Question

You know when it s late in the night and your brain is fried? I m having one of those nights right now, and my function so far is not working as it should, so please take a look at it: (I should note that I m using the PHP 5.2.9, and the function / method DateTime:Diff() is not available until PHP 5.3.0.

<?php
    function time_diff($ts1, $ts2) {
        # Find The Bigger Number
        if ($ts1 == $ts2) {
            return  0 Seconds ;
        } else if ($ts1 > $ts2) {
            $large = $ts1;
            $small = $ts2;
        } else {
            $small = $ts1;
            $large = $ts2;
        }
        # Get the Diffrence
        $diff = $large - $small;
        # Setup The Scope of Time
        $s = 1;         $ss = 0;
        $m = $s * 60;   $ms = 0;
        $h = $m * 60;   $hs = 0;
        $d = $h * 24;   $ds = 0;
        $n = $d * 31;   $ns = 0;
        $y = $n * 365;  $ys = 0;
        # Find the Scope
        while (($diff - $y) > 0) { $ys++; $diff -= $y; }
        while (($diff - $n) > 0) { $ms++; $diff -= $n; }
        while (($diff - $d) > 0) { $ds++; $diff -= $d; }
        while (($diff - $h) > 0) { $hs++; $diff -= $h; }
        while (($diff - $m) > 0) { $ms++; $diff -= $m; }
        while (($diff - $s) > 0) { $ss++; $diff -= $s; }
        # Print the Results
        return "$ys Years, $ns Months, $ds Days, $hs Hours, $ms Minutes & $ss Seconds.";
    }
    // Test the Function:
    ediff(strtotime( December 16, 1988 ), time());
    # Output Should be:
    # 20 Years, 11 Months, 8 Days, X Hours, Y Minutes & Z Seconds.
?>

Answer 1

如何做到这一点:

function time_diff($t1, $t2)
{
   $totalSeconds = abs($t1-$t2);
   $date = getdate($totalSeconds); 
   $firstYear = getdate(0);
   $years = $date[ year ]-$firstYear[ year ];
   $months = $date[ mon ];
   $days = $date[ mday ];
   $hours = $date[ hour ];
   $minutes = $date[ minutes ];
   $seconds = $date[ seconds ];

   return "$years Years, $months Months, $days Days, $hours Hours, $minutes Minutes & $seconds Seconds.";
}

这把特定时间的区别作为日期。然后,你可以让“目标”为你们做一切工作。唯一的挑战是年数——这只是年复一年(差异)减去“Unix epoch年”(1970年)。

如果你不喜欢使用一个实际月,你也可以将“年”时间按12个月的平均天数分开。

$months = $date[ yday ] / (365/12);

同样,在剩下的几天里,可以用模块组合列出。

$days = $date[ yday ] % (365/12);

Answer 2

这是对你的问题的回答,但我只是想指出......。

while (($diff - $y) > 0) { $ys++; $diff -= $y; }

书写方式非常低

$ys = $diff / $y;
$diff = $diff % $y;

此外,

       else if ($ts1 > $ts2) {
                $large = $ts1;
                $small = $ts2;
        } else {
                $small = $ts1;
                $large = $ts2;
        }
        # Get the Diffrence
        $diff = $large - $small;

易于改写

$diff = abs($ts1 - $ts2);

我感到,如果问题不那么尖锐的话,你守则中的问题就会更加明显。

Answer 3

如何简单地简化第一部分

$diff = abs($ts2 - $ts1);

然后,当你这样做时:

 $n = $d * 31;   $ns = 0;
 $y = $n * 365;  $ys = 0;

实际上,你说,一年是365天,长31个月。这实际上长达36年。也许不是你想要的东西。

最后,我们都在这里成长。请使用可变名称,即YEAR_IN_SECONDS而不是美元。你可以清楚地看到,你可以一劳永逸地写法,但另外20个cks子将不得不阅读。

Answer 4

就特定时段内所有几个月的需要而言,我们利用了以下编码:

function MonthsBetweenTimeStamp($t1, $t2) {
   $monthsYear = array();
   $lastYearMonth  = strtotime(gmdate( F-Y , $t2));
   $startYearMonth = strtotime(gmdate( F-Y , $t1));
    while ($startYearMonth < $lastYearMonth) {
        $monthsYear[] = gmdate("F-Y", $startYearMonth);
                    //Increment of one month directly 
        $startYearMonth = strtotime(gmdate("F-Y", $startYearMonth) .   + 1 month );
    }

    if (empty($monthsYear)) {
        $monthsYear = array($startYearMonth));
    }

    return $monthsYear;

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