How to convert year format YY into YYYY in PHP

Question

I need to convert year(YY) to YYYY. Here i am try to use date function in PHP but not getting the expected result.

$expiry_year = "30"; //year=2030
echo date('Y',strtotime($expiry_year)); //Result: 1970
echo date('Y',$expiry_year); //Result: 1970

Expected Result: 2030

Thanks To All!

Coding languages are very bad at _Making guesses_ If you know it should be 2030 concatenate a 20 on the front — RiggsFolly, Dec 20 '16 at 09:36
Your year "30" will always return 1970. This is due to the Unixtime only starting on the 1st January 1970 — scottevans93, Dec 20 '16 at 09:36
How about `30+2000` since you don't provide any other requirements? If it needs more complicated handling, add some more logic. — Sami Kuhmonen, Dec 20 '16 at 09:37
@Sunil Pachlangia Thanks! your logic is correct and I am getting the result. — Ramalingam Perumal, Dec 20 '16 at 09:40
@RamaLingam Hope my Employer know this as well (Just joking). — Sunil Pachlangia, Dec 20 '16 at 09:41

score 5 · Accepted Answer · answered Dec 20 '16 at 09:41

5

Try this, use createFromFormat

$date = "30";
$dates = DateTime::createFromFormat('y', $date);
//now to get the outpu:
$arr = $dates->format('Y'); // output : 2030

DEMO

answered Dec 20 '16 at 09:41

Dave

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Thanks! It's working fine. – Ramalingam Perumal Dec 20 '16 at 09:52

Narendrasingh Sisodia · Answer 2 · 2016-12-20T09:47:42.540

1

Note: If the number of the year is specified in a two digit format, the values between 00-69 are mapped to 2000-2069 and 70-99 to 1970-1999. See the notes below for possible differences on 32bit systems (possible dates might end on 2038-01-19 03:14:07).

For this you can use date_create_from_format Alias of the DateTime class createFromFormat method like as

$expiry_year = "30";
$date = date_create_from_format('y',$expiry_year);
echo $date->format('Y'); //2030

Docs

edited Dec 20 '16 at 09:47

answered Dec 20 '16 at 09:41

Narendrasingh Sisodia

21,247
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1

Thanks! It's working fine. – Ramalingam Perumal Dec 20 '16 at 09:52

score 1 · Answer 3 · answered Dec 20 '16 at 09:45

1

Try this :

$date = DateTime::createFromFormat('y', '30');
echo $date->format('Y');

answered Dec 20 '16 at 09:45

Steven

313
3
10

Thanks! It's working fine. – Ramalingam Perumal Dec 20 '16 at 09:53

Jicao · Answer 4 · 2016-12-20T10:53:25.410

-1

$expiry_year = "30";
$ce = substr(date('Y'),0,2); // Check the century
$ny = substr(date('Y'),2,2); // Check current year (2 digits)
if($expiry_year < $ny) {
    $next_ce = $ce+1;
    $output = $next_ce.$expiry_year;
} else {
    $output = $ce.$expiry_year;
}

Something like that ? :D

edited Dec 20 '16 at 10:53

answered Dec 20 '16 at 09:46

Jicao

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This is not my expect solution. Thnaks – Ramalingam Perumal Dec 20 '16 at 09:54
This should help poeple with php lower than 5.2 :) – Jicao Dec 20 '16 at 10:00
1

Your code return Result: 2130 – Ramalingam Perumal Dec 20 '16 at 10:42
Sorry I edited my answer with "<" instead of ">"... My bad. – Jicao Dec 20 '16 at 10:54
Assume `$expiry_year = "08";` then return `2108` Totally wrong – Ramalingam Perumal Dec 20 '16 at 10:57
Yeah right... The answer is not correct in your mind knowing what you wanted... But the logic was there and you could just modify a bit the code and it's working... Like I said, It'll help someone with PHP lower than 5.2 ! – Jicao Dec 20 '16 at 14:21
1

Ok. Thanks for your answer. – Ramalingam Perumal Dec 20 '16 at 14:25

How to convert year format YY into YYYY in PHP

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