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Trying to determine how many 'selling' days left in a month

Posted on 2013-10-26
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Last Modified: 2013-10-26
Hi,
Our business isn't open on Sundays. In my app I need to show how many selling days are left in the month (days we are open). Currently you can show how many days left in a month using something  like this:
<?php echo date('t') - date('j');?>

But how could I take into account Sundays?
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Question by:tjyoung
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Chris Stanyon earned 500 total points
ID: 39602528
Here you go:

<?php
$start = new DateTime(); //get today
$end = new DateTime('last day of this month'); //get the end of the month
$days = $start->diff($end, true)->days; //number of days between now and the end of the month
$sundays = intval($days / 7) + ($start->format('N') + $days % 7 >= 7); //number of sundays

$sellingDays = $days - $sundays; //number of days minus the Sundays

printf("The are %d selling days left this month", $sellingDays);
?>

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by:tjyoung
ID: 39602623
Sorry, I don't suppose you know how to implement the above in php5.2?
'last day of this month' for example doesn't work unless its 5.3 and above and I can't upgrade the machine unfortunately....
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by:Chris Stanyon
ID: 39602648
Not really. Don't have access to 5.2 any more!

The diff() method is also >=5.3.

You really ought to upgrade your system though - for security reasons if nothing else. If this is a hosted platform, then chase them up (or change providers)
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LVL 43

Expert Comment

by:Chris Stanyon
ID: 39602652
Hmmm. Thought a little more and maybe try this. We don't actually need the date of the end of the month, just the day number, which you can get with format('t'):

$start = new DateTime();
$days = $start->format('t') - $start->format('d');
$sundays = intval($days / 7) + ($start->format('N') + $days % 7 >= 7);

$sellingDays = $days - $sundays;
printf("The are %d selling days left this month", $sellingDays);

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by:tjyoung
ID: 39602654
Great trying it now...
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by:tjyoung
ID: 39602662
Awesome, that did the trick! I appreciate you taking another look Chris.
Thanks again,
Todd
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Expert Comment

by:Chris Stanyon
ID: 39602670
No worries ;)
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