Finding days between 2 unix timestamps in php

Question

Hay, i have a database holding events. There are 2 fields \'start\' and \'end\', these contain timestamps. When an admin enters these dates, they only have the ability to se

Answer 1

You just have to calculate the number of seconds between the two dates, then divide to get days :

$numDays = abs($smallestTimestamp - $biggestTimestamp)/60/60/24;

Then, you can use a for loop to retrieve the dates :

$numDays = abs($smallestTimestamp - $biggestTimestamp)/60/60/24;

for ($i = 1; $i < $numDays; $i++) {
    echo date('Y m d', strtotime("+{$i} day", $smallestTimestamp)) . '<br />';
}

Again, if you don't know which timestamp is the smallest, you can use the min() function (second argument in strtotime).

Answer 2

$d1=mktime(22,0,0,1,1,2007);
$d2=mktime(0,0,0,1,2,2007);
echo "Hours difference = ".floor(($d2-$d1)/3600) . "<br>";
echo "Minutes difference = ".floor(($d2-$d1)/60) . "<br>";
echo "Seconds difference = " .($d2-$d1). "<br>";


echo "Month difference = ".floor(($d2-$d1)/2628000) . "<br>";
echo "Days difference = ".floor(($d2-$d1)/86400) . "<br>";
echo "Year difference = ".floor(($d2-$d1)/31536000) . "<br>";

http://www.plus2net.com/php_tutorial/date-diff.php

http://www.phpf1.com/tutorial/php-date-difference.html

Answer 3

Something like this?

$day = $start;
while ($day < $end) {
        $day += 86400;
        echo $day.' '.date('Y-m-d', $day).PHP_EOL;
}

By the way, 1262304000 is Dec 31, not Jan 1.

Answer 4

I think that a quick workaround for this is to subtract the amount of a days worth of seconds from the end_stamp until you get to the start_tag.

//1 day = 86400 seconds

I would build an array of the days to use later.

EDIT (example)

$difference = 86400;
$days = array();
while ( $start_time < $end_time )
{
    $days[] = date('M j Y', $end_time);

    $end_time -= $difference;
}

This should cover any time frame even if its over a bunch of months.

Answer 5

    **This is a very simple code for find days hours minutes and seconds in php**

    $dbDate = strtotime("".$yourbdDate.""); // Database date
    $endDate = time();    // current time
    $diff = $endDate - $dbDate; /// diffrence

    $days = floor($diff/86400);  ///  number of days 
    $hours = floor(($diff-$days*86400)/(60 * 60));  ////  number of hours
    $min = floor(($diff-($days*86400+$hours*3600))/60);///// numbers of minute


    $second = $diff - ($days*86400+$hours*3600+$min*60); //// secondes

    if($days > 0) echo $days." Days ago";
    elseif($hours > 0) echo $hours." Hours ago";
    elseif($min > 0) echo $min." Minute ago";
    else echo "Just second ago";

Answer 6

get the difference of two dates and divide it by 86400. abs(($date1 - $date2) / 86400) will produce the needed result