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.
To do this you have to add the Days Between to the Day of the Week, and then FLOOR that value to get the highest integer result. You then divide that number by 7 to get the amount of WEEKS and then multiply that by 2 to calculate the amount of WeekEnd days. Name this calculation "Weekend days".
=NETWORKDAYS(A2,B2) Then type Enter key, and you will count the number of workdays excluding Sundays and Saturdays between the two dates. See screenshot: Note: In the above formula, A2 indicates the start date, and B2 indicates the end date.
Count days excluding Sundays with formula Here I can introduce a formula for counting days excluding Sundays during a date range. Select a blank cell, here is C2, and type this formula =B2-A2-INT((B2-A2-WEEKDAY(B2)+1)/7) into it, and then press Enter key, a date displayed.
To calculate weekend dates in Excel, you will need to use a combination of the WEEKDAY and DATE functions. The WEEKDAY function will return a number between 1 and 7, representing the day of the week. The DATE function will return a date value based on a given year, month, and day.
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!
<?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
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