[RESOLVED] getting date difference using php

[rosy]

Hi
How to get difference between two dates.
one is current todays date and another from a selection through drop down box.
I need to claculate the number of days between dates in post form.
my code for date in drop down box is

<?PHP

print "<TD>\n<select name=\"aninstallyear\" id=\"aninstallyear\" CLASS=\"text\" tabindex=\"1\">\n";

$todays_date = date("Y-m-d");

$timea = explode(" ",$todays_date);
$timea1 = explode("-",$timea[0]);
$timeron = mktime(0,0,0,$timea1[1],$timea1[2],$timea1[0]);

$installday = $timea1[2];
$installyear = $timea1[0];

// drop-down boxes for year
print "<option value=\"\"> </option>\n";
for ($x=$startyear; $x<=($startyear + 10); $x++){
$output = "<option value=\"" . sprintf("%02d", $x) . "\"";


$output .= ">" . sprintf("%02d", $x) . "</option>\n";
print $output;
}

print "</select>\n<select name=\"aninstallmonth\" id=\"aninstallmonth\" CLASS=\"text\" tabindex=\"2\">\n";

//drop-down for month
print "<option value=\"\"> </option>\n";
for ($x=1; $x<=12; $x++){
$output = "<option value=\"" . sprintf("%02d", $x) . "\"";

$output .= ">" . sprintf("%02d", $x) . "</option>\n";
print $output;
}
//print "</select>";

print "</select>\n<select name=\"aninstallday\" id=\"aninstallday\" CLASS=\"text\" tabindex=\"3\">\n";
//drop-down for day
print "<option value=\"\"> </option>\n";
for ($x=1; $x<=31; $x++){
$output = "<option value=\"" . sprintf("%02d", $x) . "\"";

$output .= ">" . sprintf("%02d", $x) . "</option>\n";
print $output;
}
print "</select>";

//print "<input type=hidden value=1 name=aninstallday>";

print "</TD>";

?>

Thanks and any help would be appreciated.

[Neumy]

Your best bet is to make good use of date('U'). Get both dates with this and making a quick subtraction, and you should get how many days between both dates.

unchecked code:

$earlyDate = date("U", mktime(0, 0, 0, 7, 1, 2000));
$laterDate = date("U", mktime(0, 0, 0, 7, 1, 2006));
$daysDifference = $laterDate - $earlyDate;
echo "there are $daysDifference days between said dates.";

[bpat1434]

@:
Your solution won't work completely since you'd have an answer in seconds, not days. You still have to calculate the number of days that's in that amount of seconds.

-------------------------------------------------------------------------------------------

Get todays UNIX Timestamp:

time();

Now, create a UNIX Timestamp of the submitted date:

$sub_time = mktime(0,0,0,$_POST['aninstallmon'],$_POST['aninstallday'],$_POST['aninstallyear']);

Now from here is some good old math:

<?php

// Get current time
$curtime = time();

// Get submitted dates time:
$subtime = mktime(0,0,0,$_POST['aninstallmon'],$_POST['aninstallday'],$_POST['aninstallyear']);

// 1.) Calculate the difference in times
$diff = abs($curtime-$subtime);

// 2.) Define Time lengths:
$sec = 60;
$min = $sec*60;
$hour = $min*60;
$day = $hour*24;
$week = $day*7; // Not needed, since you want days between

// 3.) Calculate the number of days in the difference:
$days = floor($diff / $day);

// 3a.) Update the difference
$diff = $diff - ($days*$day);

// 4.) Display the days:
echo 'There are ' . $days . ' days between now and ' . $_POST['aninstallmon'] . '/' . $_POST['aninstallday'] . '/' . $_POST['aninstallyear'];
?>

That should get you going....

[rosy]

Hi,
I tried the above coding, i get wrong calculation.
Please help.

[Neumy]

Here's a little script that I created based on my above script with a little modification thanks to bpat1434's observation.

$earlyDate = date("U", mktime(0,0,0, 1, 1, 1972)); 
$laterDate = date("U", mktime(0, 0, 0, 11, 28, 2006)); 
$seconds = $laterDate - $earlyDate; 
$minutes = $seconds / 60;
$hours = $minutes / 60;
$days = ceil($hours / 24);
$years = floor($days / 365); // approx.
print "There are $seconds seconds since I was born.\n";
print "There are $minutes minutes since I was born.\n";
print "There are $hours hours since I was born.\n";
print "There are $days days since I was born.\n";
print "There are approx. $years years since I was born.\n";

My only warning is that years is not perfect due to wonderful leap year data. You'll have to add some more calculations to get an exact figure here.

I ran a few tests and number of days seems to work correctly.

[rosy]

Thanks a lot !!!!!!!!!! Neumy:-) and bpat1434's :-)