將字符串轉換爲php中的時間

Question

如何將"2011-11-03T17:27:56Z"轉換爲php中的時間。

我希望與當前時間有所不同。

即，如果與當前時間的時差是10分鐘，我想要10分鐘。如果1天那麼我想要1天。

Answer 1

這個小片段將給你從現在到給定日期之間的秒差。

$dateString = "2011-11-03T17:27:56Z"; 
$date = strtotime($dateString); 
$diff = time() - $date; 
echo $diff; 

爲了給它你問的具體格式，你可以使用下面的功能，我發現here：

function time_diff($s) { 
    $m = 0; $hr = 0; $d = 0; $td = "now"; 
    if ($s > 59) { 
     $m = (int)($s/60); 
     $s = $s-($m*60); // sec left over 
     $td = "$m min"; 
    } 
    if ($m > 59) { 
     $hr = (int)($m/60); 
     $m = $m - ($hr*60); // min left over 
     $td = "$hr hr"; 
     if ($hr > 1) { 
      $td .= "s"; 
     } 
     if ($m > 0) { 
      $td .= ", $m min"; 
     } 
    } 
    if ($hr > 23) { 
     $d = (int) ($hr/24); 
     $hr = $hr-($d*24); // hr left over 
     $td = "$d day"; 
     if ($d > 1) { 
      $td .= "s"; 
     } 
     if ($d < 3) { 
      if ($hr > 0) { 
       $td .= ", $hr hr"; 
      } 
      if ($hr > 1) { 
       $td .= "s"; 
      } 
     } 
    } 
    return $td; 
} 

結合兩者，這是你會得到什麼：

$dateString = "2011-11-03T17:27:56Z"; 
$date = strtotime($dateString); 
$diff = time() - $date; 
echo time_diff($diff); 

輸出：

8天

Answer 2

我相信你想要的strtotime()功能：

$some_time = strtotime("2011-11-03T17:27:56Z");//outputs a UNIX TIMESTAMP 
$time_diff = (time() - $some_time); 
if ($time_diff > 86400) { 
    echo round($time_diff/86400) . " days"; 
} else if ($time_diff > 3600) { 
    echo round($time_diff/3600) . " hours"; 
} else { 
    echo round($time_diff/60) . " minutes"; 
} 

http://us.php.net/manual/en/function.strtotime.php

函數需要接受一個包含英語日期 格式的字符串，並會嘗試解析格式轉換成Unix時間戳

Answer 3

$diffInSecs = time() - strtotime('2011-11-03T17:27:56Z'); 

Answer 4

工作例如：（codepad here）

<?php 

$time_str = "2011-11-03T17:27:56Z"; 

//echo date('d/m/y H:i:s', strtotime($time_str)); 
$diff = abs(strtotime("now") - strtotime($time_str)); 

$years = floor($diff/(365*60*60*24)); 
$months = floor(($diff - $years * 365*60*60*24)/(30*60*60*24)); 
$days = floor(($diff - $years * 365*60*60*24 - $months*30*60*60*24)/ (60*60*24)); 
$hours = floor(($diff - $years * 365*60*60*24 - $months*30*60*60*24 - $days*60*60*24)/ (60*60)); 
$minuts = floor(($diff - $years * 365*60*60*24 - $months*30*60*60*24 - $days*60*60*24 - $hours*60*60)/ 60); 
$seconds = floor(($diff - $years * 365*60*60*24 - $months*30*60*60*24 - $days*60*60*24 - $hours*60*60 - $minuts*60)); 
printf("%d years, %d months, %d days, %d hours, %d minuts\n, %d seconds\n", $years, $months, $days, $hours, $minuts, $seconds); 

（時間差異了這裏：How to calculate the difference between two dates using PHP?）