I have a date $da = '18-Nov-2015'
I want month and year separate.I tried this.but it didn't work.
$month = date('F',strtotime($da));
$YEAR = date('Y',strtotime($da));
Give it try with below code:
$da = '18-Nov-2015';
$date = DateTime::createFromFormat('d-M-Y',$da);
echo $date->format("Y");
echo $date->format("F");
Note:DateTime We can create the object using arbitrary parameters like $date = DateTime::createFromFormat('d-M-Y', $weird_user_input); which can be formatted to unix timestamp or
whatever other date format We wish.
You can use the date_parse() function. It returns an array that contains the components of the date: day, month, year, hour, minute, and others.
Usage sample:
$da = '18-Nov-2015';
$dateComps = date_parse($da);
$year = $dateComps['year'];
$month = $dateComps['month'];
$day = $dateComps['day'];
//and so on, ...
use
$da=new DateTime($da);
$da->format('m-d');
hello try this hope you will get your answer what you want.
let me know if you got iy correct.
<?php
$dateValue = Date('Y-m-d');
$time=strtotime($dateValue);
$year=date("Y",$time);
$month=date("F",$time);
$date=date("d",$time);
?>
Almost you have done.This is how you can do it.
$year = date('Y', strtotime($da));
$month = date('m', strtotime($da));
F - A full textual representation of a month, such as January or March January through December
m - Numeric representation of a month, with leading zeros 01 through 12
M - A short textual representation of a month, three letters Jan through Dec
n - Numeric representation of a month, without leading zeros 1 through 12
Or else simply you can use explode method
$dateArray = explode('-', $da);
$dateArray[0] //date
$dateArray[1] //Month
$dateArray[2] //Year
Date reference : http://php.net/manual/en/function.date.php
i tried this in http://phpfiddle.org/
$da='18-Nov-2015';
$dateValue = strtotime($da);
$year = date('Y',$dateValue);
$monthName = date('F',$dateValue);
$day = date('d',$dateValue);
echo $monthName;
echo $year;
How do you go in php from a nth day in the year to the date like:
getdatefromday(275, 2012)
and it outputs a date (better if an object).
And I'd like to do the opposite too, like getdayoftheyear("21 oct 2012")
This is pretty easy all around. You should read up on the DateTime object's createFromFormat static method here, the date function here and the strtotime function here.
// This should get you a DateTime object from the date and year.
function getDateFromDay($year, $dayOfYear) {
$date = DateTime::createFromFormat('z Y', strval($year) . ' ' . strval($dayOfYear));
return $date;
}
// This should get you the day of the year and the year in a string.
date('z Y', strtotime('21 oct 2012'));
Try(days starts from 0 not 1):
$date = DateTime::createFromFormat( 'Y z' , '2012 275');
var_dump($date);
and that:
echo date('z', strtotime('21 oct 2012'));
$todayid = date("z"); // to get today's day of year
function dayofyear2date( $tDay, $tFormat = 'd-m-Y' ) {
$day = intval( $tDay );
$day = ( $day == 0 ) ? $day : $day - 1;
$offset = intval( intval( $tDay ) * 86400 );
$str = date( $tFormat, strtotime( 'Jan 1, ' . date( 'Y' ) ) + $offset );
return( $str );
}
echo dayofyear2date($todayid);
day of year
I know it's a bit old and the answer has already been accepted, but I wanted to put here an alternative way to do it, trying to use the exact formats required by the question owner:
// This gets the day of the year number
// given a formatted date string like
// "21 oct 2012"
function getdayoftheyear($dateString) {
date('z', strtotime($dateString));
}
The other way around:
// This gets the date formatted in $dateFormat way
// like "Y-m-d"
// given the $dayOfTheYear and the $year in numeric form
function getdatefromday($dateFormat, $dayOfTheYear, $year) {
date($dateFormat, mktime(0, 0, 0, 1, ($dayOfTheYear + 1), $year));
}
This second function uses a feature of mktime(), allowing to set whatever numbers in parameters list, because it manages overflows finding by itself the right month.
So if you call mktime(0, 0, 0, 1, 32, 2015) it actually knows that the 32nd day of the 1st month is the 1st day of the 2dn month, and so on.
