How to generate an array of weekdays in Y-m-d format?

Question

The array $weekDays contains 7 days that starts today.

2017-06-01/ 2017-06-02 / 2017-06-03 / 2017-06-04 / 2017-06-05 / 2017-06-06 / 2017-06-07

$weekDays[0] = date('Y-m-d');
for ($i=1; $i<7; $i++){
    $weekDays[$i] = date('Y-m-d', strtotime('+' . $i . ' days'));
}

I only need the weekdays so I tried this:

for ($j=0; $j < 7; $j++) {
        $weekDays[$j] = strtotime($weekDays[$j]);
        $weekday = date('w', $weekDays[$j]);

        if ($weekday == 0 || $weekday == 6) {
            echo "Saturday or Sunday<br>";
        } else {
                echo $weekday . "<br>";
        }
    }

But I want to return the array with only the weekdays and in Y-m-d format.

Thanks in advance!

score 3 · Accepted Answer · answered Jun 01 '17 at 13:35

strtotime() is pretty smart, you can ask it to only return weekdays.

for($d=0; $d<5; ++$d){
    $result[]=date('Y-m-d',strtotime("+$d weekdays"));
}
var_export($result);

No post loop filtering required.

Output:

array (
  0 => '2017-06-01',
  1 => '2017-06-02',
  2 => '2017-06-05',
  3 => '2017-06-06',
  4 => '2017-06-07',
)

score 0 · Answer 2 · answered Jun 01 '17 at 13:24

You're one step closer to get your solution, you can use unset to remove non-weekdays from your array:

for ($j=0; $j < 7; $j++)
    {
    $weekDays[$j] = strtotime($weekDays[$j]);
    $weekday = date('w', $weekDays[$j]);

    if ($weekday == 0 || $weekday == 6)
    {
            unset($weekDays[$j]);
    }
}
$weekDays = array_values($weekDays); //reorder after all unsets

How to generate an array of weekdays in Y-m-d format?

2 Answers2