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Here’s a simple function that calculates the number of days between two dates.
<?php
//day's seconds = 86400
function days_between($day_i,$month_i,$year_i,$day_f,$month_f,$year_f){
$days_in_between = (mktime(0,0,0,$month_f,$day_f,$year_f) - mktime(0,0,0,$month_i,$day_i,$year_i))/86400;
return $days_in_between;
}
?>
//If we want to calculate the days between 21/8/2009 and 1/9/2009 then
echo days_between(21,8,2009,1,9,2009);
//would give us 11
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When I echo:
echo days_between(21,8,2009,1,9,2009);
or another manually entered date(s) this function works fine. However, I need to use variables and then it gives a Warning for ‘Missing Arguement’. I tried:
echo days_between($lt,$today);
where:
$today=date(“j,n,Y”);
and $lt comes from a database result:
$lt=date(“j,n,Y”,strtotime($row[‘life_time’]));
If I echo $lt I get the correct database result displayed and likewise $today displays the correct day’s date.
Any ideas what I’m doing wrong?
Thanks!
Craig.
OK I guessed that it wanted day, month, year as separate values, so I did this:
$tdd=date(“j”);
$tdm=date(“n”);
$tdy=date(“Y”);
…
$std=date(“j”,strtotime($row[‘life_time’]));
$stm=date(“n”,strtotime($row[‘life_time’]));
$sty=date(“Y”,strtotime($row[‘life_time’]));
echo days_between($std,$stm,$sty,$tdd,$tdm,$tdy);
NOW it works 🙂 I don’t know if there’s a shorter way though lol
check out this one:
http://kenno.wordpress.com/2009/07/04/php-number-of-days-between-two-dates/
greetz