How to get date after 3 months in php?

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.

Pops

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asked Mar 26, 2012 at 15:33

user979331user979331

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6

Change it to this will give you the expected format:

$effectiveDate = date('Y-m-d', strtotime("+3 months", strtotime($effectiveDate)));

answered Mar 26, 2012 at 15:41

5

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;

answered Jun 5, 2018 at 10:11

SadeeSadee

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0

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

answered Mar 26, 2012 at 15:41

NickNick

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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.

answered Mar 26, 2012 at 15:44

JohnnyQJohnnyQ

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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.)

answered Dec 8, 2015 at 16:44

gleechgleech

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The following should work,Please Try this:

$effectiveDate = strtotime("+1 months", strtotime(date("y-m-d")));
echo $time = date("y/m/d", $effectiveDate);

mkl

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answered Apr 4, 2017 at 7:29

Following should work

$d = strtotime("+1 months",strtotime("2015-05-25"));
echo   date("Y-m-d",$d); // This will print **2015-06-25** 

answered May 25, 2017 at 10:44

RickyRicky

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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 "
";
    echo "Months : ".$months = date('M Y', "+$x months");
    echo '
';
    echo "Years : ".$years = date('Y', "+$y years");
    echo '
';
}

answered Aug 8, 2018 at 10:10

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)));

answered Mar 26, 2012 at 15:40

Brendon DuganBrendon Dugan

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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))); 
}

answered Sep 1 at 16:14

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