比较今天和开店的日子，看看是开了还是关了？

Question

嗨，我试图在php中创建一个函数，检查当前的一天，如果商店打开或关闭，返回。这是我到目前为止所拥有的，它工作，但它是很长的，所以如果有人知道最短的路径，那么请帮助！

$today = date('w'); // today 

$starting = "THUR"; //opening day
$closing = "MON"; //closing day


$openday = date("w", strtotime($starting));  //coverting to number 
$closeday = date("w", strtotime($closing));  // coverting to number 


//checking if closing day is greater than opening day 

if($openday < $closeday ){ 

    //checking if today lies between starting and closing day

      if($today > $openday && $today <= $closeday ){ 
        
        $stat = 'Opened';  //if today lies between starting day and ending day then Opened
      }
      else{  
        $stat = 'Closed'; //else Closed
      }
}

// if closing day is less than opening day 
else{  

  //checking if today lies between opening day and friday
  if($today > $openday && $today <= 6){ 

    $stat = 'Opened';   

  }
  //checking if today lies between sunday and closeday
  else if( $today > 0 &&  $today <= $closeday ){   

    $stat = 'Opened';    

  }
  else{
    $stat = 'Closed';  // else Closed
  }
}

如果周四至周一开盘休市，这一天就会回来。如果你知道怎么做，那么请帮帮我。

Answer 1

如果你知道你关门的日子(这是你必须做的)，为什么不这样做呢？

    $closed=array(
        date('w',strtotime('Tuesday')),
        date('w',strtotime('Wednesday'))
    );
    $status=in_array( date('w'),$closed ) ? 'Closed' : 'Open';
    printf('We are "%s" on %s', $status, date('l') );

今天的产出是：

We are "open" on Saturday