PHP date conversion mm/dd/yy to yyyy/mm/dd -error

Question

My input date format is dd/mm/yy. I am using the following code to convert it into yyyy-mm-dd.

    $cal_date= $fileop[1];
    $date = str_replace('/', '-', $cal_date);
    $invdate=date('Y-m-d',strtotime($date));

Result i got is 24/10/13-->2024-10-13. Can anyone correct me.

Here two conversions 2digit year to 4 digit year and /replaced by -dash

`strtotime()` has a [strict list of formats](http://www.php.net/manual/en/datetime.formats.date.php) that it will understand, and `dd-mm-y` or `dd/mm/y` are not part of that list. — Sammitch, Oct 25 '13 at 16:55

aynber · Answer 1 · 2013-10-25T17:26:59.863

3

Using the DateTime object might be better

try{
   $date = DateTime::createFromFormat('d/m/y',$cal_date);
   $invdate = $date->format('Y-m-d');
} catch (Exception $e) {
   echo $e->getMessage(); // or
   print_r(DateTime::getLastErrors());
}

edited Oct 25 '13 at 17:26

answered Oct 25 '13 at 16:47

aynber

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@Anthony either function will work. http://php.net/manual/en/function.date-create-from-format.php – aynber Oct 25 '13 at 16:58
One is a function, the other is a method. They both point to the same thing, but it's more clear that it's using the class when you use the method. And you had an error in your code I fixed. – Anthony Oct 25 '13 at 17:10
This code returns the fatal error Call to a member function format() on a non-object in – Indra Oct 25 '13 at 17:11
Except that, as the title states, the current format is m/d/Y. – aynber Oct 25 '13 at 17:12
And the text of the post says d/m/Y. Talk about confusing. :/ – aynber Oct 25 '13 at 17:18
@aynber I am getting an error - Call to a member function format() on a non-object in – Indra Oct 25 '13 at 17:22
@Indra Something failed somewhere, then. I updated the code with how to get the error message. – aynber Oct 25 '13 at 17:27
@aynber it doesnt work – Indra Oct 25 '13 at 17:31
What happened when you tried? – aynber Oct 25 '13 at 17:33
this is the error i got.....Call to a member function format() on a non-object in – Indra Oct 25 '13 at 17:34
Even with the try block? – aynber Oct 25 '13 at 17:35
Yes aynber. i am working on it for more than 3 hrs – Indra Oct 25 '13 at 17:36
Actually this code---$invdate = date_format(date_create_from_format('d/m/y', $fileop[1]), 'Y-m-d');--- returns required results as well as an error...."date_format() expects parameter 1 to be DateTimeInterface, boolean given in" – Indra Oct 25 '13 at 17:38
Add `print_r(DateTime::getLastErrors());` right after the createFromFormat line, and see if it prints anything out. – aynber Oct 25 '13 at 17:45

Yanki Twizzy · Answer 2 · 2013-10-25T18:03:16.757

0

$date = DateTime::createFromFormat('d/m/y',$cal_date);
$break = explode("/", $date);
$yearComp = $break[2];
if($yearComp < 70)
    $invdate = $date->format('Y-m-d'); 
else {
    // code that would add 100 years to the date generated }

But just know that 69 would print out 2069 but 70 would be 1970. If you won't get to 70, then this should be fine

edited Oct 25 '13 at 18:03

answered Oct 25 '13 at 17:03

Yanki Twizzy

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whats the difference between your code and my code – Indra Oct 25 '13 at 17:14
If it is not working it means $fileop[1] is not returning the correct thing. – Yanki Twizzy Oct 25 '13 at 17:27
$fileop[1] is 24/10/13. Here 13 is a two digit yr. – Indra Oct 25 '13 at 17:30

score -1 · Accepted Answer · answered Oct 25 '13 at 16:49

-1

Another way - See The Demo Here

<?php

$date= "02/05/2013"; //dd/mm/yy
$dd=substr($date, 0, 2);
$mm=substr($date, 3, 2);
$yy=substr($date, 6, 4);

echo $yy."/".$mm."/".$dd; // yyyy-mm-dd.

?>

answered Oct 25 '13 at 16:49

underscore

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it works. but How can i convert two digit yr to 4 digit yr – Indra Oct 25 '13 at 17:40
make it as the answer – underscore Oct 26 '13 at 04:20

evalarezo · Answer 4 · 2013-12-01T11:57:01.170

-1

In one shoot:

$invdate = date_format(date_create_from_format('m/d/y', $fileop[1]), 'Y-m-d');

If you are not confident with the imput date format:

if ($invdate = date_create_from_format('m/d/y', $fileop[1])) {
    $invdate = date_format($invdate, 'Y-m-d'); //date format yyyy-mm-dd
} else {
    echo "ERROR: '{$fileop[1]}' has not the date format mm/dd/yy";
}

edited Dec 01 '13 at 11:57

answered Oct 25 '13 at 17:17

evalarezo

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Excellent. It works but returns an error "date_format() expects parameter 1 to be DateTimeInterface, boolean given in" – Indra Oct 25 '13 at 17:26
That is because you are using a different date format than the expected. Consider fix your question example, so it must use the same date format specified at the question title. – evalarezo Dec 01 '13 at 12:14

PHP date conversion mm/dd/yy to yyyy/mm/dd -error

4 Answers4