两次相差（Unix时代）

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

This isn't an answer to your question, but I just wanted to point out...

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

is a very inefficient way of writing

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

Also, this

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

can easily be rewritten as

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

I have a feeling that the problem in your code would be more apparent if it was less verbose. :)

Answer 2

how about simplifying the first part with a simple

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

Then, when you do this:

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

you are actually saying that a year is composed of 365 31 day long months. which is actually about 36 year long years. Probably not what you want.

Finally, we are all grown ups here. Please use grown up variable names ie $YEAR_IN_SECONDS instead of $ys. As you can clearly see, you may write code once, but 20 other schmucks are going to have to read it a lot of times.

Answer 3

In the case of needed all months during the given times-stamp then we have use of the following coding in php :

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; 

Answer 4

How about this:

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.";
}

This uses the difference of the given times as a date. Then you can let the "getdate" do all the work for you. The only challenge is the number years - which is simply the getdate year (of the difference) minus the Unix epoch year (1970).

If you don't like using an actual month, you could also divide the "year" day by the number of days in 12 equal months

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

Similarly days could be figured out the remaining days with modulus

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