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I want to display the first day of the month in 'day of the week' format.
For example, The below code should select the month of August and say the day of the week 01 falls on which is Thursday, but for some reason, the below code outputs Friday which is wrong, on some months such as May it correctly says Wednesday.
$monthNumber = 3;
$base = strtotime(date('Y-m',time()) . '-01 00:00:01');
$dateTest = date("Ym" . 1, strtotime($monthNumber . " month", $base));
$unixTimestamp = strtotime($dateTest, $base);
echo date("l", $unixTimestamp);
Does anyone have any ideas to make it show the correct day?
$base fixes a bug with it not showing the correct month.
And why won't you use the dateTime?
echo (new \DateTime('first day of august 2019'))->format('l');
Read more about Relative formats.
You have to use date() at first:
$base = date('Y-m',time()) . '-01 00:00:00';
And dont have to set the first second :)
To get the day of the week falling on the first day of any month you can use the prefix first day of e.g.
echo (new \DateTime('first day of August'))->format('l');
This will show correctly Thursday for August 2019.
If you want to do it dynamically using numbers for months, you can create a dummy date like you were trying to do.
$timeString = sprintf('01.%02d.%04d', 8, 2019); // replace the numbers with your variables
echo (new \DateTime($timeString))->format('l');
What I want to do:
Getting a certain day from a certain month directly via the DateTime methods (no mktime stunts :)), like
$day = new DateTime('15th of next month');
but it's not possible to set a fixed day by it's number.
Can anybody help ?
EDIT: I've changed the DateTime from this month to next month to make the problem more clear.
You can use DateTime::createFromFormat(). If you only pass the day value, it'll default to the current month and year.
$date = DateTime::createFromFormat('d', '15');
echo $date->format('Y-m-d'); // 2018-10-15
Edit for the new question requirements:
$date = DateTime::createFromFormat('d', 15)->add(new DateInterval('P1M'));
echo $date->format('Y-m-d'); // 2018-11-15
I'm trying to set up a few variables to then pass into my code to add some classes. I know how to do most of it, but it's the setting of the variables that's difficult.
I need to set up three variables that will be compared with the current date.
<?php $current_date = date('Ymd'); ?>
The three variables are: this week, this month and next month.
I had originally set my this week variable as so:
<?php $current_date_week = date('Ymd', strtotime($current_date . ' +7 days')); ?>
But I'm getting this wrong. It shouldn't be from today's date + 7, it should get the date of the end of the current week using the Ymd date format. So if today's date is 20140326 then my $current_date_week should bring back 20140330 and so on.
This month should do the same, get the last day of the current month in Ymd format and next month should get the last day of the next month.
I hope this made sense.
$current_date_week
$current_date_month
$current_date_month_next
Thanks,
R
Do this way:
$current_date_week = date('Ymd', strtotime('this Sunday')); //20140330
$current_date_month = date('Ymd', strtotime('last day of this month'));//20140331
$next_date_month = date('Ymd', strtotime('last day of next month'));//20140430
check available list for formats.
Use these as second arguments to date():
mktime(1,2,3,date('m'),date('d')+7-date('N'),date('Y'));
mktime(1,2,3,date('m')+1,0,date('Y'));
mktime(1,2,3,date('m')+2,0,date('Y'));
I need to get previous month and year, relative to current date.
However, see following example.
// Today is 2011-03-30
echo date('Y-m-d', strtotime('last month'));
// Output:
2011-03-02
This behavior is understandable (to a certain point), due to different number of days in february and march, and code in example above is what I need, but works only 100% correctly for between 1st and 28th of each month.
So, how to get last month AND year (think of date("Y-m")) in the most elegant manner as possible, which works for every day of the year? Optimal solution will be based on strtotime argument parsing.
Update. To clarify requirements a bit.
I have a piece of code that gets some statistics of last couple of months, but I first show stats from last month, and then load other months when needed. That's intended purpose. So, during THIS month, I want to find out which month-year should I pull in order to load PREVIOUS month stats.
I also have a code that is timezone-aware (not really important right now), and that accepts strtotime-compatible string as input (to initialize internal date), and then allows date/time to be adjusted, also using strtotime-compatible strings.
I know it can be done with few conditionals and basic math, but that's really messy, compared to this, for example (if it worked correctly, of course):
echo tz::date('last month')->format('Y-d')
So, I ONLY need previous month and year, in a strtotime-compatible fashion.
