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I have this code
<?php
$date =date(Y-m-d);
$day = 5;
$newdate= $date+$day
echo "today is:"$date;
echo "<br> and after 5 days is :"$newdate;
?>
I want the result is
today is :2016-11-2
and after 5 days is : 2016-11-7
Try this
$date = new DateTime(); // Creates new DatimeTime for today
$newdate = $date->modify( '+ 5 days' ); // Adds 5 days
echo $newdate->format( 'Y-m-d' ); // Echo and format the newdate to the wanted format
It should help you:
echo date('Y-m-d', strtotime($date. ' + 5 days'));
So it will be like follows:
<?php
$date = date('Y-m-d');
$newdate = date('Y-m-d', strtotime($date.' + 5 days'));
echo "today is: $date";
echo "<br> and after 5 days is: $newdate";
?>
You can use strtotime() function to add days to current date. Please see the below :
<?php
$date =date("Y-m-d");
$day = 5;
$newdate=date('Y-m-d', strtotime("+$day days"));
echo "today is:".$date;
echo "<br> and after 5 days is :".$newdate;
?>
it may help you
$date = "Mar 03, 2016";
$date = strtotime($date);
$date = strtotime("+7 day", $date);
echo date('M d, Y', $date);
I have this PHP code:
$end=date('Y-m-d');
I use it to get the current date, and I need the date 5 years in the future, something like:
$end=date('(Y + 5)-m-d');
How can I do this?
Try with:
$end = date('Y-m-d', strtotime('+5 years'));
Modifying dates based on this post
strtotime() is really powerful and allows you to modify/transform dates easily with it’s relative expressions too:
Procedural
$dateString = '2011-05-01 09:22:34';
$t = strtotime($dateString);
$t2 = strtotime('-3 days', $t);
echo date('r', $t2) . PHP_EOL; // returns: Thu, 28 Apr 2011 09:22:34 +0100
DateTime
$dateString = '2011-05-01 09:22:34';
$dt = new DateTime($dateString);
$dt->modify('-3 days');
echo $dt->format('r') . PHP_EOL; // returns: Thu, 28 Apr 2011 09:22:34 +0100
The stuff you can throw at strtotime() is quite surprising and very human readable. Have a look at this example looking for Tuesday next week.
Procedural
$t = strtotime("Tuesday next week");
echo date('r', $t) . PHP_EOL; // returns: Tue, 10 May 2011 00:00:00 +0100
DateTime
$dt = new DateTime("Tuesday next week");
echo $dt->format('r') . PHP_EOL; // returns: Tue, 10 May 2011 00:00:00 +0100
Note that these examples above are being returned relative to the time now.
The full list of time formats that strtotime() and the DateTime constructor takes are listed on the PHP Supported Date and Time Formats page.
Another example, suitable for your case could be: based on this post
<?php
//How to get the day 3 days from now:
$today = date("j");
$thisMonth = date("n");
$thisYear = date("Y");
echo date("F j Y", mktime(0,0,0, $thisMonth, $today+3, $thisYear));
//1 week from now:
list($today,$thisMonth,$thisYear) = explode(" ", date("j n Y"));
echo date("F j Y", mktime(0,0,0, $thisMonth, $today+7, $thisYear));
//4 months from now:
list($today,$thisMonth,$thisYear) = explode(" ", date("j n Y"));
echo date("F j Y", mktime(0,0,0, $thisMonth+4, $today, $thisYear));
//3 years, 2 months and 35 days from now:
list($today,$thisMonth,$thisYear) = explode(" ", date("j n Y"));
echo date("F j Y", mktime(0,0,0, $thisMonth+2, $today+35, $thisYear+3));
?>
Use this code to add years or months or days or hours or minutes or seconds to a given date
echo date("Y-m-d H:i:s", strtotime("+1 years", strtotime('2014-05-22 10:35:10'))); //2015-05-22 10:35:10
echo date("Y-m-d H:i:s", strtotime("+1 months", strtotime('2014-05-22 10:35:10')));//2014-06-22 10:35:10
echo date("Y-m-d H:i:s", strtotime("+1 days", strtotime('2014-05-22 10:35:10')));//2014-05-23 10:35:10
echo date("Y-m-d H:i:s", strtotime("+1 hours", strtotime('2014-05-22 10:35:10')));//2014-05-22 11:35:10
echo date("Y-m-d H:i:s", strtotime("+1 minutes", strtotime('2014-05-22 10:35:10')));//2014-05-22 10:36:10
echo date("Y-m-d H:i:s", strtotime("+1 seconds", strtotime('2014-05-22 10:35:10')));//2014-05-22 10:35:11
You can also subtract replacing + to -
$date = strtotime($row['timestamp']);
$newdate = date('d-m-Y',strtotime("+1 year",$date));
Its very very easy with Carbon.
