我有一些活动。根据插入day.like,每周在应用程序中显示事件
event one > Friday
event two > sat
event three > sun因此,每周五凌晨2点到2点在应用程序中显示事件1
我搞不懂如何管理凌晨2点到凌晨2点我已经创建了一个逻辑,但它无法给我正确的计算
$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;
}如何根据插入日期显示每个事件在凌晨2点到2点之间的准确时间
假设现在是中午12点,所以今天是星期五,但事件一将从凌晨2点播放到下一个1.59AM,然后事件2将从凌晨2点播放到下午1.59AM(星期六)。
通过这种方式,将显示下一个弱自动事件
发布于 2015-05-02 00:20:23
试着这样做,让您的代码看起来更优雅一点,而不是那么笨拙
$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);如果这样的东西对你有效,请让我知道。
这是一个粘贴在CodePad上的文件,如果你想使用http://codepad.org/SLcTeGEt,你可以稍微修改一下代码
发布于 2015-05-02 02:17:52
试试这个程序..。也许就是你想要的。(你的问题非常令人困惑。)
/* 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 );
}
?>https://stackoverflow.com/questions/29990403
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