Question

Starting with a date in this format: 2011-05-01 09:00:00, how can I create an array that contains all office hours (09:00 to 17:00) for all working days of the year (so excluding all Saturday and Sundays). What I want to arrive to is something like this:

2011-05-01 09:00:00
2011-05-01 10:00:00
2011-05-01 11:00:00
2011-05-01 12:00:00
2011-05-01 13:00:00
2011-05-01 14:00:00
2011-05-01 15:00:00
2011-05-01 16:00:00
2011-05-01 17:00:00
//next day, starting at 09:00 and ending at 17:00
2011-05-02 09:00:00
...
2011-05-02 17:00:00
//until the last day of the year from 09:00 to 17:00
2011-12-31 09:00:00
...
2011-12-31 17:00:00

The start date will be the first of the current month at with 09:00 as time and the very last date (last element of the array) will always be 17:00 on the last day of the year.

Again, weekends should be excluded.

Pseudocode idea:
I thought of something like strtotime($start, "+1 one hour") with a check for "if smaller than 17:00" but it doesn’t seem to be that simple.

Answer 1

How about this:

$start = strtotime('2011-05-01');
$end = strtotime('2011-12-31');

$times = array();

for ($i = $start; $i <= $end; $i += 24 * 3600)
{
    if (date("D", $i) == "Sun" || date("D", $i) == "Sat")
    {
        continue;
    }

    for ($j = 9; $j <= 17; $j++)
    {
        $times []= date("Y-m-d $j:00:00", $i);
    }
}

The outer loop iterates through all the days in the given time period. In the outer loop, we check to see if the day is either Saturday or Sunday (a weekend), and if it is, we skip that day. If it’s not a weekend, we loop through all the valid hours, adding the full date and time to the array as we go.