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I'm trying to get a date that is one year from the date I specify.
My code looks like this:
$futureDate=date('Y-m-d', strtotime('+one year', $startDate));
It's returning the wrong date. Any ideas why?
$futureDate=date('Y-m-d', strtotime('+1 year'));
$futureDate is one year from now!
$futureDate=date('Y-m-d', strtotime('+1 year', strtotime($startDate)) );
$futureDate is one year from $startDate!
To add one year to todays date use the following:
$oneYearOn = date('Y-m-d',strtotime(date("Y-m-d", mktime()) . " + 365 day"));
For the other examples you must initialize $StartingDate with a timestamp value
for example:
$StartingDate = mktime(); // todays date as a timestamp
Try this
$newEndingDate = date("Y-m-d", strtotime(date("Y-m-d", strtotime($StaringDate)) . " + 365 day"));
or
$newEndingDate = date("Y-m-d", strtotime(date("Y-m-d", strtotime($StaringDate)) . " + 1 year"));
//1 year from today's date
echo date('d-m-Y', strtotime('+1 year'));
//1 year from from specific date
echo date('22-09-Y', strtotime('+1 year'));
hope this simpler bit of code helps someone in future :)
Try: $futureDate=date('Y-m-d',strtotime('+1 year',$startDate));
just had the same problem, however this was the simplest solution:
<?php (date('Y')+1).date('-m-d'); ?>
// Declare a variable for this year
$this_year = date("Y");
// Add 1 to the variable
$next_year = $this_year + 1;
$year_after = $this_year + 2;
// Check your code
echo "This year is ";
echo $this_year;
echo "<br />";
echo "Next year is ";
echo $next_year;
echo "<br />";
echo "The year after that is ";
echo $year_after;
I prefer the OO approach:
$date = new \DateTimeImmutable('today'); //'today' gives midnight, leave blank for current time.
$futureDate = $date->add(\DateInterval::createFromDateString('+1 Year'))
Use DateTimeImmutable otherwise you will modify the original date too!
more on DateTimeImmutable: http://php.net/manual/en/class.datetimeimmutable.php
If you just want from todays date then you can always do:
new \DateTimeImmutable('-1 Month');
If you are using PHP 5.3, it is because you need to set the default time zone:
date_default_timezone_set()
strtotime() is returning bool(false), because it can't parse the string '+one year' (it doesn't understand "one"). false is then being implicitly cast to the integer timestamp 0. It's a good idea to verify strtotime()'s output isn't bool(false) before you go shoving it in other functions.
From the docs:
Return Values
Returns a timestamp on success, FALSE
otherwise. Previous to PHP 5.1.0, this
function would return -1 on failure.
Try This
$nextyear = date("M d,Y",mktime(0, 0, 0, date("m",strtotime($startDate)), date("d",strtotime($startDate)), date("Y",strtotime($startDate))+1));
There is also a simpler and less sophisticated solution:
$monthDay = date('m/d');
$year = date('Y')+1;
$oneYearFuture = "".$monthDay."/".$year."";
echo"The date one year in the future is: ".$oneYearFuture."";
You can use strtotime() to get future time.
//strtotime('+1 day');
//strtotime('+1 week');
//strtotime('+1 month');
$now = date('Y-m-d');
$oneYearLaterFromNow = date('Y-m-d', strtotime('+1 year'));
$oneYearLaterFromAnyDate = date('Y-m-d', strtotime('+1 year', strtotime($anyValidDateString)));
In my case (i want to add 3 years to current date) the solution was:
$future_date = date('Y-m-d', strtotime("now + 3 years"));
To Gardenee, Treby and Daniel Lima:
what will happen with 29th February? Sometimes February has only 28 days :)
My solution is: date('Y-m-d', time()-60*60*24*365);
You can make it more "readable" with defines:
define('ONE_SECOND', 1);
define('ONE_MINUTE', 60 * ONE_SECOND);
define('ONE_HOUR', 60 * ONE_MINUTE);
define('ONE_DAY', 24 * ONE_HOUR);
define('ONE_YEAR', 365 * ONE_DAY);
date('Y-m-d', time()-ONE_YEAR);
I want to calculate EXACT past 30 days time period in php from now (for example 30 aug 14 23:06) to 30 days back (for example 1 aug 14 23:06). I wrote this where current datetime goes in $d1 and past 30 days datetime goes in $d2 but somehow i am not getting correct results. Any idea?
