I have three pieces of information extracted from my database, the year, the number of week (1-52), and the day of week (mon, tue, wed). And I need to convert it in to a date.
How would I go about doing that in php?
Try this :
$yr="2014";
$week="01";
$day = "Mon";
echo date("Y-m-d", strtotime($yr."W".$week."D ".$day));
PHP fiddle:http://phpfiddle.org/main/code/nzg-ws6
Did a little digging and it looks like the day will need to be in numerical format for the strtotime conversion to work properly. Using a string such as 'Monday' won't work.
$yr = "2014";
$week = "01";
$day = "1"; //Monday
echo date("Y-m-d", strtotime($yr . "W" . $week . $day)); // "2014W011"
Note that I didn't include the 'D' prefix for the day as this looks to be following ISO8601.
Swapping your days to numbers can be done a few ways:
You could make an array Monday to Sunday and use the array key, or maybe a little easier you could do something like this:
$day date("N", strtotime("Monday")); // will echo 1
So the full code could look something like this:
$yr = "2014";
$week = "01";
$day = date("N", strtotime("Monday")); //
echo date("Y-m-d", strtotime($yr . "W" . $week . $day)); // "2014W011"
Here is a demo to check: http://3v4l.org/EllgQ
This converts you $day value to desired day format.
try this
$yr="2011";
$week="01";
$day="Monday";
$day=date("D",strtotime($day));
echo date("Y-m-d", strtotime($yr."W".$week."D ".$day));;
Demo
Related
I'm trying make a function to return the exact date of previous months.
That is a example of my code:
// Dates in TimeStamp
$ts_now = strtotime('now');
$ts_month1 = strtotime("-1 month", $ts_now);
$ts_month2 = strtotime("-2 month", $ts_now);
$ts_month3 = strtotime("-3 month", $ts_now);
// Dates Formated
$date_now = date('Y-m-d', $ts_now);
$date_month1 = date('Y-m-d', $ts_month1);
$date_month2 = date('Y-m-d', $ts_month2);
$date_month3 = date('Y-m-d', $ts_month3);
//Output
echo $date_now; //2020-04-30
echo $date_month1; //2020-03-30
echo $date_month2; //2020-03-01
echo $date_month3; //2020-01-30
The problem is in $date_month2 that represents February, the output is 2020-03-01 instead 2020-02-29 and I suppose that problem will happen in months who have 30 days when present date have 31 days.
What is the best way to resolve that?
As you can see working with the end of the month can be problematic because of how PHP works with dates. Your best bet is to go back to the beginning of the month, do your date math (i.e. go backwards in time), and then go to the date you want. That way you can check to see if the current day is greater than the number of days in month. If so, use the last day of the month instead.
function getMonthsAgo(int $n): string {
$date = new DateTime();
$day = $date->format('j');
$date->modify('first day of this month')->modify('-' . $n . ' months');
if ($day > $date->format('t')) {
$day = $date->format('t');
}
$date->setDate($date->format('Y'), $date->format('m'), $day);
return $date->format('Y-m-d');
}
// Dates Formated
$date_now = date('Y-m-d');
$date_month1 = getMonthsAgo(1);
$date_month2 = getMonthsAgo(2);
$date_month3 = getMonthsAgo(3);
//Output
echo $date_now;
echo $date_month1;
echo $date_month2;
echo $date_month3;
Output:
2020-04-30
2020-03-30
2020-02-29
2020-01-30
I'm trying to get the date from the week number, day number and year.
For eg:
week number = 52
day number = 4 (of week 52)
year = 2013
In this case, the date should be 26-12-2013.
How can I do it using PHP? I've already tried with strtotime(), but I'm confused about the formats. Can someone help?
Make use of setISODate()
<?php
$gendate = new DateTime();
$gendate->setISODate(2013,52,4); //year , week num , day
echo $gendate->format('d-m-Y'); //"prints" 26-12-2013
Try this code.
<?php
function change_date($week_num, $day) {
$timestamp = strtotime(date('Y') . '-W' . $week_num . '-' . $day);
return $timestamp;
}
$timestamp = change_date(52, 4);
echo date('d-m-Y', $timestamp);
?>
You can also use the strtotime function. In your example, you can write:
date("Y-m-d", strtotime("Y2013W52-4")) // outputs: 2013-12-26
The strtotime will give you a timestamp that you can use in combination withe the date function.
According to the ISO standar, the 52nd week of 2013, when the week starts on Saturday, starts 2013-12-28 and ends 2014-01-03 (inclusive).
Code:
echo date('Y-m-d', strtotime("2013-W52-6"));//prints 2013-12-28
Now, I am trying to figure out the inverse using PHP. If I have the date 2014-01-03, how to know the ISO weeknumber of that date if the week starts on Saturday?
Use the DateTime class:
createFromFormat() to create the object based on the string you have
then format('W')
$date = "2013-12-28";
$dayofyear = date("z", $date);
$week = floor($dayofyear / 7) + 1;
// I assume week starts with Monday (I think as ISO does?)
