php function to add number of days to a date

Question

I have found several posts regarding this topic but I cannot get my function to work. If the amount of days to add > 0, the returning date = 70-01-01

Example: $date='12-07-11' $days='5'

Should return 17-07-11 instead of 70-01-01
I need to use the variables because $date and $days are results from an oracle query.

function addDayswithdate($date,$days)
{
  $originalDate = $date;
  $newDate = date("y-m-d", strtotime($originalDate));

  if ($days>0)
  {
    $var="+". $days ." days";
    $date = strtotime($var, strtotime($newDate));
  }

  $originalDate = $date;
  $newDate = date("y-m-d", strtotime($date));

  return $newDate;
}

Try following the advice [in this post](http://stackoverflow.com/questions/3727615/adding-days-to-date-in-php). — askrich, Apr 10 '14 at 11:05
I suggest you avoid handling dates as strings. In PHP you have timestamps and objects to choose from. — Álvaro González, Apr 10 '14 at 11:05
I have tried all the other posts, but the trick is in the date format, which is d-m-y — Mascha Oostenbrug, Apr 10 '14 at 11:15
Split your task in two parts (1.- convert from string to date 2.- add one day) and solve them separately. — Álvaro González, Apr 10 '14 at 11:21

score 4 · Accepted Answer · answered Apr 10 '14 at 11:08

4

Use Datetime::createFromFormat and DateInterval

$date = DateTime::createFromFormat('d-m-y','12-07-11');
$date->add(new DateInterval('P5D'));
echo $date->format('d-m-Y') . "\n";

answered Apr 10 '14 at 11:08

Abhik Chakraborty

44,654
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Thanks, I have changed it to this and it is working! $date = DateTime::createFromFormat('d-m-y',$_date); $date->add(new DateInterval('P'. $days . 'D')); return $date->format('d-m-Y') . "\n"; – Mascha Oostenbrug Apr 10 '14 at 11:23
Now this is the answer I was looking for. I have got date saved as string but every other answer on similar questions on stack overflow assumes that date is in date format! There are many such questions and your answer should definitely come at the top of all. Thanks. – gegobyte Feb 05 '16 at 10:25

score 0 · Answer 2 · answered Apr 10 '14 at 11:05

0

Why not keep it simpler?

function addDays($date, $days)
{
   $time = strtotime($date);
   $new_time = $time + (86400 * $days) //86400 seconds per day
   return $new_time;
}

answered Apr 10 '14 at 11:05

Ben Poulson

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Because some days have 23 or 25 hours (apart from leap seconds). – Álvaro González Apr 10 '14 at 11:09

InsaneSVK · Answer 3 · 2014-04-10T11:11:10.637

0

Why not do it like this:

echo date('Y-m-d', strtotime($date + $days .'days'));

edited Apr 10 '14 at 11:11

answered Apr 10 '14 at 11:05

InsaneSVK

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score 0 · Answer 4 · answered Apr 10 '14 at 11:06

0

function addDays($date, $days)
{
    $date_uts = strtotoime($date);
    $add_days = $days*86400;

    $final_date = date('Y-m-d', $date_uts+$add_days);
    return $final_date;
}

answered Apr 10 '14 at 11:06

Parag Tyagi

8,780
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If $date = 23-06-11 and $days = 1, the returning value is 2023-06-12. Not correc. Should be 24-06-11 – Mascha Oostenbrug Apr 10 '14 at 11:12
Assuming that days always have 86400 seconds is like assuming months always have 30 days, except that the error is harder to spot. Don't get into this path: date calculations are hard, let the computer do them for you. – Álvaro González Apr 10 '14 at 11:19

score 0 · Answer 5 · answered Apr 10 '14 at 11:18

0

In addition of using vanilla php DateTime class, you can use Carbon if you have to work often with dates.

Carbon let you do anything like this

$date = Carbon::now()->addDays(5);

answered Apr 10 '14 at 11:18

Fed03

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php function to add number of days to a date

5 Answers5