You can use strtotime() to get the amount of seconds for the year value ( http://de2.php.net/manual/en/function.strtotime.php). Than at the amount of days in seconds (day * 24 * 60 * 60). Now you can use this value with date() (see first answer)
function DayToTimestamp($day, $year = null)
{
isset($year) or $year = date('Y');
return strtotime("1 Jan $year +$day day");
}
function getDateFromDayOfYear($dayOfYear,$year){
return date('Y-m-d', strtotime('January 1st '.$year.' +'.$dayOfYear.' days'));
}
Getting the day of the year is easy. Just use the date function with the correct parameter as documented in the manual (it returns 0 for Jan-1 to 365 for Dec-31 on a leap year).
Going the other way will need a bit of creativity.
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
If I have a string which represents a date, like "2011/07/01" (which is 1st July 2011) , how would I output that in more readable forms, like:
1 July 2011
1 Jul 2011 (month as three letters)
And also, how could I make it intelligently show date ranges like "2011/07/01" to "2011/07/11" as
1 - 11 July 2001
(without repeating the 'July' and '2011' in this case)
You can convert your date to a timestamp using strtotime() and then use date() on that timestamp. On your example:
$date = date("j F Y", strtotime("2011/07/01")); // 1 July 2011
$date = date("j M Y", strtotime("2011/07/01")); // 1 Jul 2011
As NullUserException mentioned, you can use strtotime to convert the date strings to timestamps. You can output 'intelligent' ranges by using a different date format for the first date, determined by comparing the years, months and days:
$date1 = "2011/07/01";
$date2 = "2011/07/11";
$t1 = strtotime($date1);
$t2 = strtotime($date2);
// get date and time information from timestamps
$d1 = getdate($t1);
$d2 = getdate($t2);
// three possible formats for the first date
$long = "j F Y";
$medium = "j F";
$short = "j";
// decide which format to use
if ($d1["year"] != $d2["year"]) {
$first_format = $long;
} elseif ($d1["mon"] != $d2["mon"]) {
$first_format = $medium;
} else {
$first_format = $short;
}
printf("%s - %s\n", date($first_format, $t1), date($long, $t2));
As for the second one:
$time1 = time();
$time2 = $time1 + 345600; // 4 days
if( date("j",$time1) != date("j",$time2) && date("FY",$time1) == date("FY",$time2) ){
echo date("j",$time1)." - ".date("j F Y",$time2);
}
Can be seen in action here
Just make up more conditions
I would use strtotime AND strftime. Is a much simpler way of doing it.
By example, if a have a date string like "Oct 20 18:29:50 2001 GMT" and I want to get it in format day/month/year I could do:
$mystring = "Oct 20 18:29:50 2001 GMT";
printf("Original string: %s\n", $mystring);
$newstring = strftime("%d/%m/%Y", strtotime($mystring));
printf("Data in format day/month/year is: %s\n", $newstring);
Let's say I have a date in the following format: 2010-12-11 (year-mon-day)
With PHP, I want to increment the date by one month, and I want the year to be automatically incremented, if necessary (i.e. incrementing from December 2012 to January 2013).
Regards.
$time = strtotime("2010.12.11");
$final = date("Y-m-d", strtotime("+1 month", $time));
// Finally you will have the date you're looking for.
I needed similar functionality, except for a monthly cycle (plus months, minus 1 day). After searching S.O. for a while, I was able to craft this plug-n-play solution:
function add_months($months, DateTime $dateObject)
{
$next = new DateTime($dateObject->format('Y-m-d'));
$next->modify('last day of +'.$months.' month');
if($dateObject->format('d') > $next->format('d')) {
return $dateObject->diff($next);
} else {
return new DateInterval('P'.$months.'M');
}
}
function endCycle($d1, $months)
{
$date = new DateTime($d1);
// call second function to add the months
$newDate = $date->add(add_months($months, $date));
// goes back 1 day from date, remove if you want same day of month
$newDate->sub(new DateInterval('P1D'));
//formats final date to Y-m-d form
$dateReturned = $newDate->format('Y-m-d');
return $dateReturned;
}
Example:
$startDate = '2014-06-03'; // select date in Y-m-d format
$nMonths = 1; // choose how many months you want to move ahead
$final = endCycle($startDate, $nMonths); // output: 2014-07-02
Use DateTime::add.