$url=$row["url"];
$pageid=getPageID($url);
$date=date('y-m-d g:i');
$d1=strtotime($date);
$d2=date(strtotime('today - 30 days'));
Thanks
The problem is likely caused by the malformed date() call. The first argument passed to date() should be the format (as shown in the Docs) and the second should be an optional timestamp.
Try this:
$d2 = date('c', strtotime('-30 days'));
PHPFiddle
As a short aside, the whole snippet can be simplified as follows:
$url = $row["url"];
$pageid = getPageID($url);
$date = date('y-m-d g:i');
$d1 = time();
$d2 = date('y-m-d g:i', strtotime('-30 days'));
You can also use the DateTime class's sub() method together with an DateInterval:
$now = new DateTime();
$back = $now->sub(DateInterval::createFromDateString('30 days'));
echo $back->format('y-m-d g:i');
if you would like to get out put as 2014-08-01 then try the below code. thanks
$date = '2014-08-30 23:06';
$new_date = date('Y-m-d G:i', strtotime($date.' - 29 days'));
echo "30 days back is " . $new_date;
From your brief description and example given, I believe that you want the date to be 30 days back and time to be the same as of now. The below code will serve this purpose. Thanks.
<?php
$date=date('y-m-d g:i');
$time=date('g:i');
echo "Todays date:" . $date. "<br>";
$d2 = date('y-m-d', strtotime('-30 days'));
echo "30 days back:" . $d2 . ' ' .$time;
?>
Try:
echo date("Y-m-d h:i:s",strtotime('-30 days'));
For more detail click here
Very simple two lines of code
$date = new DateTime();
echo $date->modify('-30 day')->format('y-m-d g:i');
I know you said with PHP, however, I can't imagine not getting the records from a DB. If you want to do so from the DB,use:
$sql='SELECT * FROM myTable WHERE date > CURRENT_DATE - INTERVAL 30 DAY';
$pdo->query($sql);
Very simple one lines of code:
echo (new DateTime())->modify('-30 day')->format('y-m-d g:i');
In the example below, it makes no sense if the variable $date is not
used anywhere else!
$date = new DateTime();
echo $date->modify('-30 day')->format('y-m-d g:i');
Sample answer is
$dateBack30Days=date('Y-m-d g:i', strtotime('-30 days'));
I have a variable called $effectiveDate containing the date 2012-03-26.
I am trying to add three months to this date and have been unsuccessful at it.
Here is what I have tried:
$effectiveDate = strtotime("+3 months", strtotime($effectiveDate));
and
$effectiveDate = strtotime(date("Y-m-d", strtotime($effectiveDate)) . "+3 months");
What am I doing wrong? Neither piece of code worked.
Change it to this will give you the expected format:
$effectiveDate = date('Y-m-d', strtotime("+3 months", strtotime($effectiveDate)));
This answer is not exactly to this question. But I will add this since this question still searchable for how to add/deduct period from date.
$date = new DateTime('now');
$date->modify('+3 month'); // or you can use '-90 day' for deduct
$date = $date->format('Y-m-d h:i:s');
echo $date;
I assume by "didn't work" you mean that it's giving you a timestamp instead of the formatted date, because you were doing it correctly:
$effectiveDate = strtotime("+3 months", strtotime($effectiveDate)); // returns timestamp
echo date('Y-m-d',$effectiveDate); // formatted version
You need to convert the date into a readable value. You may use strftime() or date().
