CODEWITHSUNDEEP
phpdatetime
Jan Alexa
Jan Alexa

Reputation: 5

how may days are between two dates in specific year

I'm solving following task>

I have two dates - $start and $end and target year as $year.

dates are php DateTime objects, year is string.

add:dates comes acutaly from MySql field from this format 2017-02-01 15:00:00 ... add2: if end date is null, I use todays date ...

I need to figure out how many days are between these two dates for specific year.

Also I need to round it for whole days, even if one minute in day should be counted as whole day ...

I can solve it by many many following ifs.

Expected results for values I used in example are 2016 is 0 days 2017 is 31 days 2018 is 32 days 2019 is 0 days

But are there any elegant php functions which can help me with this ? What I did it seems to be wrong way and giving bad results - seems it counts full days only ...

Please see my code here >

<?php
$diff = True;
$start = DateTime::createFromFormat('Y-m-d H:i:s','2017-12-01 23:05:00');
$end = DateTime::createFromFormat('Y-m-d H:i:s','2017-12-03 00:05:00');
$year = '2017';

// start date
if ($start->format('Y')<$year)
{
    $newstart = new DateTime('first day of January '. $year);
}

if ($start->format('Y')==$year)
{
    $newstart = $start;
}

if ($start->format('Y')>$year)
{
    $result = 0;
    $diff = False;
}

// end date
if ($end->format('Y')>$year)
{
    $newend = new DateTime('last day of December '. $year);
}

if ($end->format('Y')==$year)
{
    $newend = $end;
}

if ($end->format('Y')<$year)
{
    $result = 0;
    $diff = False;
}               

// count if diff is applicable
if ($diff)
{
    $result  = $newend->diff($newstart)->format("%a");
}

echo $result;
?>

Upvotes: 0

Views: 77

Answers (2)

axiac
axiac

Reputation: 72256

But are there any elegant php functions which can help me with this ?

Read about DateTime::diff(). It returns a DateInterval object that contains the number of days (in $days) and by inspecting the values of $h, $i and $s you can tell if you have to increment it to round the result. You can also use min() and max() to crop the time interval to the desired year.

function getDays(DateTimeInterface $start, DateTimeInterface $end, $year)
{
    // Extend the start date and end date to include the entire day
    $s = clone $start;            // Don't modify $start and $end, use duplicates
    $s->setTime(0, 0, 0);
    $e = clone $end;
    $e->setTime(0, 0, 0)->add(new DateInterval('P1D'));  // start of the next day

    // Crop the input interval to the desired year
    $s = min($s, new DateTime("$year-01-01 00:00:00"));
    $year ++;
    $e = max(new DateTime("$year-01-01 00:00:00"), $end);     // start of the next year

    if ($e <= $s) {
        // The input interval does not span across the desired year
        return 0;
    }

    // Compute the difference and return the number of days
    $diff = $e->diff($s);

    return $diff->days;
}

Upvotes: 1

Oliver Williams
Oliver Williams

Reputation: 6354

    $d1 = strtotime('2017-05-15');
    $d2 = strtotime('2017-05-31');
    $div = 24 * 3600;
    echo abs(($d2 - $d1) / $div); // 16 days

Just make sure and ONLY have the date part and you shouldn't have to deal with rounding.

Upvotes: 0

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