PHP printing wrong date

Question

PHP:

<?php echo date("d-m-Y",strtotime($pet['pet_lost_date']));?>

Variable Value:-

[pet_lost_date] => 16/04/2017

Output:

01-01-1970

Please help me why I searched a lot but I found nothing helpful.

Why is it appearing the wrong date? Where is an Error?

And How can I count from given date to toady total days month and years?

if `$pet['pet_lost_date']` is coming from a query, wouldn't you consider using [MySQL Date and Time Functions](https://dev.mysql.com/doc/refman/5.7/en/date-and-time-functions.html) and make sure `$pet['pet_lost_date']` isn't empty before use — OldPadawan, Apr 17 '17 at 09:43

score 0 · Answer 1 · answered Apr 17 '17 at 09:37

0

You need to format the date a bit differently:

$pet['pet_lost_date'] = "16/04/2017";
$pet['pet_lost_date'] = str_replace('/', '-', $pet['pet_lost_date']);
echo date("d-m-Y",strtotime($pet['pet_lost_date']));

Output:

16-04-2017

answered Apr 17 '17 at 09:37

Marcin

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Alive to die - Anant · Accepted Answer · 2017-04-17T10:10:23.843

0

Php date() doesn't recognise dd/mm/yy format

Do like below:-

<?php 
if(!empty($pet['pet_lost_date'])){
  echo date("d-m-Y",strtotime(str_replace('/','-',$pet['pet_lost_date'])));
}
?>

Output:- https://eval.in/777593

Using date_diff():-

https://eval.in/777610

edited Apr 17 '17 at 10:10

answered Apr 17 '17 at 09:37

Alive to die - Anant

70,531
10
51
98

1

but empty `$pet['pet_lost_date']` will return `01-01-1970` I believe, so he has to make sure his var isn't empty, wich can actually be, don't you think ? – OldPadawan Apr 17 '17 at 09:46
1

@OldPadawan you are right, a empty check also needed – lazyCoder Apr 17 '17 at 09:48
I'm pretty sure is var is empty though :) – OldPadawan Apr 17 '17 at 09:49
How to count days different? – Shah Rushabh Apr 17 '17 at 09:50
@Shah Rushabh get it here http://stackoverflow.com/questions/676824/how-to-calculate-the-difference-between-two-dates-using-php – lazyCoder Apr 17 '17 at 09:51

PHP printing wrong date

2 Answers2