如何定义星期日和平日.按日期使用开关 - How to define day is sunday and Normal day. using switch case by date?

How to define day is sunday and Normal day. using switch case by date?

我在这段代码中有一个错误。请帮我解决。

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);

试试看：-

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');

输出=Noraml日

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

输出=新年

下面是holiday（）函数的一个工作实现：

function holiday($today) {    
$date = strtotime($today);     
//check if Sunday
if (date('l', $date) == 'Sunday') {
    return 'Sunday';
}
//check if New Year
if ((date('j', $date) == 1) && (date('n', $date) == 1)) {
    return 'New Year';
}
//else, just return Normal Day
return 'Normal Day';
}
//$today is in YYY/MM/DD format
echo $tday = holiday($today);

此外，PHP的date引用在这种情况下也很有用：http://php.net/manual/en/function.date.php

不确定为什么要使用switch，请阅读如何使用switch。以下功能将起作用：

function holiday($today) {
    // z returns the number of the day in the year, 0 being first of January
    if(date("z", $today) == 0) {
         return "New Year";
    }
    // w returns the number of the day in the week 0-6 where 0 is Sunday
    if(date("w", $today) == 0) {
         return "Sunday";
    }
    return "Normal Day";
}
$today = date();
echo holiday($today);