如何在PHP中檢查4天前是否有東西？

Question

我想寫一個函數來檢查一個「完成的課程」是否在四天前。例如，如何檢查課程是否在該時間範圍內。如果它是在昨天，2天前，3天前，4天前完成的，那麼它是在「4天前」的時間範圍內。

如何檢查？

到目前爲止，我已經做了：

$time = time(); 

$fourDays = 345600; 
$threeDays = 259200; 
$lastLesson = $ml->getLesson($cid, $time, true); 

$lastLessonDate = $lastLesson['deadline']; 
$displayLastLesson = false; 
if ($lastLessonDate + $fourDays < $time) 
{ 
    $displayLastLesson = true; 
    //We print lesson that was finished less than 4 days ago 
} 
else 
{ 
    //We print lesson that is in the next 3 days 

} 

眼下，如果語句保持擊球真的，是不是我想要的，因爲我有一個在5月3日結束的一課。 5月7日結束的課程應該是真的嗎？

Answer 1

$time = time(); 
$fourDays = strtotime('-4 days'); 
$lastLesson = $ml->getLesson($cid, $time, true); 

$lastLessonDate = $finishedLesson['deadline']; 
$displayLastLesson = false; 
if ($lastLessonDate >= $fourDays && $lastLessonDate <= $time) 
{ 
    $displayLastLesson = true; 
    //We print lesson that was finished less than 4 days ago 
} 
else 
{ 
    //We print lesson that is in the next 3 days 

} 

Answer 2

所有的計算應相對於在今天上午12點來計算，而不是time()現在給你當前時間（例如下午6:00）。這是因爲，當你做到這一點，1天前（現在 - 24小時）的問題的手段時間是在昨天下午6點到今天下午6點之間。相反，昨天應該是昨天早上12點到今天早上12點之間的時間。

下面是一個簡單的計算來說明這個想法：

$lastLessonDate = strtotime($lastLessonDate); 
$today = strtotime(date('Y-m-d')); // 12:00am today , you can use strtotime('today') too 
$day = 24* 60 * 60; 
if($lastLessonDate > $today) // last lesson is more than 12:00am today, meaning today 
echo 'today'; 
else if($lastLessonDate > ($today - (1 * $day)) 
echo 'yesterday'; 
else if($lastLessonDate > ($today - (2 * $day)) 
echo '2 days ago'; 
else if($lastLessonDate > ($today - (3 * $day)) 
echo '3 days ago'; 
else if($lastLessonDate > ($today - (4 * $day)) 
echo '4 days ago'; 
else 
echo 'more than 4 days ago';