$date = "2016-02-16"; // Or Your date
$newDate = Carbon::createFromFormat('Y-m-d', $date)->addYear(1);
Using Carbon:
$dt = Carbon::now();
echo $dt->addYears(5);
To add one year to todays date use the following:
$oneYearOn = date('Y-m-d',strtotime(date("Y-m-d", mktime()) . " + 365 day"));
You may use DateInterval for this purpose;
$currentDate = new \DateTime(); //creates today timestamp
$currentDate->add(new \DateInterval('P5Y')); //this means 5 Years
and you can now format it;
$currentDate->format('Y-m-d');
Try below code, i hope it will be helpful for you
<?php
$current_date=strtotime(date('Y-m-d'));
echo $end = date('Y-m-d', strtotime('+5 years',$current_date));
?>
try this ,
$presentyear = '2013-08-16 12:00:00';
$nextyear = date("M d,Y",mktime(0, 0, 0, date("m",strtotime($presentyear )), date("d",strtotime($presentyear )), date("Y",strtotime($presentyear ))+5));
echo $nextyear;
try this:
$yearnow= date("Y");
$yearnext=$yearnow+1;
echo date("Y")."-".$yearnext;
Try this code and add next Days, Months and Years
// current month: Aug 2018
$n = 2;
for ($i = 0; $i <= $n; $i++){
$d = strtotime("$i days");
$x = strtotime("$i month");
$y = strtotime("$i year");
echo "Dates : ".$dates = date('d M Y', "+$d days");
echo "<br>";
echo "Months : ".$months = date('M Y', "+$x months");
echo '<br>';
echo "Years : ".$years = date('Y', "+$y years");
echo '<br>';
}
I want to add 5 minutes to this date: 2011-04-8 08:29:49
$date = '2011-04-8 08:29:49';
When I use strtotime I am always getting 1970-01-01 08:33:31
How do I add correctly 5 minutes to 2011-04-8 08:29:49?
$date = '2011-04-8 08:29:49';
$currentDate = strtotime($date);
$futureDate = $currentDate+(60*5);
$formatDate = date("Y-m-d H:i:s", $futureDate);
Now, the result is 2011-04-08 08:34:49 and is stored inside $formatDate
Enjoy! :)
Try this:
echo date('Y-m-d H:i:s', strtotime('+5 minutes', strtotime('2011-04-8 08:29:49')));
$expire_stamp = date('Y-m-d H:i:s', strtotime("+5 min"));
$now_stamp = date("Y-m-d H:i:s");
echo "Right now: " . $now_stamp;
echo "5 minutes from right now: " . $expire_stamp;
Results in:
2012-09-30 09:00:03
2012-09-30 09:05:03
$date = '2011-04-8 08:29:49';
$newDate = date("Y-m-d H:i:s",strtotime($date." +5 minutes"))
For adding
$date = new DateTime('2014-02-20 14:20:00');
$date->add(new DateInterval('P0DT0H5M0S'));
echo $date->format('Y-m-d H:i:s');
It add 5minutes
For subtracting
$date = new DateTime('2014-02-20 14:20:00');
$date->sub(new DateInterval('P0DT0H5M0S'));
echo $date->format('Y-m-d H:i:s');
It subtract 5 minutes
If i'm right in thinking.
If you convert your date to a unix timestamp via strtotime(), then just add 300 (5min * 60 seconds) to that number.