// so only weeks that start on sunday will count as "week 0" instead of "week 1"
// I could be wildly wrong, adjust according to your needs
$first_day = date("w", date("Y", $date) . "-01-01");
if ($first_day == 6)
--$week;
echo $week;
$date = "2013-12-28";
$date_array = explode("-", $date);
$date = mktime(0, 0, 0, $date_array[1], $date_array[2], $date_array[0]);
$week_number = date('W', $date);
echo $week_number;
Something like this.
So, in PHP i'm trying to return a date value for last year based on the same day of week.
EX: (Monday) 2011-12-19 inputted should return (Monday) 2010-12-20.
I was just doing it simply by -364 but then that was failing on leap years. I came across another function :
$newDate = $_POST['date'];
$newDate = strtotime($newDate);
$oldDate = strtotime('-1 year',$newDate);
$newDayOfWeek = date('w',$oldDate);
$oldDayOfWeek = date('w',$newDate);
$dayDiff = $oldDayOfWeek-$newDayOfWeek;
$oldDate = strtotime("$dayDiff days",$oldDate);
echo 'LAST YEAR DAY OF WEEK DATE = ' . date('Ymd', $oldDate);
however, that is failing when you try to input a Sunday date, as it does a 0 (sunday) minus 6 (saturday of last year date), and returns with a value T-6. IE inputting 2011-12-25 gets you 2010-12-19 instead of 2011-12-26.
I'm kind of stumped to find a good solution in php that will work for leap years and obviously all days of the week.
Any suggestions?
Thanks!
How about this, using PHP's DateTime functionality:
$date = new DateTime('2011-12-25'); // make a new DateTime instance with the starting date
$day = $date->format('l'); // get the name of the day we want
$date->sub(new DateInterval('P1Y')); // go back a year
$date->modify('next ' . $day); // from this point, go to the next $day
echo $date->format('Ymd'), "\n"; // ouput the date
$newDate = '2011-12-19';
date_default_timezone_set('UTC');
$newDate = strtotime($newDate);
$oldDate = strtotime('last year', $newDate);
$oldDate = strtotime(date('l', $newDate), $oldDate);
$dateFormat = 'Y-m-d l w W';
echo "This date: ", date($dateFormat, $newDate), "\n";
echo "Old date : ", date($dateFormat, $oldDate);
That gives:
This date: 2011-12-19 Monday 1 51
Old date : 2010-12-20 Monday 1 51
Use strtotime() to get a date, for the same week last year.
Use the format {$year}-W{$week}-{$weekday}, like this:
echo date("Y-m-d", strtotime("2010-W12-1"));
And you can do that for as long back you wan't:
<?php
for($i = 2011; $i > 2000; $i--)
echo date("Y-m-d", strtotime($i."-W12-1"));
?>
Make it easier :)
echo date('Y-m-d (l, W)').<br/>;
echo date('Y-m-d (l, W)', strtotime("-52 week"));
Edit: I forgot to write output: :)
2015-05-06 (Wednesday, 19)
2014-05-07 (Wednesday, 19)
<?php
$date = "2020-01-11";
$newdate = date("Y-m-d",strtotime ( '-1 year' , strtotime ( $date ) )) ;
echo $newdate;
?>
ref https://www.nicesnippets.com/blog/how-to-get-previous-year-from-date-in-php
How can I find first day of the next month and the remaining days till this day from the present day?
Thank you
Create a timestamp for 00:00 on the first day of next month:
$firstDayNextMonth = strtotime('first day of next month');
The number of days til that date is the number of seconds between now and then divided by (24 * 60 * 60).
$daysTilNextMonth = ($firstDayNextMonth - time()) / (24 * 3600);
$firstDayNextMonth = date('Y-m-d', strtotime('first day of next month'));
For getting first day after two months from current
$firstDayAfterTwoMonths = date('Y-m-d', strtotime('first day of +2 month'));
You can use DateTime object like this to find out the first day of next month like this:
$date = new DateTime('first day of next month');
You can do this to know how many days left from now to the first day of next month:
$date = new DateTime('now');
$nowTimestamp = $date->getTimestamp();
$date->modify('first day of next month');
$firstDayOfNextMonthTimestamp = $date->getTimestamp();
echo ($firstDayOfNextMonthTimestamp - $nowTimestamp) / 86400;
The easiest and quickest way is to use strtotime() which recognizes 'first day next month';
$firstDayNextMonth = date('Y-m-d', strtotime('first day next month'));
Since I googled this and came to this answer, I figured I'd include a more modern answer that works for PHP 5.3.0+.
//Your starting date as DateTime
$currentDate = new DateTime(date('Y-m-d'));
//Add 1 month
$currentDate->add(new DateInterval('P1M'));
//Get the first day of the next month as string
$firstDayOfNextMonth = $currentDate->format('Y-m-1');
You can get the first of the next month with this:
$now = getdate();
$nextmonth = ($now['mon'] + 1) % 13 + 1;
$year = $now['year'];
if($nextmonth == 1)
$year++;
$thefirst = gmmktime(0, 0, 0, $nextmonth, $year);
With this example, $thefirst will be the UNIX timestamp for the first of the next month. Use date to format it to your liking.
