CODEWITHSUNDEEP
php
user1444027
user1444027

Reputation: 5251

Show list of days of week except Sunday

I'm attempting to show a list of the days of week excluding Sunday. The first day on the list should be whatever-day-it-is-today with a CSS class of 'active'. If today = Sunday, the first list item should be Monday.

Below is the code I'm using. The problem is, this outputs the first 6 days from today (including Sunday) and then misses out the-7th-day-from-today.

Any idea what I need to change to get it to skip Sunday instead?

function daysofweek() {

$datetime = new \DateTime();

$listItem = array('<li class="active"><a href="#', '" data-toggle="tab">', '</a></li>');

$i = 0;

echo '<ul class="nav nav-tabs">';

    while (true) {
        if ($i === 6) break;

        if ($datetime->format('N') === '7' && $i === 0) {
            $datetime->add(new \DateInterval('P1D'));
            continue;        
        }

        $the_day = $datetime->format('D');

        echo $listItem[0] . $the_day . $listItem[1] . $the_day . $listItem[2];
        $listItem = array('<li><a href="#', '" data-toggle="tab">', '</a></li>');    

        $datetime->add(new \DateInterval('P1D'));
        $i++;
    }

echo '</ul>';

}

Upvotes: 0

Views: 310

Answers (3)

Choerun Asnawi
Choerun Asnawi

Reputation: 181

I prefer to create a function that returns an array, instead of writing the output inside it. Here my explanation first:

  • To get the current date, use the time function.
  • To get the numeric representaion of current day of the week, use date function with value 'w' as first argument. It will return 0 for Sunday through 6 for Saturday.
  • To get the next date, add the current date with (24 * 60 * 60). That would be 86,400 seconds or 1 days from current time. The value 24 is for the hour, the first 60 is the minute, and the last 60 is the second.
  • To get the name of current day of the week, use date function with value 'l' (lowercase of letter L) as first argument.

Nuff said, here's the function:

function weekDaysFromToday() {
  $days = array();
  $date = time();
  while (count($days) < 7) {
    $date += (24 * 60 * 60);
    if (!date('w', $date)) continue;
    $days[] = date('l', $date);
  }
  return $days;
}

If you want that the current day is not repeated at the end of the list, you should change the while (count($days) < 7) into while (count($days) < 6).

And here's the code to render the array using your code above:

$days = weekDaysFromToday();
$active = ' class="active"';
echo "<ul class=\"nav nav-tabs\">\n";
foreach ($days as $day) {
  echo "<li$active><a href=\"#$day\" data-toggle=\"tab\">$day</a></li>\n";
  $active = '';
}
echo '</ul>';

There you are, hope it helps.

Upvotes: 1

cronoklee
cronoklee

Reputation: 6722

This is the same thing in much more concise formatting:

$today = date('w');
$first = 1;
for($i=$today; $i<$today+7; $i++){
    if($i%7) echo "<li ".($first ? "class='active'" : '').">".date('D', strtotime("Sunday +".($i%7)." days"))."<li>";
    $first=0;
}

It works by using the modulo %7 to cycle through the week, starting from today.

Upvotes: 0

Machavity
Machavity

Reputation: 31644

This is a bad way to code

while(true)

This ALWAYS evaluates to true. Even with your break this is poor syntax. Your while should contain the logic. So here's your code, restructured. I removed the duplicate call to add() and made it run for every iteration. The loop naturally terminates itself as well.

$i = 0;
while($i < 7) {
    $the_day = $datetime->format('N');
    if($the_day != 7) {
         echo $listItem[0] . $the_day . $listItem[1] . $the_day . $listItem[2];
         $listItem = array('<li><a href="#', '" data-toggle="tab">', '</a></li>');      
    }
    $datetime->add(new \DateInterval('P1D'));
    $i++;
}

Upvotes: 2

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