PHP：分鐘前，秒前：這是正確的編碼？

Question

if($timestamp > time() - 60){ 
    // Count seconds ago 
    $deff = time() - $timestamp; 
    $seconds = $deff/60; 
    echo intval($seconds)." seconds ago"; 
} elseif($timestamp > time() - 3600){ 
    // Count minutes ago 
    $deff = time() - $timestamp; 
    $minuts = $deff/60; 
    echo intval($minuts)." minutes ago"; 
} 

我覺得分鐘前是正確的，但不是秒？我應該如何做「秒前」的權利？

Answer 1

不要除以60 該值以秒爲代表的已經，你不需要把它轉換成秒。

這段代碼有點亂。考慮使用DateTime包;它具有內置的差異和格式化功能。

Answer 2

我認爲你的'$秒'應該等於'$ deff'，不是嗎？

Answer 3

DateInterval :: format（）提供了一個更好的方法。

http://php.net/manual/en/dateinterval.format.php

Answer 4

我使用這個...它處理過去和未來的時間。

/** 
* Get a string representing the duration difference in two timestamps 
* 
* @param int $time The timestamp to compare 
* @param int $timeBase The base timestamp to compare against, defaults to time() 
* @return string Returns the duration summary 
*/ 
public function getTimeSummary ($time, $timeBase = false) { 
    if (!$timeBase) { 
     $timeBase = time(); 
    } 

    if ($time <= time()) { 
     $dif = $timeBase - $time; 

     if ($dif < 60) { 
      if ($dif < 2) { 
       return "1 second ago"; 
      } 

      return $dif." seconds ago"; 
     } 

     if ($dif < 3600) { 
      if (floor($dif/60) < 2) { 
       return "A minute ago"; 
      } 

      return floor($dif/60)." minutes ago"; 
     } 

     if (date("d n Y", $timeBase) == date("d n Y", $time)) { 
      return "Today, ".date("g:i A", $time); 
     } 

     if (date("n Y", $timeBase) == date("n Y", $time) && date("d", $timeBase) - date("d", $time) == 1) { 
      return "Yesterday, ".date("g:i A", $time); 
     } 

     if (date("Y", $time) == date("Y", time())) { 
      return date("F, jS g:i A", $time); 
     } 
    } else { 
     $dif = $time - $timeBase; 

     if ($dif < 60) { 
      if ($dif < 2) { 
       return "1 second"; 
      } 

      return $dif." seconds"; 
     } 

     if ($dif < 3600) { 
      if (floor($dif/60) < 2) { 
       return "Less than a minute"; 
      } 

      return floor($dif/60)." minutes"; 
     } 

     if (date("d n Y", ($timeBase + 86400)) == date("d n Y", ($time))) { 
      return "Tomorrow, at ".date("g:i A", $time); 
     } 
    } 

    return date("F, jS g:i A Y", $time); 
} 

Answer 5

我用類似這樣的一個小功能，可能不是最好的，並且不應對未來的時間，但它的罰款，如果你想要的是過去。

function ago($when) { 
     $diff = date("U") - $when; 

     // Days 
     $day = floor($diff/86400); 
     $diff = $diff - ($day * 86400); 

     // Hours 
     $hrs = floor($diff/3600); 
     $diff = $diff - ($hrs * 3600); 

     // Mins 
     $min = floor($diff/60); 
     $diff = $diff - ($min * 60); 

     // Secs 
     $sec = $diff; 

     // Return how long ago this was. eg: 3d 17h 4m 18s ago 
     // Skips left fields if they aren't necessary, eg. 16h 0m 27s ago/10m 7s ago 
     $str = sprintf("%s%s%s%s", 
       $day != 0 ? $day."d " : "", 
       ($day != 0 || $hrs != 0) ? $hrs."h " : "", 
       ($day != 0 || $hrs != 0 || $min != 0) ? $min."m " : "", 
       $sec."s ago" 
     ); 

     return $str; 
} 

將一個unix時間戳傳遞給它（你的時間在過去），它會說明它是多久以前。如果您希望返回值略有不同，或者可能希望在其中添加數月或數年，則可輕鬆修改。