This will give you the remaining days in the month:
$now = getdate();
$months = array(
31,
28 + ($now['year'] % 4 == 0 ? 1 : 0), // Support for leap years!
31,
30,
31,
30,
31,
31,
30,
31,
30,
31
);
$days = $months[$now['mon'] - 1];
$daysleft = $days - $now['mday'];
The number of days left will be stored in $daysleft.
Hope this helps!
$firstDayNextMonth = date('Y-m-d', mktime(0, 0, 0, date('m')+1, 1, date('Y')));
As another poster has mentioned the DateInterval object does not give accurate results for the next month when you use dates such as 1/31/2016 or 8/31/2016 as it skips the next month. My solution was to still use the DateInterval object but reformat your current date to be the first day of the current month prior to utilizing the DateInterval.
$currentDate = '8/31/2016';
$date = new DateTime(date("n", strtotime($currentDate))."/1/".date("Y", strtotime($currentDate)));
//add 1 month
$date->add(new DateInterval('P1M'));
$currentDate=$date->format('m/1/Y');
echo($currentDate);
easiest way to get the last day of the month
date('Y-m-d', mktime(0, 0, 0, date('m')+1, 1, date('Y')));
I took mattbasta's approach because it's able to get the 'first day of next month' with a given date, but there is a tiny problem in calculating the $nextmonth. The fix is as below:
$now = getdate();
$nextmonth = ($now['mon'] + 1) % 13 + 1;
$year = $now['year'];
if($nextmonth == 1)
$year++;
else
$nextmonth--;
$thefirst = gmmktime(0, 0, 0, $nextmonth, $year);
I initially thought about using a DateInterval object (as discussed above in another answer) but it is not reliable. For example, if the current DateTime is 31 January and then we add on a month (to get the next month) then it will skip February completely!
Here is my solution:
function start_of_next_month(DateTime $datetime)
{
$year = (int) $datetime->format('Y');
$month = (int) $datetime->format('n');
$month += 1;
if ($month === 13)
{
$month = 1;
$year += 1;
}
return new DateTime($year . '-' . $month . '-01');
}
Even easier way to get first and last day of next month
$first = strttotime("first day of next month");
$last = strttotime("last day of next month");
You could do something like this. To have this functionality, you need to make use of a php library available in https://packagist.org/packages/ishworkh/navigable-date.
With that is really easy to do what you're asking for here.
For e.g:
$format = 'Y-m-d H:i:s';
$Date = \NavigableDate\NavigableDateFacade::create('2017-02-24');
var_dump($Date->format($format));
$resetTime = true;
$resetDays = true;
$NextMonth = $Date->nextMonth($resetTime, $resetDays);
var_dump($NextMonth->format($format));
$DayUntilFirstOfNextMonth = $NextMonth->getDifference($Date);
var_dump('Days left:' . $DayUntilFirstOfNextMonth->format('%d'));
gives you ouput:
string(19) "2017-02-24 00:00:00"
string(19) "2017-03-01 00:00:00"
string(11) "Days left:5"
Note: Additionally this library let you navigate through dates by day(s), week(s), year(s) forward or backward. For more information look into its README.
(PHP 5 >= 5.5.0, PHP 7, PHP 8)
To get the first day of next month a clean solution:
<?php
$date = new DateTimeInmutable('now');
$date->modify('first day of next month');//here the magic occurs
echo $date->format('Y-m-d') . '\n';
Since you just want to calculate it I suggest using DateTimeInmutable class.
using the class DateTime and $date->modify('first day of next month'); will modify your original date value.
$month = date('m')+1;
if ($month<10) {
$month = '0'.$month;
}
echo date('Y-').$month.'-01';
Simplest way to achieve this. You can echo or store into variable.
I came up with this for my needs:
if(date('m') == 12) { $next_month = 1; } else { $next_month = date('m')+1; }
if($next_month == 1) { $year_start = date('Y')+1; } else { $year_start = date('Y'); }
$date_start_of_next_month = $year_start . '-' . $next_month . '-01 00:00:00';
if($next_month == 12) { $month_after = 1; } else { $month_after = $next_month+1; }
if($month_after == 1) { $year_end = date('Y')+1; } else { $year_end = date('Y'); }
$date_start_of_month_after_next = $year_end . '-' . $month_after . '-01 00:00:00';
Please note that instead of getting $date_end_of_next_month I chose to go with a $date_start_of_month_after_next date, it avoids the hassles with leap years and months containing different number of days.
You can simply use the >= comparaision sign for $date_start_of_next_month and the < one for $date_start_of_month_after_next.
If you prefer a timestamp format for the date, from there you will want to apply the strtotime() native function of PHP on these two variables.
You can use the php date method to find the current month and date, and then you would need to have a short list to find how many days in that month and subtract (leap year would require extra work).