Try this:
$effectiveDate = strtotime("+3 months", strtotime($effectiveDate));
$effectiveDate = strftime ( '%Y-%m-%d' , $effectiveDate );
echo $effectiveDate;
This should work. I like using strftime better as it can be used for localization you might want to try it.
Tchoupi's answer can be made a tad less verbose by concatenating the argument for strtotime() as follows:
$effectiveDate = date('Y-m-d', strtotime($effectiveDate . "+3 months") );
(This relies on magic implementation details, but you can always go have a look at them if you're rightly mistrustful.)
The following should work,Please Try this:
$effectiveDate = strtotime("+1 months", strtotime(date("y-m-d")));
echo $time = date("y/m/d", $effectiveDate);
Following should work
$d = strtotime("+1 months",strtotime("2015-05-25"));
echo date("Y-m-d",$d); // This will print **2015-06-25**
Add nth Days, months and years
$n = 2;
for ($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>';
}
As of PHP 5.3, DateTime along with DateInterval could be a feasible option to achieve the desired result.
$months = 6;
$currentDate = new DateTime();
$newDate = $currentDate->add(new DateInterval('P'.$months.'M'));
echo $newDate->format('Y-m-d');
If you want to subtract time from a date, instead of add, use sub.
Here are more examples on how to use DateInterval:
$interval = new DateInterval('P1Y2M3DT4H5M6S');
// This creates an interval of 1 year, 2 months, 3 days, 4 hours, 5 minutes, and 6 seconds.
$interval = new DateInterval('P2W');
// This creates an interval of 2 weeks (which is equivalent to 14 days).
$interval = new DateInterval('PT1H30M');
// This creates an interval of 1 hour and 30 minutes (but no days or years, etc.).
The following should work, but you may need to change the format:
echo date('l F jS, Y (m-d-Y)', strtotime('+3 months', strtotime($DateToAdjust)));
public function getCurrentDate(){
return date("Y-m-d H:i:s");
}
public function getNextDateAfterMonth($date1,$monthNumber){
return date('Y-m-d H:i:s', strtotime("+".$monthNumber." months", strtotime($date1)));
}
I'm really stuck with adding X minutes to a datetime, after doing lots of google'ing and PHP manual reading, I don't seem to be getting anywhere.
The date time format I have is:
2011-11-17 05:05: year-month-day hour:minute
Minutes to add will just be a number between 0 and 59
I would like the output to be the same as the input format with the minutes added.
Could someone give me a working code example, as my attempts don't seem to be getting me anywhere?
$minutes_to_add = 5;
$time = new DateTime('2011-11-17 05:05');
$time->add(new DateInterval('PT' . $minutes_to_add . 'M'));
$stamp = $time->format('Y-m-d H:i');
The ISO 8601 standard for duration is a string in the form of P{y}Y{m1}M{d}DT{h}H{m2}M{s}S where the {*} parts are replaced by a number value indicating how long the duration is.
For example, P1Y2DT5S means 1 year, 2 days, and 5 seconds.
In the example above, we are providing PT5M (or 5 minutes) to the DateInterval constructor.
PHP's DateTime class has a useful modify method which takes in easy-to-understand text.
$dateTime = new DateTime('2011-11-17 05:05');
$dateTime->modify('+5 minutes');
You could also use string interpolation or concatenation to parameterize it:
$dateTime = new DateTime('2011-11-17 05:05');
$minutesToAdd = 5;
$dateTime->modify("+{$minutesToAdd} minutes");
$newtimestamp = strtotime('2011-11-17 05:05 + 16 minute');
echo date('Y-m-d H:i:s', $newtimestamp);
result is
2011-11-17 05:21:00
Live demo is here
If you are no familiar with strtotime yet, you better head to php.net to discover it's great power :-)
You can do this with native functions easily:
strtotime('+59 minutes', strtotime('2011-11-17 05:05'));
I'd recommend the DateTime class method though, just posted by Tim.