$start = new DateTime("2010-12-11", new DateTimeZone("UTC"));
$month_later = clone $start;
$month_later->add(new DateInterval("P1M"));
I used clone because add modifies the original object, which might not be desired.
strtotime( "+1 month", strtotime( $time ) );
this returns a timestamp that can be used with the date function
You can use DateTime::modify like this :
$date = new DateTime('2010-12-11');
$date->modify('+1 month');
See documentations :
https://php.net/manual/en/datetime.modify.php
https://php.net/manual/en/class.datetime.php
UPDATE january 2021 : correct mistakes raised by comments
This solution has some problems for months with 31 days like May etc.
Exemple : this jumps from 31st May to 1st July which is incorrect.
To correct that, you can create this custom function
function addMonths($date,$months){
$init=clone $date;
$modifier=$months.' months';
$back_modifier =-$months.' months';
$date->modify($modifier);
$back_to_init= clone $date;
$back_to_init->modify($back_modifier);
while($init->format('m')!=$back_to_init->format('m')){
$date->modify('-1 day') ;
$back_to_init= clone $date;
$back_to_init->modify($back_modifier);
}
}
Then you can use it like that :
$date = new DateTime('2010-05-31');
addMonths($date, 1);
print_r($date);
//DateTime Object ( [date] => 2010-06-30 00:00:00.000000 [timezone_type] => 3 [timezone] => Europe/Berlin )
This solution was found in PHP.net posted by jenspj : https://www.php.net/manual/fr/datetime.modify.php#107592
(date('d') > 28) ? date("mdY", strtotime("last day of next month")) : date("mdY", strtotime("+1 month"));
This will compensate for February and the other 31 day months. You could of course do a lot more checking to to get more exact for 'this day next month' relative date formats (which does not work sadly, see below), and you could just as well use DateTime.
Both DateInterval('P1M') and strtotime("+1 month") are essentially blindly adding 31 days regardless of the number of days in the following month.
2010-01-31 => March 3rd
2012-01-31 => March 2nd (leap year)
Please first you set your date format as like 12-12-2012
After use this function it's work properly;
$date = date('d-m-Y',strtotime("12-12-2012 +2 Months");
Here 12-12-2012 is your date and +2 Months is increment of the month;
You also increment of Year, Date
strtotime("12-12-2012 +1 Year");
Ans is 12-12-2013
I use this way:-
$occDate='2014-01-28';
$forOdNextMonth= date('m', strtotime("+1 month", strtotime($occDate)));
//Output:- $forOdNextMonth=02
/*****************more example****************/
$occDate='2014-12-28';
$forOdNextMonth= date('m', strtotime("+1 month", strtotime($occDate)));
//Output:- $forOdNextMonth=01
//***********************wrong way**********************************//
$forOdNextMonth= date('m', strtotime("+1 month", $occDate));
//Output:- $forOdNextMonth=02; //instead of $forOdNextMonth=01;
//******************************************************************//
Just updating the answer with simple method for find the date after no of months. As the best answer marked doesn't give the correct solution.
<?php
$date = date('2020-05-31');
$current = date("m",strtotime($date));
$next = date("m",strtotime($date."+1 month"));
if($current==$next-1){
$needed = date('Y-m-d',strtotime($date." +1 month"));
}else{
$needed = date('Y-m-d', strtotime("last day of next month",strtotime($date)));
}
echo "Date after 1 month from 2020-05-31 would be : $needed";
?>
If you want to get the date of one month from now you can do it like this
echo date('Y-m-d', strtotime('1 month'));
If you want to get the date of two months from now, you can achieve that by doing this
echo date('Y-m-d', strtotime('2 month'));
And so on, that's all.
