Date difference showing wrong values

Question

I have used a function to calculate date difference between 2 dates.

Here is my function

function date_difference ($date_1, $date_2) {   


    $val_1 = new DateTime($date_1);
    $val_2 = new DateTime($date_2);

    $interval = $val_1->diff($val_2);
    $year     = $interval->y;
    $month    = $interval->m;
    $day      = $interval->d;

    $output   = '';

    if($year > 0){
        if ($year > 1){
            $output .= $year." years ";     
        } else {
            $output .= $year." year ";
        }
    }

    if($month > 0){
        if ($month > 1){
            $output .= $month." months ";       
        } else {
            $output .= $month." month ";
        }
    }

    if($day > 0){
        if ($day > 1){
            $output .= $day." days ";       
        } else {
            $output .= $day." day ";
        }
    }
    if($day == 0)
        $output.=' Almost Over';
    if($day < 0)
        $output.= ' Expired';
    return $output;
}

I am using it like this

echo date_difference(date('m/d/Y'),'02/06/2013');

It shows the result as 25 days where as it should show expired. Can anyone point where i am doing wrong.

I find it [hard to believe that you wrote that function](http://stackoverflow.com/a/13786899/562459). — Mike Sherrill 'Cat Recall', Mar 03 '13 at 04:03
Yes i haven't written the function just modified it to work like i want it. — Nirmal Ram, Mar 03 '13 at 04:05
Just a personal little thing but I would use the ternary operator instead of the `$[timelength] > 1` if statements as it would be much cleaner. — James Coyle, Mar 03 '13 at 04:06
@jimjimmy1995 I'd actually use a quick little helper function: `$p = function($n) {return $n==1?'':'s';};` and then just do things like `$output .= $day." day".$p($day)." ";` — Niet the Dark Absol, Mar 03 '13 at 04:15

score 2 · Accepted Answer · edited Feb 08 '17 at 14:39

2

As soon as I saw this XKCD page I wanted an opportunity to post it, and here it is!

xkcd

When your code tries to parse 02/06/2013, how can it know whether you mean "February 2nd", or "June 6th"? You should ALWAYS use the YYYY-MM-DD format when giving a date to parse, or better yet hardcode the actual numeric timestamp (in this case 1360126800)

edited Feb 08 '17 at 14:39

Community

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1

answered Mar 03 '13 at 04:11

Niet the Dark Absol

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Tried like this but still shows 25 days. `echo time_difference(date('Y-m-d'),'2013-02-06');` – Nirmal Ram Mar 03 '13 at 04:18
You need to check `if( $interval['invert']) return 'Expired';` – Niet the Dark Absol Mar 03 '13 at 04:21
Fatal error: Cannot use object of type DateInterval as array in /home/crowdfun/public_html/wp-content/themes/vantage-child/functions.php on line 87 – Nirmal Ram Mar 03 '13 at 04:32
1

My mistake, it should be `$interval->invert`. – Niet the Dark Absol Mar 03 '13 at 04:34
Yes thanks that worked but why i need to `return 'Expired'` can't i append it to `$output` ? – Nirmal Ram Mar 03 '13 at 04:38
Well you could, but then you'd get strings like "1 year 2 months Almost Over Expired" – Niet the Dark Absol Mar 03 '13 at 04:39

score 2 · Answer 2 · answered Mar 03 '13 at 04:17

2

DateInterval won't have nagative values, you need to compare the two DateTime object.

Change to

if($val_1 < $val_2 && $day == 0)
    $output.=' Almost Over';
if($val_1 > $val_2)
    $output.= ' Expired';
return $output;

answered Mar 03 '13 at 04:17

xdazz

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That also didn't work. I got 25 days expired. and for almost over i got blank. – Nirmal Ram Mar 03 '13 at 04:28
I think i can modify your code to make it work. Almost over shows correct result for me. So only expired needs to be modified. I did like this. `if($val_1 > $val_2) return ' Expired';` – Nirmal Ram Mar 03 '13 at 04:41

score 1 · Answer 3 · answered Mar 03 '13 at 04:09

1

just use the UNIX time stamp, that way it should be a very easy calculation.

it can be shown in Y-D-M and you can even make a count down clock if you feel a bit fancy.

most MMO's and management systems use it to register the date & time of registration and to show how long the member has been on the community.

hope it helped!.

answered Mar 03 '13 at 04:09

LUX

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Already tried that but it gives 2 months 30 days `echo time_difference(date('Y-d-m'),'2013-06-02');;` – Nirmal Ram Mar 03 '13 at 04:20
so simply say'd you want the code to subtract 1 until the value is 25 and then return "expired", am i understanding it correct? – LUX Mar 03 '13 at 04:26
Yes. if the difference between the dates is less than 0 then it should return expired – Nirmal Ram Mar 03 '13 at 04:35

Date difference showing wrong values

3 Answers3