$timestamp = strtotime($date) + (5*60)
Hope this helps
more illustrative for simple and clear solution
$date = '2011-04-8 08:29:49';
$newtimestamp = strtotime($date. ' + 5 minute');//gets timestamp
//convert into whichever format you need
$newdate = date('Y-m-d H:i:s', $newtimestamp);//it prints 2011-04-08 08:34:49
How can I get the timestamp of 12 o'clock of today, yesterday and the day before yesterday by using strtotime() function in php?
12 o'clock is a variable and would be changed by user.
$hour = 12;
$today = strtotime($hour . ':00:00');
$yesterday = strtotime('-1 day', $today);
$dayBeforeYesterday = strtotime('-1 day', $yesterday);
strtotime supports a number of interesting modifiers that can be used:
$hour = 12;
$today = strtotime("today $hour:00");
$yesterday = strtotime("yesterday $hour:00");
$dayBeforeYesterday = strtotime("yesterday -1 day $hour:00");
echo date("Y-m-d H:i:s\n", $today);
echo date("Y-m-d H:i:s\n", $yesterday);
echo date("Y-m-d H:i:s\n", $dayBeforeYesterday);
It works as predicted:
2011-01-24 12:00:00
2011-01-23 12:00:00
2011-01-22 12:00:00
OO Equivalent
$iHour = 12;
$oToday = new DateTime();
$oToday->setTime($iHour, 0);
$oYesterday = clone $oToday;
$oYesterday->modify('-1 day');
$oDayBefore = clone $oYesterday;
$oDayBefore->modify('-1 day');
$iToday = $oToday->getTimestamp();
$iYesterday = $oYesterday->getTimestamp();
$iDayBefore = $oDayBefore->getTimestamp();
echo "Today: $iToday\n";
echo "Yesterday: $iYesterday\n";
echo "Day Before: $iDayBefore\n";
You can easily find out any date using DateTime object, It is so flexible
$yesterday = new DateTime('yesterday');
echo $yesterday->format('Y-m-d');
$firstModayOfApril = new DateTime('first monday of april');
echo $firstModayOfApril->format('Y-m-d');
$nextMonday = new DateTime('next monday');
echo $nextMonday->format('Y-m-d');
to get start of day yesterday
$oDate = new DateTime();
$oDate->modify('-1 day');
echo $oDate->format('Y-m-d 00:00:00');
result
2014-11-05 00:00:00
All the answers here are too long and bloated, everyone loves one-lines ;)
$yesterday = Date('Y-m-d', strtotime('-1 day'));
(Or if you are American you can randomize the date unit order to m/d/y (or whatever you use) and use Cups, galloons, feet and horses as units...)
As of PHP 7 you can write something like this:
$today = new \DateTime();
$yesterday = (clone $today)->modify('-1 day');
$dayBefore = (clone $yesterday)->modify('-1 day');
// Then call ->format('Y-m-d 00:00:00'); on each objects
you can also use new DateTime("now") for today new DateTime("1 day ago") for yesterday or all can be parse by strtotime php function.
Then format as you want.
$timeStamp = time();
// $timeStamp = time() - 86400;
if (date('d.m.Y', $timeStamp) == date('d.m.Y')) {
echo 'Today';
} elseif (date('d.m.Y', $time) == date('d.m.Y', strtotime('-1 day'))) {
echo 'Yesterday';
}
I have a date returned as part of a MySQL query in the form 2010-09-17.
I would like to set the variables $Date2 to $Date5 as follows:
$Date2 = $Date + 1
$Date3 = $Date + 2
etc., so that it returns 2010-09-18, 2010-09-19, etc.
I have tried
date('Y-m-d', strtotime($Date. ' + 1 day'))
but this gives me the date before $Date.
What is the correct way to get my Dates in the format form 'Y-m-d' so that they may be used in another query?
All you have to do is use days instead of day like this:
<?php
$Date = "2010-09-17";
echo date('Y-m-d', strtotime($Date. ' + 1 days'));
echo date('Y-m-d', strtotime($Date. ' + 2 days'));
?>
And it outputs correctly:
2010-09-18
2010-09-19
If you're using PHP 5.3, you can use a DateTime object and its add method:
$Date1 = '2010-09-17';
$date = new DateTime($Date1);
$date->add(new DateInterval('P1D')); // P1D means a period of 1 day
$Date2 = $date->format('Y-m-d');
Take a look at the DateInterval constructor manual page to see how to construct other periods to add to your date (2 days would be 'P2D', 3 would be 'P3D', and so on).
