如何在PHP中计算两个日期时间之间的分钟差异?


当前回答

这就是我如何显示“xx次前”在php > 5.2 ..这里是DateTime对象的更多信息

//Usage:
$pubDate = $row['rssfeed']['pubDates']; // e.g. this could be like 'Sun, 10 Nov 2013 14:26:00 GMT'
$diff = ago($pubDate);    // output: 23 hrs ago

// Return the value of time different in "xx times ago" format
function ago($timestamp)
{

    $today = new DateTime(date('y-m-d h:i:s')); // [2]
    //$thatDay = new DateTime('Sun, 10 Nov 2013 14:26:00 GMT');
    $thatDay = new DateTime($timestamp);
    $dt = $today->diff($thatDay);

    if ($dt->y > 0){
        $number = $dt->y;
        $unit = "year";
    } else if ($dt->m > 0) {
        $number = $dt->m;
        $unit = "month";
    } else if ($dt->d > 0) {
        $number = $dt->d;
        $unit = "day";
    } else if ($dt->h > 0) {
        $number = $dt->h;
        $unit = "hour";
    } else if ($dt->i > 0) {
        $number = $dt->i;
        $unit = "minute";
    } else if ($dt->s > 0) {
        $number = $dt->s;
        $unit = "second";
    }
    
    $unit .= $number  > 1 ? "s" : "";
 
    $ret = $number." ".$unit." "."ago";
    return $ret;
}

其他回答

一个更通用的版本,返回日,小时,分钟或秒的结果,包括分数/小数:

function DateDiffInterval($sDate1, $sDate2, $sUnit='H') {
//subtract $sDate2-$sDate1 and return the difference in $sUnit (Days,Hours,Minutes,Seconds)
    $nInterval = strtotime($sDate2) - strtotime($sDate1);
    if ($sUnit=='D') { // days
        $nInterval = $nInterval/60/60/24;
    } else if ($sUnit=='H') { // hours
        $nInterval = $nInterval/60/60;
    } else if ($sUnit=='M') { // minutes
        $nInterval = $nInterval/60;
    } else if ($sUnit=='S') { // seconds
    }
    return $nInterval;
} //DateDiffInterval

DateTime::diff很酷,但对于这种需要单个单元结果的计算来说很尴尬。手动减去时间戳效果更好:

$date1 = new DateTime('2020-09-01 01:00:00');
$date2 = new DateTime('2021-09-01 14:00:00');
$diff_mins = abs($date1->getTimestamp() - $date2->getTimestamp()) / 60;
<?php
$start = strtotime('12:01:00');
$end = strtotime('13:16:00');
$mins = ($end - $start) / 60;
echo $mins;
?>

输出:

75

它在我的程序上工作,我使用date_diff,你可以在这里检查date_diff手册。

$start = date_create('2015-01-26 12:01:00');
$end = date_create('2015-01-26 13:15:00');
$diff=date_diff($end,$start);
print_r($diff);

你得到你想要的结果。

用未来最大的1减去过去最大的1,然后除以60。

时间是Unix格式的,所以它们只是一个大数字,显示了从格林尼治时间1970年1月1日00:00:00开始的秒数