查找php中2个unix时间戳之间的天数

8 回答

• 2

  您只需计算两个日期之间的秒数，然后除以得到天数：

  $numDays = abs($smallestTimestamp - $biggestTimestamp)/60/60/24;
  

  然后，您可以使用for循环来检索日期：

  $numDays = abs($smallestTimestamp - $biggestTimestamp)/60/60/24;
  
  for ($i = 1; $i < $numDays; $i++) {
      echo date('Y m d', strtotime("+{$i} day", $smallestTimestamp)) . '
  ';
  }
  

  同样，如果您不知道哪个时间戳是最小的，则可以使用min（）函数（strtotime中的第二个参数） .

  回复于 2024-04-25T18:18:41+08:00
• 1

  我认为快速解决此问题的方法是从end_stamp中减去几天的秒数，直到你到达start_tag .

  //1 day = 86400 seconds
  

  我会 Build 一个日后使用的数组 .

  编辑（示例）

  $difference = 86400;
  $days = array();
  while ( $start_time < $end_time )
  {
      $days[] = date('M j Y', $end_time);
  
      $end_time -= $difference;
  }
  

  这应该涵盖任何时间范围，即使它超过一个月 .

  回复于 2024-04-25T18:18:41+08:00
• 0

  试试这个：

  while($date_start <= $date_end) {
      echo date('M d Y', $date_start) . '<br>';
      $date_start = $date_start + 86400;
  }
  

  希望这可以帮助 ！

  回复于 2024-04-25T18:18:41+08:00
• 1

  $d1=mktime(22,0,0,1,1,2007);
  $d2=mktime(0,0,0,1,2,2007);
  echo "Hours difference = ".floor(($d2-$d1)/3600) . "<br>";
  echo "Minutes difference = ".floor(($d2-$d1)/60) . "<br>";
  echo "Seconds difference = " .($d2-$d1). "<br>";
  
  
  echo "Month difference = ".floor(($d2-$d1)/2628000) . "<br>";
  echo "Days difference = ".floor(($d2-$d1)/86400) . "<br>";
  echo "Year difference = ".floor(($d2-$d1)/31536000) . "<br>";
  

  http://www.plus2net.com/php_tutorial/date-diff.php

  http://www.phpf1.com/tutorial/php-date-difference.html

  回复于 2024-04-25T18:18:41+08:00
• 2

  **This is a very simple code for find days hours minutes and seconds in php**
  
      $dbDate = strtotime("".$yourbdDate.""); // Database date
      $endDate = time();    // current time
      $diff = $endDate - $dbDate; /// diffrence
  
      $days = floor($diff/86400);  ///  number of days 
      $hours = floor(($diff-$days*86400)/(60 * 60));  ////  number of hours
      $min = floor(($diff-($days*86400+$hours*3600))/60);///// numbers of minute
  
  
      $second = $diff - ($days*86400+$hours*3600+$min*60); //// secondes
  
      if($days > 0) echo $days." Days ago";
      elseif($hours > 0) echo $hours." Hours ago";
      elseif($min > 0) echo $min." Minute ago";
      else echo "Just second ago";
  

  回复于 2024-04-25T18:18:41+08:00
• 42

  像这样的东西？

  $day = $start;
  while ($day < $end) {
          $day += 86400;
          echo $day.' '.date('Y-m-d', $day).PHP_EOL;
  }
  

  顺便说一句，1262304000是12月31日，而不是1月1日 .

  回复于 2024-04-25T18:18:41+08:00
• 0

  得到两个日期的差异并除以86400.abs（（$ date1 - $ date2）/ 86400）将产生所需的结果

  回复于 2024-04-25T18:18:41+08:00
• 0

  $daysInBetween = range($startTs, $endTs, 86400);
  $secondDay = date('M d Y', $daysInBetween[1]);
  /*
  $thirdDay = date('M d Y', $daysInBetween[2]);
  ...
  */
  

  请注意 range() 函数包含在内 .

  回复于 2024-04-25T18:18:41+08:00