Answer (thanks, #dnagirl):
// Today is 2011-03-30
echo date('Y-m-d', strtotime('first day of last month')); // Output: 2011-02-01
Have a look at the DateTime class. It should do the calculations correctly and the date formats are compatible with strttotime. Something like:
$datestring='2011-03-30 first day of last month';
$dt=date_create($datestring);
echo $dt->format('Y-m'); //2011-02
if the day itself doesn't matter do this:
echo date('Y-m-d', strtotime(date('Y-m')." -1 month"));
I found an answer as I had the same issue today which is a 31st. It's not a bug in php as some would suggest, but is the expected functionality (in some since). According to this post what strtotime actually does is set the month back by one and does not modify the number of days. So in the event of today, May 31st, it's looking for April-31st which is an invalid date. So it then takes April 30 an then adds 1 day past it and yields May 1st.
In your example 2011-03-30, it would go back one month to February 30th, which is invalid since February only has 28 days. It then takes difference of those days (30-28 = 2) and then moves two days past February 28th which is March 2nd.
As others have pointed out, the best way to get "last month" is to add in either "first day of" or "last day of" using either strtotime or the DateTime object:
// Today being 2012-05-31
//All the following return 2012-04-30
echo date('Y-m-d', strtotime("last day of -1 month"));
echo date('Y-m-d', strtotime("last day of last month"));
echo date_create("last day of -1 month")->format('Y-m-d');
// All the following return 2012-04-01
echo date('Y-m-d', strtotime("first day of -1 month"));
echo date('Y-m-d', strtotime("first day of last month"));
echo date_create("first day of -1 month")->format('Y-m-d');
So using these it's possible to create a date range if your making a query etc.
If you want the previous year and month relative to a specific date and have DateTime available then you can do this:
$d = new \DateTimeImmutable('2013-01-01', new \DateTimeZone('UTC'));
$firstDay = $d->modify('first day of previous month');
$year = $firstDay->format('Y'); //2012
$month = $firstDay->format('m'); //12
date('Y-m', strtotime('first day of last month'));
strtotime have second timestamp parameter that make the first parameter relative to second parameter. So you can do this:
date('Y-m', strtotime('-1 month', time()))
if i understand the question correctly you just want last month and the year it is in:
<?php
$month = date('m');
$year = date('Y');
$last_month = $month-1%12;
echo ($last_month==0?($year-1):$year)."-".($last_month==0?'12':$last_month);
?>
Here is the example: http://codepad.org/c99nVKG8
ehh, its not a bug as one person mentioned. that is the expected behavior as the number of days in a month is often different. The easiest way to get the previous month using strtotime would probably be to use -1 month from the first of this month.
$date_string = date('Y-m', strtotime('-1 month', strtotime(date('Y-m-01'))));
I think you've found a bug in the strtotime function. Whenever I have to work around this, I always find myself doing math on the month/year values. Try something like this:
$LastMonth = (date('n') - 1) % 12;
$Year = date('Y') - !$LastMonth;
date("m-Y", strtotime("-1 months"));
would solve this
Perhaps slightly more long winded than you want, but i've used more code than maybe nescessary in order for it to be more readable.
That said, it comes out with the same result as you are getting - what is it you want/expect it to come out with?
//Today is whenever I want it to be.
$today = mktime(0,0,0,3,31,2011);
$hour = date("H",$today);
$minute = date("i",$today);
$second = date("s",$today);
$month = date("m",$today);
$day = date("d",$today);
$year = date("Y",$today);
echo "Today: ".date('Y-m-d', $today)."<br/>";
echo "Recalulated: ".date("Y-m-d",mktime($hour,$minute,$second,$month-1,$day,$year));
If you just want the month and year, then just set the day to be '01' rather than taking 'todays' day:
$day = 1;
That should give you what you need. You can just set the hour, minute and second to zero as well as you aren't interested in using those.
date("Y-m",mktime(0,0,0,$month-1,1,$year);
Cuts it down quite a bit ;-)
This is because the previous month has less days than the current month. I've fixed this by first checking if the previous month has less days that the current and changing the calculation based on it.
If it has less days get the last day of -1 month else get the current day -1 month:
if (date('d') > date('d', strtotime('last day of -1 month')))
{
$first_end = date('Y-m-d', strtotime('last day of -1 month'));
}
else
{
$first_end = date('Y-m-d', strtotime('-1 month'));
}
If a DateTime solution is acceptable this snippet returns the year of last month and month of last month avoiding the possible trap when you run this in January.