Thanks Jason, your post was very helpful. I reformatted it and added more comments to help me understand it all. In case that helps anyone, I have posted it here:
function cycle_end_date($cycle_start_date, $months) {
$cycle_start_date_object = new DateTime($cycle_start_date);
//Find the date interval that we will need to add to the start date
$date_interval = find_date_interval($months, $cycle_start_date_object);
//Add this date interval to the current date (the DateTime class handles remaining complexity like year-ends)
$cycle_end_date_object = $cycle_start_date_object->add($date_interval);
//Subtract (sub) 1 day from date
$cycle_end_date_object->sub(new DateInterval('P1D'));
//Format final date to Y-m-d
$cycle_end_date = $cycle_end_date_object->format('Y-m-d');
return $cycle_end_date;
}
//Find the date interval we need to add to start date to get end date
function find_date_interval($n_months, DateTime $cycle_start_date_object) {
//Create new datetime object identical to inputted one
$date_of_last_day_next_month = new DateTime($cycle_start_date_object->format('Y-m-d'));
//And modify it so it is the date of the last day of the next month
$date_of_last_day_next_month->modify('last day of +'.$n_months.' month');
//If the day of inputted date (e.g. 31) is greater than last day of next month (e.g. 28)
if($cycle_start_date_object->format('d') > $date_of_last_day_next_month->format('d')) {
//Return a DateInterval object equal to the number of days difference
return $cycle_start_date_object->diff($date_of_last_day_next_month);
//Otherwise the date is easy and we can just add a month to it
} else {
//Return a DateInterval object equal to a period (P) of 1 month (M)
return new DateInterval('P'.$n_months.'M');
}
}
$cycle_start_date = '2014-01-31'; // select date in Y-m-d format
$n_months = 1; // choose how many months you want to move ahead
$cycle_end_date = cycle_end_date($cycle_start_date, $n_months); // output: 2014-07-02
function dayOfWeek($date){
return DateTime::createFromFormat('Y-m-d', $date)->format('N');
}
Usage examples:
echo dayOfWeek(2016-12-22);
// "4"
echo dayOfWeek(date('Y-m-d'));
// "4"
$date = strtotime("2017-12-11");
$newDate = date("Y-m-d", strtotime("+1 month", $date));
If you want to increment by days you can also do it
$date = strtotime("2017-12-11");
$newDate = date("Y-m-d", strtotime("+5 day", $date));
For anyone looking for an answer to any date format.
echo date_create_from_format('d/m/Y', '15/04/2017')->add(new DateInterval('P1M'))->format('d/m/Y');
Just change the date format.
//ECHO MONTHS BETWEEN TWO TIMESTAMPS
$my_earliest_timestamp = 1532095200;
$my_latest_timestamp = 1554991200;
echo '<pre>';
echo "Earliest timestamp: ". date('c',$my_earliest_timestamp) ."\r\n";
echo "Latest timestamp: " .date('c',$my_latest_timestamp) ."\r\n\r\n";
echo "Month start of earliest timestamp: ". date('c',strtotime('first day of '. date('F Y',$my_earliest_timestamp))) ."\r\n";
echo "Month start of latest timestamp: " .date('c',strtotime('first day of '. date('F Y',$my_latest_timestamp))) ."\r\n\r\n";
echo "Month end of earliest timestamp: ". date('c',strtotime('last day of '. date('F Y',$my_earliest_timestamp)) + 86399) ."\r\n";
echo "Month end of latest timestamp: " .date('c',strtotime('last day of '. date('F Y',$my_latest_timestamp)) + 86399) ."\r\n\r\n";
$sMonth = strtotime('first day of '. date('F Y',$my_earliest_timestamp));
$eMonth = strtotime('last day of '. date('F Y',$my_earliest_timestamp)) + 86399;
$xMonth = strtotime('+1 month', strtotime('first day of '. date('F Y',$my_latest_timestamp)));
while ($eMonth < $xMonth) {
echo "Things from ". date('Y-m-d',$sMonth) ." to ". date('Y-m-d',$eMonth) ."\r\n\r\n";
$sMonth = $eMonth + 1; //add 1 second to bring forward last date into first second of next month.
$eMonth = strtotime('last day of '. date('F Y',$sMonth)) + 86399;
}
I find the mtkime() function works really well for this:
$start_date="2021-10-01";
$start_date_plus_a_month=date("Y-m-d", mktime(0, 0, 0, date("m",strtotime($start_date))+1, date("d",strtotime($start_date)), date("Y",strtotime($start_date))));
result: 2021-11-01
I like to subtract 1 from the 'day' to produce '2021-10-31' which can be useful if you want to display a range across 12 months, e.g. Oct 1, 2021 to Sep 30 2022
$start_date_plus_a_year=date("Y-m-d", mktime(0, 0, 0, date("m",strtotime($start_date))+12, date("d",strtotime($start_date))-1, date("Y",strtotime($start_date))));
result: 2022-09-30
The correct answer to the exact question asked is Giuseppe Canale's answer from earlier. I'm going to answer a slightly more generic question of how to increment the date by an arbitrary number of months, however.
