javascript - PHP build_calendar not displaying table
one text
I am trying to use the build_calendar() function on PHP to display a simple calendar but when I run the following code I don??�t get any error messages but I just get a blank page.
I??�m not sure where I went wrong. Do I need to add a MySQL database first? I am running this using XAMPP on a Mac on Google Chrome
<?php
function build_calendar($month, $year){
$daysOfWeek = array('Sunday', 'Monday','Tuesday','Wednesday','Thursday','Friday','Saturday');
question?
$firstDayOfMonth = mktime(0,0,0,$month,1,$year);
$numberDays = date('t',$firstDayOfMonth);
$dateComponents = getdate($firstDayOfMonth);
$monthName = $dateComponents['month'];
$dayOfWeek = $dateComponents['wday'];
$datetoday = date('Y-m-d');
$calendar = "<table class='table table-bordered'>";
$calendar .= "<center><h2>$monthName $year</h2>";
$calendar .= "<tr>";
foreach($daysOfWeek as $day) {
$calendar .= "<th class='header'>$day</th>";
}
$currentDay = 1;
$calendar .= "</tr><tr>";
if($dayOfWeek > 0) {
for($k=0;$k<$dayOfWeek;$k++){
$calendar .= "<td class='empty'></td>";
}
}
$month = str_pad($month, 2, "0", STR_PAD_LEFT);
while ($currentDay <= $numberDays) {
if ($dayOfWeek == 7) {
$dayOfWeek = 0;
$calendar .= "</tr><tr>";
}
$currentDayRel = str_pad($currentDay, 2, "0", STR_PAD_LEFT);
$date = "$year-$month-$currentDayRel";
$dayname = strtolower(date('l', strtotime($date)));
$eventNum = 0;
$today = $date==date('Y-m-d')? "today" : "";
$calendar.="<td><h4>$currentDay</h4>";
$calendar .="</td>";
$currentDay++;
$dayOfWeek++;
}
if ($dayOfWeek != 7) {
$remainingDays = 7 - $dayOfWeek;
for($l=0;$l<$remainingDays;$l++){
$calendar .= "<td class='empty'></td>";
}
}
$calendar .= "</tr>";
$calendar .= "</table>";
}
?>
<!DOCTYPE html>
<html>
<body>
<div class="container">
<div class="row">
<div class="col-md-12">
<div id="calendar">
<?php
$dateComponents = getdate();
$month = $dateComponents['mon'];
$year = $dateComponents['year'];
echo build_calendar($month,$year);
?>
</div>
</div>
</div>
</div>
</body>
</html>
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