如何定义日是星期日和正常日。 按日期使用开关盒？ [关闭]

Question

I have an error in this code. Please help me solve it.

function holiday($today) {    

    $year = substr($today, 0, 4);     

    switch($today) {

      case $year.'-01-01':
          $holiday = 'New Year';
          break;

      case $today:
          $today11 = new DateTime($today);
          $R= $today11->format('l') . PHP_EOL;
          $Sunday='0';

          if($R == 0) {
              $holiday = 'Sunday';
          } else {
              $holiday = 'Normal Day';  
          }
      }

      return $holiday;
}

echo $tday= holiday($today); 

Answer 1

Try it:-

function holyday($today)
{
    $start_date = strtotime($today);


    if(date('z',$start_date) ==0)
    {
         return 'New Year';

    }else{

        if(date('l',$start_date) =='Sunday')
        {
             return 'Sunday';

        }else{

            return "Noraml Day";
        }
    }
}

echo holyday('2012-09-06');

Output = Noraml Day

echo holyday('2013-01-01');

Output = New year