I am taking week number from given date, but I observed date('W',strtotime("2015-12-31")); and date('W',strtotime("2016-01-01")); returns week number 53, because year is changed but week doesn't , so is there any way to reset week counter when crossing the new year?
Note: I am using php5.3
While technically correct, the answer PHP gives you may not be the one you were looking for. I suppose you could keep compatibility by calling that half week "week 0"
You would have to target the next Monday and then subtract 1 to the result. This for example should work for you:
<?php
echo date('W', strtotime('NEXT MONDAY', strtotime('2016-01-01'))) - 1;
As you can see, for the first of January you get the result int(0) which doesn't break compatibility the rest of the year, but works nicely for the year 2016.
NOTE: You will have to fix it for the last week of December yourself. But I hope this gives you a direction you can work with.
echo date('m') == 1? date('W', strtotime('NEXT MONDAY', time())) - 1 : date('W', time())
Could be an example.
here is your solution
This checks for month and checks get week as per your requirement
$w=(int)date('W', strtotime('2016-01-01'));
$m=(int)date('n', strtotime('2016-01-01'));
$week = $w==1?($m==12?53:1):($w>=51?($m==1?0:$w):$w);
echo $week;
Related
I am using a date() format to return the starting weekday of a month. The code I have below is how I am attempting to achieve this. For the current year (2018) this works as normal. For example This month is august and the starting weekday is a Wednesday so it will return a 3 for Wednesday. (It works so far)
As we advance the year to 2019 it starts to get the starting weekday wrong.
For example January 2019 starts on a Tuesday so it should return 2 but returns 1. (one day out)
This error seems to be cumulative so if we go to 2020 then it is 2 days out etc.
I have tried so hard to format this Date() correctly but to no avail. Is this even the correct way to do this?
Code:
$future_month = 5 /*for January 2019*/
$starting_weekday = date('N',mktime(0, 0, 0, date('m', strtotime('+'.$future_month.' months', strtotime(date('Y-m-01')))), 1));
Many Thanks
Cameron
Your code makes this much more complicated than it needs to be.
$dt = new DateTime('first day of +5 months')
$dt->format('N'); // "2"
i want to get the day of the next week with php..i need something like this
If today is monday, i want to display the days (es 18-19-20) of the next week, not of this week..the week start on monday...but it will work also if today is sunday..always the days of the next week..i can't found anything because i tried to use jddayofweek but i don't understand it maybe..i haven't found an italian guide..
I have a table which display the days in stringn form..so sunday monday etc, fir my problem i tried date() function but if i use +1 on the day i will have problem with february for example
Not sure exactly what you're doing, but maybe something like:
echo date("d", strtotime("monday next week"))
This will give you next Monday:
$date = date_create()->modify('monday next week');
echo date_format($date, "d");
The following gives me October
echo date('F');
I'm using the following code to give me the name of the coming month (the month after this one), so I'm hoping to see November.
$nextmonth = date("F",strtotime("+1 months"));
Today is 31st Oct, but the above gives 'December' What am I doing wrong? How can I get the month after the current month?
Use relative formats to get the first day of the next month:
echo (new DateTime('first day of next month'))->format('F');
The reason why you see this problem is when adding time to a date at the end of the month you run into issues with months having fewer than 31 days. This can cause you to skip a month. The best bet is to always start your date math at the beginning of the month before adding time to it or just relative formats as demonstrated above.
PHP 5.3:
$nextMonth = new DateTime('first day of next month');
echo $nextMonth->format('F');
I'm preparing a query for mySQL to grab record from the previous week, but I have to treat weeks as Monday - Sunday. I had originally done this:
WHERE YEARWEEK(contactDate) = YEARWEEK(DATE_SUB(CURDATE(),INTERVAL 7 DAY))
to discover that mySQL treats weeks as Sunday - Monday. So instead I'm parsing getting the begin & end dates in php like this:
$i = 0;
while(date('D',mktime(0,0,0,date('m'), date('d')-$i, date('y'))) != "Mon") {
$i++;
}
$start_date = date('Y-n-j', mktime(0,0,0,date('m'), date('d')-($i+7), date('y')));
$end_date = date('Y-n-j', mktime(0,0,0,date('m'), date('d')-($i+1), date('y')));
This works - it gets the current week's date for monday (walking backwards until a monday is hit) then calculates the previous week's dates based on that date.
