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this code is use for date with day name

$month = 12;
$year = 2019;
$dayCount = date('t', strtotime('01-' . $month . '-' . $year));
$days = [];
for ($i = 1; $i <= $dayCount; $i++) {
    $tempDate = $year . '-' . $month . '-' . $i;
    $days[] = sprintf('%04d-%02d-%02s', $year, $month, date('D', strtotime($tempDate)));
}
echo json_encode($days, JSON_PRETTY_PRINT);
Fenz
  • 139
  • 11

1 Answers1

2

I think that is what you looking for:

<?php
$month = 12;
$year = 2019;

$dayCount = date('t', strtotime('01-'. $month. '-' . $year));

$days = [];

for($i = 1; $i <= $dayCount; $i++) {
    $tempDate = $year . '-' . $month . '-' . $i;
    $days[] = date('D', strtotime($tempDate));
}

echo json_encode($days);

Output:

["Sun","Mon","Tue","Wed","Thu","Fri","Sat","Sun","Mon","Tue","Wed","Thu","Fri","Sat","Sun","Mon","Tue","Wed","Thu","Fri","Sat","Sun","Mon","Tue","Wed","Thu","Fri","Sat","Sun","Mon","Tue"]
Serghei Leonenco
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