32 hours ago excluding weekends with php - php

Question

So I have a script that does multiple checks for 32, 48 and 72 hours ago.
Basically I check my database for entries that are at least x hours old.
Now this works fine like this:
$date = date('Y-m-d H:i:s',strtotime('-32 hours'));
$q = "SELECT * FROM `table` WHERE `date` <= '".$date."'";
Now I want this to exclude weekends. I know you can use weekdays within strtotime to get this effect however this doesn't work for hours.
For 48 hours it's easy because I can simply do the following:
echo date('Y-m-d H:i:s',
strtotime(date("Y-m-d H:i:s").
" -2 weekdays ".
date('H:i:s')));
For 72 hours it's also easy because it's 3 days. However 32 hours poses a problem because it's ±1.3 days.
In conclusion, how do I get the datetime of 32 hours ago excluding weekends.

Answer 1

Use strtotime as you had initially:
$time = strtotime('-32 hours');
Then do the weekend/weekday calculation manually.
// If the day is Sunday or Saturday subtract a full day.
while (date('w', $time) % 6 == 0) {
$time = strtotime('-1 day', $time);
}
$date = date('Y-m-d H:i:s', $time);

Answer 2

I am not sure if this is correct or the best way to do it but something like:
function getDateBackExcludingWeekend( $hours ) {
$now = time();
$secondsBack = $hours * 3600;
$actual = $now - $secondsBack;
$monday = strtotime("last monday");
if( $actual < $monday ) {
$diff = ($secondsBack - ($now - $monday));
$backthen = ($monday - 172800 /* two days */) - $diff;
return date("Y-m-d H:i:s", $backthen);
}
return date("Y-m-d H:i:s", $actual);
}

Answer 3

Why not just remove two days and add 16 hours semi-manually to make up for it?
$DateTMP = date('Y-m-d h:i:s',(strtotime(date(Y-m-d)." -2 weekdays") + (60 * 60 * 16)));