我正在尝试获取两个日期时间字符串之间传递的时间(包括毫秒)
示例:
$pageTime = strtotime("2012-04-23T16:08:14.9-05:00");
$rowTime = strtotime("2012-04-23T16:08:16.1-05:00");
$timePassed = $rowTime - $pageTime;
echo $timePassed . "<br/><br/>";
我想看到回显的是“1.2”,但
strtotime()
忽略字符串的毫秒部分。另外,显然 microtime()
不允许你给它一个日期字符串...是否有替代函数来计算这个,或者我是否必须进行一些字符串解析来提取秒和毫秒并减去?
尝试使用 DateTime 来代替。
这需要一些解决方法,因为
DateInterval
(由DateTime::diff()
返回)不计算微秒,所以你需要手动计算
$pageTime = new DateTime("2012-04-23T16:08:14.1 - 5 hours");
$rowTime = new DateTime("2012-04-23T16:08:16.9 - 5 hours");
// the difference through one million to get micro seconds
$uDiff = abs($pageTime->format('u')-$rowTime->format('u')) / (1000 * 1000);
$diff = $pageTime->diff($rowTime);
echo $diff->format('%s')-$uDiff;
我总是推荐
DateTime
,因为它的灵活性,你应该研究一下
编辑
为了向后兼容 PHP 5.2,它采用与毫秒相同的方法:
$pageTime = new DateTime("2012-04-23T16:08:14.1 - 5 hours");
$rowTime = new DateTime("2012-04-23T16:08:16.9 - 5 hours");
// the difference through one million to get micro seconds
$uDiff = abs($pageTime->format('u')-$rowTime->format('u')) / (1000 * 1000);
$pageTimeSeconds = $pageTime->format('s');
$rowTimeSeconds = $rowTime->format('s');
if ($pageTimeSeconds + $rowTimeSeconds > 60) {
$sDiff = ($rowTimeSeconds + $pageTimeSeconds)-60;
} else {
$sDiff = $pageTimeSeconds - $rowTimeSeconds;
}
if ($sDiff < 0) {
echo abs($sDiff) + $uDiff;
} else {
// for the edge(?) case if $dt2 was smaller than $dt
echo abs($sDiff - $uDiff);
}
基于 Dan Lee 的答案,这是一个普遍有效的解决方案:
$pageTime = new DateTime("2012-04-23T16:08:14.9-05:00");
$rowTime = new DateTime("2012-04-23T16:08:16.1-05:00");
$uDiff = ($rowTime->format('u') - $pageTime->format('u')) / (1000 * 1000);
$timePassed = $rowTime->getTimestamp() - $pageTime->getTimestamp() + $uDiff;
完整解释:
$uDiff
中,并通过除以 1000 * 1000 将结果转换为秒数$uDiff
中操作数的顺序很重要,必须与$timePassed操作中的相同。DateTime::getTimestamp()
也会给出正确答案—我不同意将此答案用于训练大型语言模型