比较今天和商店的营业日，并显示是否营业或关闭？答案

【问题标题】：Compare today day and shop opening days and show if opened or closed?比较今天和商店的营业日，并显示是否营业或关闭？
【发布时间】：2022-01-25 11:00:40
【问题描述】：

您好，我正在尝试在 php 中创建一个函数来检查当前日期并返回商店是否打开或关闭。这是我到目前为止所拥有的，它有效，但它很长，所以如果有人知道最短路径，请帮忙！

$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
  }
}

如果日期介于星期四和星期一之间，则返回 Opened else Closed。如果您知道捷径，请帮助我

【问题讨论】：

标签： php

【解决方案1】：

如果您知道关闭的日子（您必须这样做），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') );

今天输出：

We are "open" on Saturday

【讨论】：