PHP add days on to a specific date? - php

Question

I am feeling a bit thick today and maybe a little tired..
I am trying to add days on to a string date...
$startdate = "18/7/2011";
$enddate = date(strtotime($startdate) . " +1 day");
echo $startdate;
echo $enddate;
My heads not with it... where am i going wrong ?
Thanks
Lee

Answer 1

Either
$enddate = date(strtotime("+1 day", strtotime($startdate)));
or
$enddate = date(strtotime($startdate . "+1 day"));
should work. However, neither is working with the 18/7/2011 date. They work fine with 7/18/2011: http://codepad.viper-7.com/IDS0gI . Might be some localization problem.
In the first way, using the second parameter to strtotime says to add one day relative to that date. In the second way, strtotime figures everything out. But apparently only if the date is in the USA's date format, or in the other format using dashes: http://codepad.viper-7.com/SKJ49r

Answer 2

try this one, (tested and worked fine)
date('d-m-Y',strtotime($startdate . ' +1 day'));
date('d-m-Y',strtotime($startdate . ' +2 day'));
date('d-m-Y',strtotime($startdate . ' +3 day'));
date('d-m-Y',strtotime($startdate . ' +30 day'));

Answer 3

first parameter of date() is format
d.m.Y G:i:s
for example
additionally, your $startdate is invalid

Answer 4

You're probably looking for strtotime($startdate . "+ 1 day") or something

Answer 5

First you have to change the date format
by calling changeDateFormat("18/7/2011"): returns: 2011-07-18
if your parsing argument
function changeDateFormat($vdate){
$pos = strpos($vdate, '/');
if ($pos === false) return $vdate;
$pieces = explode("/", $vdate);
$thisday = str_pad($pieces[0], 2, "0", STR_PAD_LEFT);
$thismonth = str_pad($pieces[1], 2, "0", STR_PAD_LEFT);
$thisyear = $pieces[2];
$thisdate = "$thisyear-$thismonth-$thisday";
return $thisdate;
}
And this..
$startdate = changeDateFormat($startdate);
$enddate = date('Y-m-d', strtotime($startdate . "+".$noOfDays." day"));

Answer 6

This will work
$startdate = "18/7/2011";
$enddate = date('d/m/Y', strtotime($startdate) + strtotime("+1 day", 0));
echo $startdate;
echo $enddate;
First, the start date is parsed into integer, then the relative time is parsed.
You might also utilize the second parametr of strToTime:
$startdate = "18/7/2011";
$enddate = date('d/m/Y', strtotime("+1 day", strtotime($startdate)));