<?php
/**
* Will return a timestamp corresponding to first day of the month that is N months into the future.
* #param int $months_later number of months into the future: 0 for current one
* #param string $today if supplied will be used as the "now" time
* #return int
*/
function rel_month_to_time($months_later, $today=null) {
if ($months_later===0) {
return is_null($today) ? time() : strtotime($today);
}
return strtotime('first day of next month', rel_month_to_time($months_later-1, $today));
}
As is many times the case, you can use recursion for these "human problems" like calendars. The above can be used to return a timestamp corresponding to "next month" -- the way we humans think of it.
<?php echo date('Y-m-d', rel_month_to_time(1, '2023-01-30'));
// 2023-02-01
As pointed by #NetVicious i corrected the code, it should work with all dates, some example:
2013-01-30 will be 2013-02-28
2013-05-15 will be 2013-05-15
2013-05-31 will be 2013-06-30
This code uses the DateTime class to create a new date object, then it adds 1 month to the date using the modify method. Next, it gets the day of the next month using the format method. If the next month's day doesn't match the original day, it modifies the date to the last day of the previous month using the modify method.
$original_date = "2013-01-30";
$original_day = date("d", strtotime($original_date));
$date = new DateTime($original_date);
$date->modify('+1 month');
$next_month_day = $date->format('d');
if ($next_month_day != $original_day) {
$date->modify('last day of previous month');
}
$new_date = $date->format('Y-m-d');
echo $new_date;
All presented solutions are not working properly.
strtotime() and DateTime::add or DateTime::modify give sometime invalid results.
Examples:
- 31.08.2019 + 1 month gives 01.10.2019 instead 30.09.2019
- 29.02.2020 + 1 year gives 01.03.2021 instead 28.02.2021
(tested on PHP 5.5, PHP 7.3)
Below is my function based on idea posted by Angelo that solves the problem:
// $time - unix time or date in any format accepted by strtotime() e.g. 2020-02-29
// $days, $months, $years - values to add
// returns new date in format 2021-02-28
function addTime($time, $days, $months, $years)
{
// Convert unix time to date format
if (is_numeric($time))
$time = date('Y-m-d', $time);
try
{
$date_time = new DateTime($time);
}
catch (Exception $e)
{
echo $e->getMessage();
exit;
}
if ($days)
$date_time->add(new DateInterval('P'.$days.'D'));
// Preserve day number
if ($months or $years)
$old_day = $date_time->format('d');
if ($months)
$date_time->add(new DateInterval('P'.$months.'M'));
if ($years)
$date_time->add(new DateInterval('P'.$years.'Y'));
// Patch for adding months or years
if ($months or $years)
{
$new_day = $date_time->format("d");
// The day is changed - set the last day of the previous month
if ($old_day != $new_day)
$date_time->sub(new DateInterval('P'.$new_day.'D'));
}
// You can chage returned format here
return $date_time->format('Y-m-d');
}
Usage examples:
echo addTime('2020-02-29', 0, 0, 1); // add 1 year (result: 2021-02-28)
echo addTime('2019-08-31', 0, 1, 0); // add 1 month (result: 2019-09-30)
echo addTime('2019-03-15', 12, 2, 1); // add 12 days, 2 months, 1 year (result: 2019-09-30)
put a date in input box then click the button get day from date in jquery
$(document).ready( function() {
$("button").click(function(){
var day = ["Sunday","Monday","Tuesday","Wednesday","Thursday","Friday","Saturday"];
var a = new Date();
$(".result").text(day[a.getDay()]);
});
});
<?php
$selectdata ="select fromd,tod from register where username='$username'";
$q=mysqli_query($conm,$selectdata);
$row=mysqli_fetch_array($q);
$startdate=$row['fromd'];
$stdate=date('Y', strtotime($startdate));
$endate=$row['tod'];
$enddate=date('Y', strtotime($endate));
$years = range ($stdate,$enddate);
echo '<select name="years" class="form-control">';
echo '<option>SELECT</option>';
foreach($years as $year)
{ echo '<option value="'.$year.'"> '.$year.' </option>'; }
echo '</select>'; ?>