Question

我有两个时间戳由time（）提供，并作为varchar存储在数据库中。

现在我要计算两次之间的差异;一个是1366627990，另一个是$time2=time();。

我使用了一个脚本

function dateDiff($time1, $time2, $precision = 3) {
        date_default_timezone_set("UTC");
    // If not numeric then convert texts to unix timestamps
    if (!is_int($time1)) {
      $time1 = strtotime($time1);
    }
    if (!is_int($time2)) {
      $time2 = strtotime($time2);
    }

    // If time1 is bigger than time2
    // Then swap time1 and time2
    if ($time1 > $time2) {
      $ttime = $time1;
      $time1 = $time2;
      $time2 = $ttime;
    }

    // Set up intervals and diffs arrays
    $intervals = array('year','month','day','hour','minute','second');
    $diffs = array();

    // Loop thru all intervals
    foreach ($intervals as $interval) {
      // Set default diff to 0
      $diffs[$interval] = 0;
      // Create temp time from time1 and interval
      $ttime = strtotime("+1 " . $interval, $time1);
      // Loop until temp time is smaller than time2
      while ($time2 >= $ttime) {
    $time1 = $ttime;
    $diffs[$interval]++;
    // Create new temp time from time1 and interval
    $ttime = strtotime("+1 " . $interval, $time1);
      }
    }

    $count = 0;
    $times = array();
    // Loop thru all diffs
    foreach ($diffs as $interval => $value) {
      // Break if we have needed precission
      if ($count >= $precision) {
    break;
      }
      // Add value and interval 
      // if value is bigger than 0
      if ($value > 0) {
    // Add s if value is not 1
    if ($value != 1) {
      $interval .= "s";
    }
    // Add value and interval to times array
    $times[] = $value . " " . $interval;
    $count++;
      }
    }

    // Return string with times
    return implode(", ", $times);
  }

我将该函数称为

echo dateDiff("1366627990",$time2, 3);

但是它给了我

43 years, 3 months, 22 days

虽然时间不应超过10 days的最大值。

我的剧本有问题吗？

Answer 1

好吧，首先抓住两个时间戳的intval（），从另一个中减去一个并取结果的绝对值。你最终会得到两次之间的总秒数。

然后使用其中一个示例将该数字（秒）转换为人类可读的内容：

seconds to minutes and days to weeks

Answer 2

首先，不要指望差异小于10天。这是31年：

echo date("r", 366627990); // your first time. Fri, 14 Aug 1981 09:06:30 +0000
echo date("r", 1366971706); // $time2 in my case. Fri, 26 Apr 2013 10:21:46 +0000

更新

好的，现在您已将日期更正为1366627990。这是4天的差异：

echo date("r", 1366627990); // updated time. Mon, 22 Apr 2013 10:53:10 +0000

但是这个答案中提供的解决方案仍然是正确的，并且适用于您的新旧时间戳（以及任何其他两个时间戳:-)）。

现在让我在您的脚本中解决问题，然后向您展示更好的计算时间和日期差异的替代方案。

修复脚本

问题是您的“如果不是数字然后将文本转换为unix时间戳”，请在开头查看。因为如果将UNIX时间戳作为字符串传递，这将无效，您显然会这样做。修复此问题的黑客将在函数的开头包含以下内容：

function dateDiff($time1, $time2, $precision = 3) {
    // hack for passing timestamps as strings
    if ($time1 == intval($time1)) {
        $time1 = intval($time1);
    }
    if ($time2 == intval($time2)) {
        $time2 = intval($time2);
    }

现在，您的函数会产生正确的输出：

31 years, 8 months, 12 days

计算时间间隔的更好方法

从版本5.2.0开始，PHP具有超级整洁DateTime class和a diff method，这正是您所需要的：

$d1 = new DateTime();
$d1->SetTimestamp($time1);
$d2 = new DateTime();

$diff = $d1->diff($d2);

echo $diff->format("%y years, %m months, %d days");

这个也给你了

31 years, 8 months, 12 days

Answer 3

尝试

<?php
    $time1 = (int)1366971673; //The value of time() when I wrote this.
    $time2 = (int)1366627990; //Your time value to compare.

    $diff = (int)$time1-$time2;

    echo round(($diff/60/60/24)).' days between dates.';
?>

应输出;

4 days between dates.

http://phpfiddle.org/main/code/ahc-mg3

两个时间戳之间的时间计算

3 个答案:

更新

修复脚本

计算时间间隔的更好方法