I don't know why the approach set as solution didn't work for me.
So I'm posting here what worked for me in hope it can help anybody:
$startTime = date("Y-m-d H:i:s");
//display the starting time
echo '> '.$startTime . "<br>";
//adding 2 minutes
$convertedTime = date('Y-m-d H:i:s', strtotime('+2 minutes', strtotime($startTime)));
//display the converted time
echo '> '.$convertedTime;
I thought this would help some when dealing with time zones too. My modified solution is based off of #Tim Cooper's solution, the correct answer above.
$minutes_to_add = 10;
$time = new DateTime();
**$time->setTimezone(new DateTimeZone('America/Toronto'));**
$time->add(new DateInterval('PT' . $minutes_to_add . 'M'));
$timestamp = $time->format("Y/m/d G:i:s");
The bold line, line 3, is the addition. I hope this helps some folks as well.
A bit of a late answer, but the method I would use is:
// Create a new \DateTime instance
$date = DateTime::createFromFormat('Y-m-d H:i:s', '2015-10-26 10:00:00');
// Modify the date
$date->modify('+5 minutes');
// Output
echo $date->format('Y-m-d H:i:s');
Or in PHP >= 5.4
echo (DateTime::createFromFormat('Y-m-d H:i:s', '2015-10-26 10:00:00'))->modify('+5 minutes')->format('Y-m-d H:i:s')
If you want to give a variable that contains the minutes.
Then I think this is a great way to achieve this.
$minutes = 10;
$maxAge = new DateTime('2011-11-17 05:05');
$maxAge->modify("+{$minutes} minutes");
Use strtotime("+5 minute", $date);
Example:
$date = "2017-06-16 08:40:00";
$date = strtotime($date);
$date = strtotime("+5 minute", $date);
echo date('Y-m-d H:i:s', $date);
As noted by Brad and Nemoden in their answers above, strtotime() is a great function. Personally, I found the standard DateTime Object to be overly complicated for many use cases. I just wanted to add 5 minutes to the current time, for example.
I wrote a function that returns a date as a string with some optional parameters:
1.) time:String | ex: "+5 minutes" (default = current time)
2.) format:String | ex: "Y-m-d H:i:s" (default = "Y-m-d H:i:s O")
Obviously, this is not a fully featured method. Just a quick and simple function for modifying/formatting the current date.
function get_date($time=null, $format='Y-m-d H:i:s O')
{
if(empty($time))return date($format);
return date($format, strtotime($time));
}
// Example #1: Return current date in default format
$date = get_date();
// Example #2: Add 5 minutes to the current date
$date = get_date("+5 minutes");
// Example #3: Subtract 30 days from the current date & format as 'Y-m-d H:i:s'
$date = get_date("-30 days", "Y-m-d H:i:s");
one line mysql datetime format
$mysql_date_time = (new DateTime())->modify('+15 minutes')->format("Y-m-d H:i:s");
One more example of a function to do this: (changing the time and interval formats however you like them according to this for function.date, and this for DateInterval):
(I've also written an alternate form of the below function.)
// Return adjusted time.
function addMinutesToTime( $dateTime, $plusMinutes ) {
$dateTime = DateTime::createFromFormat( 'Y-m-d H:i', $dateTime );
$dateTime->add( new DateInterval( 'PT' . ( (integer) $plusMinutes ) . 'M' ) );
$newTime = $dateTime->format( 'Y-m-d H:i' );
return $newTime;
}
$adjustedTime = addMinutesToTime( '2011-11-17 05:05', 59 );
echo '<h1>Adjusted Time: ' . $adjustedTime . '</h1>' . PHP_EOL . PHP_EOL;
Without using a variable:
$yourDate->modify("15 minutes");
echo $yourDate->format( "Y-m-d H:i");
With using a variable:
$interval= 15;
$yourDate->modify("+{$interval } minutes");
echo $yourDate->format( "Y-m-d H:i");
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);