Without PHP 5.3, you should be able to use strtotime the way you did it (I've tested it and it works in both 5.1.6 and 5.2.10):
$Date1 = '2010-09-17';
$Date2 = date('Y-m-d', strtotime($Date1 . " + 1 day"));
// var_dump($Date2) returns "2010-09-18"
From PHP 5.2 on you can use modify with a DateTime object:
http://php.net/manual/en/datetime.modify.php
$Date1 = '2010-09-17';
$date = new DateTime($Date1);
$date->modify('+1 day');
$Date2 = $date->format('Y-m-d');
Be careful when adding months... (and to a lesser extent, years)
Here is a small snippet to demonstrate the date modifications:
$date = date("Y-m-d");
//increment 2 days
$mod_date = strtotime($date."+ 2 days");
echo date("Y-m-d",$mod_date) . "\n";
//decrement 2 days
$mod_date = strtotime($date."- 2 days");
echo date("Y-m-d",$mod_date) . "\n";
//increment 1 month
$mod_date = strtotime($date."+ 1 months");
echo date("Y-m-d",$mod_date) . "\n";
//increment 1 year
$mod_date = strtotime($date."+ 1 years");
echo date("Y-m-d",$mod_date) . "\n";
You can also use the following format
strtotime("-3 days", time());
strtotime("+1 day", strtotime($date));
You can stack changes this way:
strtotime("+1 day", strtotime("+1 year", strtotime($date)));
Note the difference between this approach and the one in other answers: instead of concatenating the values +1 day and <timestamp>, you can just pass in the timestamp as the second parameter of strtotime.
Here has an easy way to solve this.
<?php
$date = "2015-11-17";
echo date('Y-m-d', strtotime($date. ' + 5 days'));
?>
Output will be:
2015-11-22
Solution has found from here - How to Add Days to Date in PHP
Using a variable for Number of days
$myDate = "2014-01-16";
$nDays = 16;
$newDate = strtotime($myDate . '+ '.$nDays.' days');
echo new Date('d/m/Y', $newDate); //format new date
Here is the simplest solution to your query
$date=date_create("2013-03-15"); // or your date string
date_add($date,date_interval_create_from_date_string("40 days"));// add number of days
echo date_format($date,"Y-m-d"); //set date format of the result
This works. You can use it for days, months, seconds and reformat the date as you require
public function reformatDate($date, $difference_str, $return_format)
{
return date($return_format, strtotime($date. ' ' . $difference_str));
}
Examples
echo $this->reformatDate('2021-10-8', '+ 15 minutes', 'Y-m-d H:i:s');
echo $this->reformatDate('2021-10-8', '+ 1 hour', 'Y-m-d H:i:s');
echo $this->reformatDate('2021-10-8', '+ 1 day', 'Y-m-d H:i:s');
To add a certain number of days to a date, use the following function.
function add_days_to_date($date1,$number_of_days){
/*
//$date1 is a string representing a date such as '2021-04-17 14:34:05'
//$date1 =date('Y-m-d H:i:s');
// function date without a secrod argument returns the current datetime as a string in the specified format
*/
$str =' + '. $number_of_days. ' days';
$date2= date('Y-m-d H:i:s', strtotime($date1. $str));
return $date2; //$date2 is a string
}//[end function]
All have to use bellow code:
$nday = time() + ( 24 * 60 * 60);
echo 'Now: '. date('Y-m-d') ."\n";
echo 'Next Day: '. date('Y-m-d', $nday) ."\n";
Another option is to convert your date string into a timestamp and then add the appropriate number of seconds to it.
$datetime_string = '2022-05-12 12:56:45';
$days_to_add = 1;
$new_timestamp = strtotime($datetime_string) + ($days_to_add * 60 * 60 * 24);
After which, you can use one of PHP's various date functions to turn the timestamp into a date object or format it into a human-readable string.
$new_datetime_string = date('Y-m-d H:i:s', $new_timestamp);