function fn_LastMonthYearNumber()
{
$now = new DateTime();
$lastMonth = $now->sub(new DateInterval('P1M'));
$lm= $lastMonth->format('m');
$ly= $lastMonth->format('Y');
return array($lm,$ly);
}
//return timestamp, use to format month, year as per requirement
function getMonthYear($beforeMonth = '') {
if($beforeMonth !="" && $beforeMonth >= 1) {
$date = date('Y')."-".date('m')."-15";
$timestamp_before = strtotime( $date . ' -'.$beforeMonth.' month' );
return $timestamp_before;
} else {
$time= time();
return $time;
}
}
//call function
$month_year = date("Y-m",getMonthYear(1));// last month before current month
$month_year = date("Y-m",getMonthYear(2)); // second last month before current month
function getOnemonthBefore($date){
$day = intval(date("t", strtotime("$date")));//get the last day of the month
$month_date = date("y-m-d",strtotime("$date -$day days"));//get the day 1 month before
return $month_date;
}
The resulting date is dependent to the number of days the input month is consist of. If input month is february (28 days), 28 days before february 5 is january 8. If input is may 17, 31 days before is april 16. Likewise, if input is may 31, resulting date will be april 30.
NOTE: the input takes complete date ('y-m-d') and outputs ('y-m-d') you can modify this code to suit your needs.
I've seen some variants on this question but I believe this one hasn't been answered yet.
I need to get the starting date and ending date of a week, chosen by year and week number (not a date)
example:
input:
getStartAndEndDate($week, $year);
output:
$return[0] = $firstDay;
$return[1] = $lastDay;
The return value will be something like an array in which the first entry is the week starting date and the second being the ending date.
OPTIONAL: while we are at it, the date format needs to be Y-n-j (normal date format, no leading zeros.
I've tried editing existing functions that almost did what I wanted but I had no luck so far.
Using DateTime class:
function getStartAndEndDate($week, $year) {
$dto = new DateTime();
$dto->setISODate($year, $week);
$ret['week_start'] = $dto->format('Y-m-d');
$dto->modify('+6 days');
$ret['week_end'] = $dto->format('Y-m-d');
return $ret;
}
$week_array = getStartAndEndDate(52,2013);
print_r($week_array);
Returns:
Array
(
[week_start] => 2013-12-23
[week_end] => 2013-12-29
)
Explained:
Create a new DateTime object which defaults to now()
Call setISODate to change object to first day of $week of $year instead of now()
Format date as 'Y-m-d' and put in $ret['week_start']
Modify the object by adding 6 days, which will be the end of $week
Format date as 'Y-m-d' and put in $ret['week_end']
A shorter version (works in >= php5.3):
function getStartAndEndDate($week, $year) {
$dto = new DateTime();
$ret['week_start'] = $dto->setISODate($year, $week)->format('Y-m-d');
$ret['week_end'] = $dto->modify('+6 days')->format('Y-m-d');
return $ret;
}
Could be shortened with class member access on instantiation in >= php5.4.
Many years ago, I found this function:
function getStartAndEndDate($week, $year) {
$dto = new DateTime();
$dto->setISODate($year, $week);
$ret['week_start'] = $dto->format('Y-m-d');
$dto->modify('+6 days');
$ret['week_end'] = $dto->format('Y-m-d');
return $ret;
}
$week_array = getStartAndEndDate(52,2013);
print_r($week_array);
We can achieve this easily without the need for extra computations apart from those inherent to the DateTime class.
function getStartAndEndDate($year, $week)
{
return [
(new DateTime())->setISODate($year, $week)->format('Y-m-d'), //start date
(new DateTime())->setISODate($year, $week, 7)->format('Y-m-d') //end date
];
}
The setISODate() function takes three arguments: $year, $week, and $day respectively, where $day defaults to 1 - the first day of the week. We therefore pass 7 to get the exact date of the 7th day of the $week.
Slightly neater solution, using the "[year]W[week][day]" strtotime format:
function getStartAndEndDate($week, $year) {
// Adding leading zeros for weeks 1 - 9.
$date_string = $year . 'W' . sprintf('%02d', $week);
$return[0] = date('Y-n-j', strtotime($date_string));
$return[1] = date('Y-n-j', strtotime($date_string . '7'));
return $return;
}
shortest way to do it:
function week_date($week, $year){
$date = new DateTime();
return "first day of the week is ".$date->setISODate($year, $week, "1")->format('Y-m-d')
."and last day of the week is ".$date->setISODate($year, $week, "7")->format('Y-m-d');
}
echo week_date(12,2014);
You can get the specific day of week from date as bellow that I get the first and last day
$date = date_create();
// get the first day of the week
date_isodate_set($date, 2019, 1);
//convert date format and show
echo date_format($date, 'Y-m-d') . "\n";
// get the last date of the week
date_isodate_set($date, 2019, 1, 7);
//convert date format and show
echo date_format($date, 'Y-m-d') . "\n";
Output =>
2018-12-31
2019-01-06
The calculation of Roham Rafii is wrong. Here is a short solution:
// week number to timestamp (first day of week number)
function wn2ts($week, $year) {
return strtotime(sprintf('%dW%02d', $year, $week));
}
if you want the last day of the week number, you can add up 6 * 24 * 3600
This is an old question, but many of the answers posted above appear to be incorrect.