My question is: Is there a better way to do this? Just seems sloppy, and I expect someone out there can give me a cleaner way to do it - or perhaps not because I need Monday - Sunday weeks.
Edit
Apparently, there is:
$start = date('Y-m-d',strtotime('last monday -7 days'));
$end = date('Y-m-d',strtotime('last monday -1 days'));
That's about a million times more readable. Thank you.
you can use strtotime for this kind of date issues
echo strtotime("last Monday");
(complementing on marvin and Stomped ) Also you can use it this way
echo date('Y-m-d',strtotime('-1 Monday')); //last Monday
echo date('Y-m-d',strtotime('-2 Monday')); //two Mondays ago
echo date('Y-m-d',strtotime('+1 Monday')); //next Monday
strtotime("previous week Monday")
Monday of the previous week.
Marvin's answer is really elegant, although if you really wanted to go for performance you could do it with a little arithmetic. You could derive a formula/method to convert an arbitrary date to "days since some starting point" (probably 01.01.0000) and then the rest of the operations would be easy with that. Getting the day of week from such a number is as simple as subtracting and getting the remainder of a division.
Actually, PHP had a Date class in its PEAR library which did exactly this.
I have to point out for all the readers here there is a big issue with marvin's answer
Here is the catch
The "last Monday" function will return date the "latest" Monday.It will be explained like this , if today is Monday, then it will return the date of "LAST WEEK" Monday, however if today is not Monday, it will return "THIS WEEK" Monday
The Question is request to return "Last Week" Monday. Therefore the result will be incorrect if today is not Monday.
I have solved the issue and wrapped in my Date Time Helper
It will be only one line after you "INCLUDE" the class
$lastWeekMonday = Model_DTHpr::getLastWeekMonday();
Check Here
https://github.com/normandqq/Date-Time-Helper
Bit stuck about how to go about this one. Given the current month, I need to to return the date of the fourth saturday of each month.
e.g. This month would be Feb 20th, next would be March 27th.
Thanks
I'm not a PHP coder, however, it seems strtotime is what you're after.
You can use strtotime("fourth Saturday") and it will return the 4th saturday.
Check out the strtotime docs.
EDIT:
Just to make the answer complete, thanks to Tom and Paul Dixon
date('dS F',strtotime('Fourth Saturday '.date('F o')));
You can use strtotime to find "next saturday" based on a starting date. If that starting date is the day before the earliest possible preceding day (21st) we get the answer...
//required year/month
$yyyymm="2009-01";
//find next saturday after earliest possible date
$t=strtotime("next saturday", strtotime("{$yyyymm}-21"));
//here you go!
echo "4th saturday of $yyyymm is ".strftime("%Y-%m-%d",$t)."\n";
Earliest possible 4th repeat of a day in any month is the 22nd (1,8,15,22), last possible 4th repeat is 28th (7,14,21,28).
EDIT: Although it's not clear in the documentation, you can request the "fourth saturday" too - use the zeroth day of the month as the basis:
$t=strtotime("fourth saturday", strtotime("{$yyyymm}-00"));
or omit the basis time and specify the month and year directly:
$t=strtotime("fourth saturday feb 2009");
Tip of the hat to Robin "I'm not a PHP coder" Day for spotting that :)
The earliest date for the fourth Saturday is the 22nd of the month. So look at the 22nd, see what day of the week it is, if it's not Saturday, add one day to the date, and check again, until you find a match (maximum you would have to check is 6 days).
Find the first Saturday of the month, and then add three weeks to that.
If you don't know when the first Saturday is (or, rather, don't know specifically a date corresponding with a day name), you might want to look at the Doomsday algorithm, which I conveniently looked at for another post with a somewhat similar issue.
function fourth_saturday($year, $month)
{
$info = localtime(mktime(0, 0, 0, $month , 1, $year));
return 28 - $info[6];
}
in PHP rather than pseudo code (think requires 5.2)
$date = getdate();
$date-> setDate($date->format('Y'), $date->format('Y'), '1'); // 1st of month.
while ($date->format('w' != 6)
$date->modify("+1 day");
$date->modify("+21 day"); // date is now on the fourth saturday