[SOLVED] The number of days between 2 dates?

[citricsquidut78696]

Hi,

I've found a couple of solutions on this, however everything I've written only works for certain date formats and it also messes up when the date goes over a year.

 

I have the date in the format Day/Month/year - So 28/09/1991 would be the 28th of September 1991.

Now, I need to work out the amount of days between 28/09/1991 and 18/02/1996 for example.

 

How would I do this? The data would be coming in from a database, so it would be coming in that date format, d/m/y.

 

Any help much appreciated

[marcus]

<?php
$date1 = mktime(0,0,0,9,28,1991);
$date2 = mktime(0,0,0,2,18,1996);

$diff = $date2 - $date1;

$day = 86400;

echo floor($diff/$day) . " days";
?>

[craygo]

try this

function days($date1, $date2){
$month1 = date("m", strtotime($date1));
$day1 = date("d", strtotime($date1));
$year1 = date("Y", strtotime($date1));
$month2 = date("m", strtotime($date2));
$day2 = date("d", strtotime($date2));
$year2 = date("Y", strtotime($date2));
$first = gregoriantojd($month1, $day1, $year1);
$second = gregoriantojd($month2, $day2, $year2);
$diff = abs($first-$second);
return $diff;
}

 

Now you can use your dates from the database

 

$days = days($date1, $date2);

 

Ray

[citricsquidut78696]

See, this is what's getting at me, everything doesn't work.

 

I'm doing;

function days($date1, $date2){
$month1 = date("m", strtotime($date1));
$day1 = date("d", strtotime($date1));
$year1 = date("Y", strtotime($date1));
$month2 = date("m", strtotime($date2));
$day2 = date("d", strtotime($date2));
$year2 = date("Y", strtotime($date2));
$first = gregoriantojd($month1, $day1, $year1);
$second = gregoriantojd($month2, $day2, $year2);
$diff = abs($first-$second);
return $diff;
}

 

Then to call it;

 

$date1 = "28/09/1992";
$date2 = "28/09/1993";
$days = days($date1, $date2);

 

And it doesn't do anything ;|

This is starting to get beyond annoying, I've spent hours trying to solve this and EVERY solution doesn't work. Where am I going wrong?

[kenrbnsn]

This function assumes that a date with slashes in it is in the form "mm/dd/yyyy" not "dd/mm/yyyy". It would be best to pass the dates in the form "yyyy-mm-dd".

 

Ken

[citricsquidut78696]

This function assumes that a date with slashes in it is in the form "mm/dd/yyyy" not "dd/mm/yyyy". It would be best to pass the dates in the form "yyyy-mm-dd".

 

Ken

 

Brilliant, thanks.

It's not perfect, but I can work with that, just involves messing with user input, but I can manage that

 

Thanks posters <3.