I came up with my own solution:
function getStartAndEndDate($week, $year){
$dates[0] = date("Y-m-d", strtotime($year.'W'.str_pad($week, 2, 0, STR_PAD_LEFT)));
$dates[1] = date("Y-m-d", strtotime($year.'W'.str_pad($week, 2, 0, STR_PAD_LEFT).' +6 days'));
return $dates;
}
First we need a day from that week so by knowing the week number and knowing that a week has seven days we are going to do so the
$pickADay = ($weekNo-1) * 7 + 3;
this way pickAday will be a day in our desired week.
Now because we know the year we can check which day is that.
things are simple if we only need dates newer than unix timestamp
We will get the unix timestamp for the first day of the year and add to that 24*3600*$pickADay and all is simple from here because we have it's timestamp we can know what day of the week it is and calculate the head and tail of that week accordingly.
If we want to find out the same thing of let's say 12th week of 1848 we must use another approach as we can not get the timestamp. Knowing that each year a day advances 1 weekday meaning (1st of november last year was on a sunday, this year is on a monday, exception for the leap years when it advances 2 days I believe, you can check that ). What I would do if the year is older than 1970 than make a difference between it and the needed year to know how many years are there, calculate the day of the week as my pickADay was part of 1970, shift it back one weekday for each. $shiftTimes = ($yearDifference + $numberOfLeapYears)%7, in the difference. shift the day backwords $shiftTimes, then you will know what day of the week was that day those years ago, then find the weekhead and weektail. Same thing can be used also for the future if it seems simpler. Try it if it works and tell me if it does not.
For documentation (since Google ranks this question first when searching for "php datetime start end this week").
If you need the startdate and enddate for the current week (using DateTime):
$dateTime = new DateTime('now');
$monday = clone $dateTime->modify(('Sunday' == $dateTime->format('l')) ? 'Monday last week' : 'Monday this week');
$sunday = clone $dateTime->modify('Sunday this week');
var_dump($monday->format('Y-m-d')); // e.g. 2018-06-25
var_dump($sunday->format('Y-m-d')); // e.g. 2018-07-01
Hope this will help.
The "first day of the week" is subjective. Some cultures use "Monday" others "Sunday", maybe others something else?
For my purposes, I want the first day of the week to be "Sunday" and the last day of the week to be "Saturday".
Also, using DateTime with no arguments will default to "now" which includes the current time. The following method will disregard the current time by specifying "today" in the DateTime constructor.
Furthermore the string "sunday this week" does not seem to be reliable. It actually will return Sunday the next week (according to my view of what a week is).
I've built a method which returns a PHP object containing two DateTime objects. One for the first day (Sunday) of the given week, the second for the last day (Saturday) of the given week.
function get_first_and_last_day_of_week( $year_number, $week_number ) {
// we need to specify 'today' otherwise datetime constructor uses 'now' which includes current time
$today = new DateTime( 'today' );
return (object) [
'first_day' => clone $today->setISODate( $year_number, $week_number, 0 ),
'last_day' => clone $today->setISODate( $year_number, $week_number, 6 )
];
}
Have you tried PHP relative dates? It might work.
Even if you dont want to use a specific date you cannot escape it. You can calculate a week based on the date ONLY.
Steps:
get the first day of the year
decide when the first week starts ( there are some rules that include first Thursday if I remember.
add some number of weeks (your first param). Zend_Date has an add() function where you can add weeks for example. This will give you the first day of the week.
offset and get the last day.
I would recommend working with a consistent dates sistem like Zend_Date or Pear Date.
function getStartAndEndDate($week, $year)
{
$week_start = new DateTime();
$week_start->setISODate($year,$week);
$return[0] = $week_start->format('d-M-Y');
$time = strtotime($return[0], time());
$time += 6*24*3600;
$return[1] = date('d-M-Y', $time);
return $return;
}
$dateParam = '2018-06-10';
$week = date('w', strtotime($dateParam));
$date = new DateTime($dateParam);
$firstWeek = $date->modify("-".$week." day")->format("Y-m-d H:i:s");
$endWeek = $date->modify("+6 day")->format("Y-m-d H:i:s");
echo $firstWeek."<br/>";
echo $endWeek;
will print
2018-06-10 00:00:00
2018-06-16 00:00:00
hopefully will help