A date x number of weekdays from today

[Anon]

I want to be able return a date that is x number of weekdays away from the current date, x being a number supplied by the user. Any tips?

Thanks --
James

[[deleted]]

make a timestamp (seconds since 1970) for 'now', add x2460*60 seconds, be happy

[Anon]

Hi!

In PHP, you should use Vincent's approach (get timestamp and add a number of seconds to it, then format with date() function).

You can also feed mktime() with 'invalid' date arguments (such as $day + number_of_days_to_add) and then use date() to format the timestamp.

$tomorrow = mktime (0,0,0,date("m") ,date("d")+1,date("Y"));
$lastmonth = mktime (0,0,0,date("m")-1,date("d"), date("Y"));
$nextyear = mktime (0,0,0,date("m"), date("d"), date("Y")+1);
// this is example from PHP docs

If you need to use date logic in SQL, then use this (MySQL):
SELECT * from YourTable WHERE DATE_ADD(YourDaateColumn, INTERVAL X DAY) >= CURDATE();

Best regards,
Stas

[Anon]

I want to add only weekdays, excluding Saturday and Sunday.

For example, if this was run on Jan. 1, 2002, a Tuesday, and I wanted to add 10 weekdays to the date, it should return Jan 15, 2002, which is 10 weekdays from Jan 1st, not Jan 11, which is 10 days from Jan 1st.

I hope that makes it more clear.

Thanks! -- 🙂
James

[lucianoes]

It makes sense to me, but isn't working for some reason. See if you can fix it and get it working.

<?php
$now = time();
$nowx = number_format($now);
$then = $now + $x*86400; // $x input by user
$thenx = number_format($then);

if ($nowx > $thenx) {echo "I said please select a FUTURE date!!!";}
else {
for ($nowx = $nowx; $nowx <= $thenx; $nowx = $nowx+86400)
{static $wd = 0; $weekday = date("l",$now);
if ($weekday == 'Saturday') {$do_nothing = 'do_nothing';}
elseif ($weekday == 'Sunday') {$do_nothing = 'do_nothing';}
else {$wd ++;}
}
}
echo "the number of working days until then is $wd";
?>

Good luck,
Luciano ES
Santos, SP - Brasil

[lucianoes]

Sorry, I just realized my code would not produce the result you want even if it worked. I'll sleep on it (literally) and let you know if I come up with the right solution. This is a very interesting challenge.

    Luciano ES
    Santos, SP - Brasil

[Anon]

Actually, Luciano, I wanted to return the date that was x number of weekdays from today, not find out how many weekdays were between two dates, but what you posted will certainly come in handy in the future.

Here's what I came up with... If anyone has any hints for optimizing this, I'd appreciate it. 🙂

<?php
// for testing purposes, we'll seed the # of days to 10
if (!$weekdays) {$weekdays=10;}

$realdays = 0;
while ($weekdays) {
$realdays = $realdays + 1;
$futuredate = mktime (0,0,0,date("m"),date("d")+$realdays,date("Y"));
$dayofweek = date("w", $futuredate);
if ($dayofweek == '0' || $dayofweek == '6') {
// do nothing since it's Saturday or Sunday
} else {
$weekdays = $weekdays - 1;
}
}
echo "The target date is " . date("l, F d, Y", $futuredate);
?>

[Anon]

SLM