 2 BC in the proleptic Gregorian calendar).
It's an known bug... or a documented feature, depending on how you look at it. See Mark Baker's answer for more details.
Workaround :
To get the date you expect, just add 01 to your input as a value for the day of the month.
$input = '02-2016';
$date = DateTime::createFromFormat('d-m-Y', '01-' . $input);
echo $date->format('Y-m-d');
// OUTPUT : 2016-02-01
I'm using php date('W'); function to get a number of the week of this year.
date('W') says "40" (that's correct).
The problem is, that my Linux machine, CentOS has a correct date, but shows incorrect week number.
date +"%W" says "39"
Does anyone know why it works this way and how could I fix it?
Okay, I found the answer:
I need to use it like this:
date +%V
%V ISO week number, with Monday as first day of week (01..53)
%W week number of year, with Monday as first day of week (00..53)
date('W');
The code above gives the ISO-8601 week number (don't know what Linux shows by default). If you want the correct year there as well, use
date('W o');
instead of (W Y), because the o gives the right year with the W-week.
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.