Question

我有一些事件。每周应用程序中显示的事件取决于插入的日期。像

event one > Friday
event two > sat
event three > sun

所以第一个事件每周五凌晨2点到凌晨2点在申请中显示

我很困惑如何管理，凌晨2点到凌晨2点。我已经创建了一个逻辑，但它不能给我正确的计算

    $input = time();
                $day = date('D', $input );

                switch ($day) {
                    case 'Sun':
                        $finalday='0';
                        break;
                    case 'Mon':
                        $finalday='1';
                        break;
                    case 'Tue':
                        $finalday='2';
                        break;
                    case 'Wed':
                        $finalday='3';
                        break;
                    case 'Thu':
                        $finalday='4';
                        break;
                    case 'Fri':
                        $finalday='5';
                        break;
                    case 'Sat':
                        $finalday='6';
                        break;
                }




                $now = time();
                $event_time = strtotime("02:00 am");

                if( ($now - $event_time) < 0) // 5 minutes * 60 seconds, replace with 300 if you'd like
                {
                    //before day
                    if($finalday=='0')
                    {

                        $query_day='6';
                    }
                    else
                    {
                        $query_day=$finalday-1;
                    }


                }
                else
                {
//current day
                    $query_day=$finalday;
                }

我如何准确地显示每个事件凌晨2点到凌晨2点取决于插入的日期

现在假设现在是上午12点所以星期五是星期五，但是事件1将从凌晨2点到下午1.59点显示，然后事件2将从凌晨2点到凌晨1.5点（坐着）显示。

通过这种方式，将显示下一个弱自动事件

Answer 1

尝试这样的方法只是为了让你的代码看起来更优雅而不那么笨重

$dayOfWeek = date('w'); //0 for Sunday through 6 for Saturday
$hourOfDay = date('H'); //0-23
$eventOne = null;
$eventTwo = null;
$eventThree = null;

//logic structure to set events
if($hourOfDay >= 0 && $hourOfDay < 2){
    $dayOfWeek -= 1; //set to previous day if earlier than 2AM
    $dayOfWeek = $dayOfWeek == 0 ? 6 : $dayOfWeek; //quick check to set to Sunday if day was on Monday
    $eventOne = $dayOfWeek;
    $eventTwo = $dayOfWeek+1;
    $eventThree = $dayOfWeek+2;

    //single line if statements to correct weekly overflow
    if($eventTwo == 7) $eventTwo = 0;
    if($eventThree == 7) $eventThree = 0;
    if($eventThree == 8) $eventThree = 1;
}else{

    $eventOne = $dayOfWeek;
    $eventTwo = $dayOfWeek+1;
    $eventThree = $dayOfWeek+2;

    //single line if statements to correct weekly overflow
    if($eventTwo == 7) $eventTwo = 0;
    if($eventThree == 7) $eventThree = 0;
    if($eventThree == 8) $eventThree = 1;
}



function getDayOfEvent($event){
    switch($event){
        case 0: return "Sunday"; break;
        case 1: return "Monday"; break;
        case 2: return "Tuesday"; break;
        case 3: return "Wednesay"; break;
        case 4: return "Thursday"; break;
        case 5: return "Friday"; break;
        case 6: return "Saturday"; break;
    }   
}

print "Event One: ". getDayOfEvent($eventOne)."\nEvent Two: ".getDayOfEvent($eventTwo)."\nEvent Three: ".getDayOfEvent($eventThree);

如果这样的话适合你，请告诉我。对不起，但我很难翻译你的英语，我尽我所能:)我希望这对你有所帮助，如果没有，请告诉我，我会帮助你解决这个问题。

这是CodePad上的一个粘贴，如果你想要http://codepad.org/SLcTeGEt

，你可以稍微使用代码

Answer 2

试试这个程序......也许就是你想要的。（你的问题很混乱。）

/* Day Of Week 0 = Sun ... 6 = Sat
 * ---------------------------------
 * Day      Hour     Result    Case
 * ---------------------------------
 *  5     00 - 02   No event    C
 *  5     02 - 24   Event 1     B
 *  6     00 - 02   Event 1     A
 *  6     02 - 24   Event 2     B
 *  0     00 - 02   Event 2     A
 *  0     02 - 24   Event 3     B
 *  1     00 - 02   Event 3     A
 *  1     02 - 24   No Event    C
 * Other  Other     No Event    C
 * ---------------------------------
 */

function getEvent( $timestamp, $eventTime ) {
    $d = (int) date( 'w', $timestamp ); // Day
    $h = (int) date( 'G', $timestamp ); // Hour
    $event = $h < $eventTime && ( $d > 5 || $d < 2 )  // Case A
        ? ( $d + 2 ) % 7                              // Case A Result
        : ( $h >= $eventTime && ( $d > 4 || $d == 0 ) // Case B
            ? ( $d + 3 ) % 7                          // Case B Result
            : null );                                 // Case C Result
    printf ( "\n%s %02d:00 :: %s",                    // ... and show
             date( 'D', strtotime( "Sunday +{$d} days" ) ),
             $h, $event ? "Event $event" : 'No event' );
}

$eventTime = 2;
echo '<pre>';
/* Testing the getEvent function */
for ( $timestamp = mktime( 23, 0, 0, 4, 30, 2015 ); // Thu at 23:00
      $timestamp <= mktime( 22, 0, 0, 5, 7, 2015 ); // Thu at 22:00
      $timestamp += 3600 * 2 ) {                    // Each 2 hours
    getEvent( $timestamp, $eventTime );
}

?>

如何改进我当前的代码以获得正确的结果

2 个答案: