CODEWITHSUNDEEP
phptime
Daman Mokha
Daman Mokha

Reputation: 372

calculate sundays between two dates

I want to calculate all Sunday's between given two dates. I tried following code. It works fine if days are less but if i enter more days. It keeps processing and Maximum execution time exceeds i changed the time but it even keeps processing even execution time is 200sec.

code is

<?php
$one="2013-01-01";
$two="2013-02-30";

$no=0;
for($i=$one;$i<=$two;$i++)
{

    $day=date("N",strtotime($i));
    if($day==7)
    {
    $no++;
    }
}
echo $no;

?>

please help.

Upvotes: 8

Views: 20005

Answers (3)

John Conde
John Conde

Reputation: 219804

<?php
$no = 0;
$start = new DateTime('2013-01-01');
$end   = new DateTime('2013-04-30');
$interval = DateInterval::createFromDateString('1 day');
$period = new DatePeriod($start, $interval, $end);
foreach ($period as $dt)
{
    if ($dt->format('N') == 7)
    {
        $no++;
    }
}
echo $no;

See it in action

Upvotes: 10

Lingo
Lingo

Reputation: 600

Here is a solution if you want the Sundays in a specific date range.

function dateRange($begin, $end, $interval = null)
{
  $begin = new DateTime($begin);
  $end = new DateTime($end);

  $end = $end->modify('+1 day');
  $interval = new DateInterval($interval ? $interval : 'P1D');

  return iterator_to_array(new DatePeriod($begin, $interval, $end));
}

/* define date range */
$dates = dateRange('2018-03-01', '2018-03-31');

/* define weekdays */
$weekends = array_filter($dates, function ($date) {
  $day = $date->format("N");
  return $day === '6' || $day === '7';
});

/* weekdays output */
foreach ($weekends as $date) {
  echo $date->format("D Y-m-d") . "</br>";
}

/* define sundays */
$sundays = array_filter($dates, function ($date) {
  return $date->format("N") === '7';
});

/* sundays output */
foreach ($sundays as $date) {
echo $date->format("D Y-m-d") . "</br>";
}

/* define mondays */
$mondays = array_filter($dates, function ($date) {
 return $date->format("N") === '1';
});

/* mondays output */
foreach ($mondays as $date) {
echo $date->format("D Y-m-d") . "</br>";
}

Just change the number for any days you want in your output:

Monday = 1
Tuesday = 2
Wednesday = 3
Thursday = 4
Friday = 5
Saturday = 6
Sunday = 7

Upvotes: 3

Quinn Strahl
Quinn Strahl

Reputation: 1818

John Conde's answer is correct, but here is a more efficient and mathy solution:

$start = new DateTime('2013-01-06');
$end = new DateTime('2013-01-20');
$days = $start->diff($end, true)->days;

$sundays = intval($days / 7) + ($start->format('N') + $days % 7 >= 7);

echo $sundays;

Let me break it down for you.

$start = new DateTime('2013-01-06');
$end = new DateTime('2013-01-20');

First, create some DateTime objects, which are powerful built-in PHP objects meant for exactly this kind of problem.

$days = $start->diff($end, true)->days;

Next, use DateTime::diff to find the difference from $start to $end (passing true here as the second parameter ensures that this value is always positive), and get the number of days between them. 

$sundays = intval($days / 7) + ($start->format('N') + $days % 7 >= 7);

Here comes the big one - but it's not so complicated, really. First, we know there is one Sunday for every week, so we have at least $days / 7 Sundays to begin with, rounded down to the nearest int with intval.

On top of that, there could be a Sunday in a span of time less than a week; for example, Friday to Monday of the next week contains 4 days; one of them is a Sunday. So, depending on when we start and end, there could be another. This is easy to account for:

  • $start->format('N') (see DateTime::format) gives us the ISO-8601 day of the week for the start date, which is a number from 1 to 7 (1 is Monday, 7 is Sunday).
  • $days % 7 gives us the number of leftover days that don't divide evenly into weeks.

If our starting day and the number of leftover days add up to 7 or more, then we reached a Sunday. Knowing that, we just have to add that expression, which will give us 1 if it's true or 0 if it's false, since we're adding it to an int value.

And there you have it! The advantage of this method is that it doesn't require iterating over every day between the given times and checking to see if it's a Sunday, which will save you a lot computation, and also it will make you look really clever. Hope that helps!

Upvotes: 35

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