Question

我有一个代码，可以节省MySQL中特定业务的营业时间和工作日。从MySQL获取数据时，输出如下：

Array ( [open_hours] => 0 10 0,1 10 0,2 10 0,3 10 0,4 10 0,5 10 0,6 10 0 )

0 10 0只是意味着Mon 10am 12am，如果它是0 15 18意味着Mon 3pm 6pm。这不是问题，因为我已经通过以下代码解决了这个问题：

$openHrs = explode(",", $openHrs['open_hours']);

        $weekdays = array('Sun','Mon','Tue','Wed','Thu','Fri','Sat');
        $res      = array();
        foreach($openHrs as &$temp) {

            $temp = explode(" ", $temp);

            $temp[1] = $temp[1] > 12 ? $temp[1] - 12 . 'pm' : $temp[1] . 'am';
            $temp[2] = $temp[2] > 12 ? $temp[2] - 12 . 'pm' : $temp[2] . 'am';
            $res[]   = $weekdays[$temp[0]] . ' ' . $temp[1] . ' - ' . $temp[2];
        }

现在我如何编码更高级的代码，以便如果今天是星期一，我只需要显示星期一的输出结果？哪个是0 10 0或Mon 10am 12am？谢谢你的帮助

Answer 1

U可以使用此代码解决此问题

     for($i=0;$i<=6;$i++) 
     {
     if($res[$i]==date('w')) //get today's week number
           {
                echo $res[$i]." ".$temp[1]." ".$temp[2];              
            } 
     }

OR

     for($i=0;$i<=6;$i++) 
     {
     if($res[$i]==date('D')) // get today's weekday
           {
                echo $res[$i]." ".$temp[1]." ".$temp[2];              
            } 
     }

Answer 2

如果您想在今天的工作日过滤掉open_hours，则可以使用以下代码。

$openHrs = explode(",", $openHrs['open_hours']);

    $weekdays = array('Sun','Mon','Tue','Wed','Thu','Fri','Sat');
    $res      = array();

    $todayWeekDay = date('D'); // get today's weekday
    $todayWeekNum = array_search($todayWeekDay, $weekdays); // get number for today's weekday

    foreach($openHrs as &$temp) {

        $temp = explode(" ", $temp);

        $temp[1] = $temp[1] > 12 ? $temp[1] - 12 . 'pm' : $temp[1] . 'am';
        $temp[2] = $temp[2] > 12 ? $temp[2] - 12 . 'pm' : $temp[2] . 'am';
        // only add the item where the weekday is equal to today's
        if ($temp[0] == $todayWeekNum) {
             $res[]   = $weekdays[$temp[0]] . ' ' . $temp[1] . ' - ' . $temp[2];
        }
    }

Answer 3

    $i; // the weekday; comes as input

    $openHrs = explode(",", $openHrs['open_hours']);

    $weekdays = array('Sun','Mon','Tue','Wed','Thu','Fri','Sat');
    $res      = array();
    $temp = explode(" ", $openHrs[$i]);

    $temp[1] = $temp[1] > 12 ? $temp[1] - 12 . 'pm' : $temp[1] . 'am';
    $temp[2] = $temp[2] > 12 ? $temp[2] - 12 . 'pm' : $temp[2] . 'am';
    $res   = $weekdays[$temp[0]] . ' ' . $temp[1] . ' - ' . $temp[2];

Answer 4

这听起来像是一个棘手的问题，但看看是否值得尝试：

$open_hours = array(Mon 10 0,Tue 10 0,Wed 10 0,Thu 10 0,Fri 10 0,Sat 10 0,Sun 10 0 );
$today = date('D');
$associate_array_key = array_keys($open_hours, $today);
$openHrs = explode(" ", $open_hours[$associate_array_key]);
foreach($openHrs as &$temp) {
  $temp[0]; 
  $temp[1] = $temp[1] > 12 ? $temp[1] - 12 . 'pm' : $temp[1] . 'am';
  $temp[2] = $temp[2] > 12 ? $temp[2] - 12 . 'pm' : $temp[2] . 'am';
}

这只是我能想到的概念，当然可能需要对代码进行一些操作！

PHP获得工作日时间

4 个答案: