Is it possible to convert string 14 into 2014 in PHP

Question

I have an array formed from a string (2014/15):

$arr = explode("/", $Years, 2);

..and then two strings:

$frstHalfYear = $arr[0];
$secHalfYear = $arr[1];

..the output is following:

for $frstHalfYear: 2014
for $secHalfYear: 15

My question is : is it possible to convert 15 into a full year like 2014. This:

date("Y", strtotime($secHalfYear))

obviously is not working. The simple way would be to add 20at the start of the string, but I was wondering if there is some more generic way..

Thank you

Does it get any more generic than what you already proposed? — John Conde, Jul 22 '14 at 14:52
Just prepend `20` and you should be fine for the next 85 years. — user703016, Jul 22 '14 at 14:53
Yeah.. it will sound ridiculous, but what if it would be say.. 2099 and 2100... — Dims, Jul 22 '14 at 14:54
@ParkYoung-Bae And people thought it was safe to assume years started with 19__... — Niet the Dark Absol, Jul 22 '14 at 14:54
*"Yeah.. it will sound ridiculous, but what if it would be say.. 2099 and 2100"* - If that's based on someone's life, or a warantee, then that is soon to be outlived ;-) I won't live that long to reap the benefits. — Funk Forty Niner, Jul 22 '14 at 14:55
`15 + 2000`. don't do `'20' . 15`, because at some point your code will be spitting out `20100` — Marc B, Jul 22 '14 at 14:56
@MarcB I'd argue that the chances that PHP is still used in 2100 are slim. But I'd consider creating something that you know will break bad practice as well. — Jonathan, Jul 22 '14 at 15:12
@Jonathan: never say never. we're already getting bit by the y2038 bug in various "future dealings" stuff. — Marc B, Jul 22 '14 at 15:37
@Jonathan To further Marc's comment about the y2038 bug, have a read here http://stackoverflow.com/q/2012589/ on the subject. — Funk Forty Niner, Jul 22 '14 at 15:39

score 5 · Accepted Answer · answered Jul 22 '14 at 14:55

5

Here's the most generic solution I can think of:

$secHalfYear = substr($frstHalfYear,0,-strlen($secHalfYear)).$secHalfYear;

Basically, this will take the 20 from 2014 and stick it before the 15 to make 2015.

It even supports 2099/100 format! But more importantly, you can write 2014/2015 and it will work just fine too.

answered Jul 22 '14 at 14:55

Niet the Dark Absol

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Works great!! Thanks!! Will check back in 2100 to see how it'll manage.. :-) – Dims Jul 22 '14 at 15:00
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@Dims You've evidently found the fountain of youth ;-) – Funk Forty Niner Jul 22 '14 at 15:01
@Dims Your secret is safe with me ;-) – Funk Forty Niner Jul 22 '14 at 15:58

PeteAUK · Answer 2 · 2014-07-22T16:10:56.030

If you want something completely over the top: http://ideone.com/LzcgA3

var_dump(getDates('2014/15'));
var_dump(getDates('1999/12'));
var_dump(getDates('14/15'));


function getDates($dateRange) {
    $exp = explode('/', $dateRange);

    $startYear = $exp[0];
    // Cater for 14/15
    if ($startYear < 100) $startYear += date('Y') - date('y');

    $endYear = $exp[1];

    // If passed the full year, just use that
    if ($endYear < 100) {
        $shortStartYear = substr($startYear, 2, 2);
        if ($shortStartYear > $endYear) {
            // Jumping millenium
            $endYear += $startYear - $shortStartYear + 100;
        } else {
            // Same millenium
            $endYear += $startYear - $shortStartYear;
        }
    }

    return array('from'=>$startYear, 'to'=>$endYear);
}

This will actually cater for most dates. But appreciate if it's a little excessive ;)

Woah... That's a massive one... But it should cover all cases... Thanks a lot Pete! — Dims, Jul 22 '14 at 15:47
`if ($startYear < 100) $startYear += 2000;` is basically a "Y2.1K" bug waiting to happen. — Niet the Dark Absol, Jul 22 '14 at 15:56

score -1 · Answer 3 · answered Jul 22 '14 at 15:07

-1

$secHalfYear = $frstHalfYear + 1;

You have to make certain assumptions to expand a date appropriately, but I don't believe I've ever seen this sort of year display when the second year didn't directly follow the first.

answered Jul 22 '14 at 15:07

Jason

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Is it possible to convert string 14 into 2014 in PHP

3 Answers3