PHP - Calculate Days From Date To Date

I have a SQL row that has a date field: ex: 2010-11-01.
When a car is sold there either is a 30 day warranty, a 60 day warranty, or 0 day warranty.
What I'm trying to do is display when the vehicles warranty expires, based on the date it was sold, or when did it expire based on the same sold date pulled from the database.

Example using last months date: 2010-10-01

60 day: "Expires 11-30-10"
30 day: "Expired 11-01-10"

I can not seem to use the date function properly...
Any help would be greatly appreciated.

Hi guys, I've hit a brick wall here and am in need of your help.

I'm pretty new to PHP and have limited knowledge to say the least. I'll explain what it is I'm trying to do.

Set start date as 01/01/2004 (dmY) $oFour
Set how many days has it been since then? $today
Set how many days it was from $ofour 30 days ago. $today -30 = $thirtyDaysAgo


But the problem is I don't know how to make date('z'); work from 2004 and not 01/01/2010.

So $today will be how many days it has been since the start of 2004 and $thirtyDaysAgo will be $today -30. I can set up $thirtyDaysAgo no problem but it's just finding out how to get the $today number...


Hope anyone can offer a little light to my situation :/

Mav

Hi fellas, this is really kicking my arse and i know its so simple!

I retrieve a date from the database, done!
I am manipulating it to display as i want, done!
How the hell do i add 365 days to this date?

$date= ($row['date']);
$subscription = strtotime($date);
echo
"<p>Subscription renewal date: ". date('l jS F Y', $subscription) . "</p>";

I was wondering if there was a way to have the MAX function NOT return a Date that is more than 2 days into the future (from the current day)?  If there is a Date that is more than 2 days into the future I would like to return the one closest to the current day.  Here is the code I have:

Code: [Select]
<?php

mysql_connect("local", "xxx", "xxx") or die(mysql_error());
mysql_select_db("pricelink") or die(mysql_error());



// Get a specific result from the "ft9_fuel_tax_price_lines" table
$query ="SELECT ItemNumber,TableCode,Cost, MAX(`Date`) as `max_date`, MAX(`Time`) as 'max_time' FROM `ft9_fuel_tax_price_lines`
         GROUP BY `ItemNumber`,`TableCode`";

$result = mysql_query($query) or die(mysql_error());

echo "<table border='1'>";
echo "<tr> <th>ItemNumber</th> <th>TableCode</th> <th>Date</th> <th>Time</th> <th>Cost</th> </tr>";


// keeps getting the next row until there are no more to get
while($row=mysql_fetch_array($result))
{

// Print out the contents of each row into a table
echo "<tr><td>"; 
echo $row['ItemNumber'];
        echo "</td><td>";
        echo $row['TableCode'];
        echo "</td><td>";
        echo $row['max_date'];
        echo "</td><td>";
        echo $row['max_time'];
        echo "</td><td>";
        echo $row['Cost'];
echo "</td></tr>"; 


} 

echo "</table>";
?>

Any help would be appreciated.
Thanks!

is there an easy way to add weekdays to a stored date ... so far i have

echo date ( 'Y-m-j' , strtotime ( '5 weekdays' ) );

this adds 5 weekdays to the current date , can i have it add 5 weekdays to say $TableDate 1;

Thanks in advance...

Hi, I am trying to get the number of days between the current date and a date in the future specified by column 'end_date'. The code I have seems to be working but it displays the number of days as a negative number, how do I change this to be a positive number? I have tried simply changing $days = $now - $end_date; to $days = $end_date - $now; but that doesn't work as I thought it would! Thanks in advance..

Code: [Select]
$now = time();
      $end_date = strtotime($row['end_date']);
      $days = $now - $end_date;
      echo floor($days/(60*60*24));

i have date of birth stored as DATE type in mysql.  i tried this so it would show the age but it comes up blank.

Code: [Select]
$getprof = mysql_query("SELECT * FROM Profile WHERE username='$search'")or die(mysql_error());

while($rowprof = mysql_fetch_assoc($getprof))
{

$username1 = $rowprof['username'];
$location = $rowprof['location'];
$gender = $rowprof['gender'];
$dateofbirth = $rowprof['dateofbirth'];
$information = $rowprof['information'];
}

function GetAge($dateofbirth)

{

        // Explode the date into meaningful variables

        list($BirthYear,$BirthMonth,$BirthDay) = explode("-", $dateofbirth);

        // Find the differences

        $YearDiff = date("Y") - $BirthYear;

        $MonthDiff = date("m") - $BirthMonth;

        $DayDiff = date("d") - $BirthDay;

        // If the birthday has not occured this year

        if ($DayDiff < 0 || $MonthDiff < 0)

          $YearDiff--;

        return $YearDiff;

}

echo $YearDiff;

Hey,

I'm using a script which allows you to click on a calendar to select the date to submit to the database.

The date is submitted like this: 2014-02-08

Is there a really simple way to prevent rows showing if the date is in the past?

Something like this:
if($currentdate < 2014-02-08 || $currentdate == 2014-02-08) {

}

Thanks very much,
Jack

Hi,

I need to calculate absent percentage but not sure how. working days from Sunday to Thursday every week so 5 working days a week. i will calculate the absent days from joining date until current date. Example if joining date is 24-3-2020 and today's date is 7-4-2020 i should get the number of absent days to 11 days since both Friday and Saturday are excluded. How to do that?

here is what I have:

$workingdays = $curdate - $joindate;
$workingdays = $workingdays - 				
$absent = ($counter / $workingdays) * 100;

the second line i missing the number of days for (Fridays and Saturdays) that should be excluded from calculations. The third line i did calculate the actual working days ($counter) for the employee so then i can get percentage of absent days. How to exclude the weekend days from calculations?

Edited April 1, 2020 by ramiwahdan
mistake in dates

Hello all,
The exact thing that i need is to calculate how much days there is in between two dates.
The only problem is that every thing that i found dont care about leap year
Anyone have a function to do that?

Hello. I'm new to pHp and I would like to know how to get my $date_posted to read as March 12, 2012, instead of 2012-12-03.
Here is the code:
Code: [Select]
<?php

$sql = " SELECT id, title, date_posted, summary FROM blog_posts ORDER BY date_posted ASC LIMIT 10 ";

$result = mysql_query($sql);

while($row = mysql_fetch_assoc($result)) {
$id = $row['id'];
$title = $row['title'];
$date_posted = $row['date_posted'];
$summary = $row['summary'];

echo "<h3>$title</h3>\n";
echo "<p>$date_posted</p>\n";
echo "<p>$summary</p>\n";
echo "<p><a href=\"post.php?id=$id\" title=\"Read More\">Read More...</a></p>\n";
}

?>
I have tried the date() function but it always updates with the current time & date so I'm a little confused on how I get this to work.

I have tried a large number of "solutions" to  this but everytime I use them I see 0000-00-00 in my date field instead of  the date even though I echoed and can see that the date looks correct.

Here's where I'm at:
I have a drop down for the month (1-12) and date fields (1-31) as well as a text input field for the year.  Using the POST array, I  have combined them into the xxxx-xx-xx format that I am using in my field as a date field in mysql. 

<code>
$date_value =$_POST['year'].'-'.$_POST['month'].'-'.$_POST['day'];
echo $date_value;
</code>

This outputs 2012-5-7 in my test echo but 0000-00-00 in the database.

I have tried unsuccessfully to use in a numberof suggested versions of:
strtotime()
mktime

Any help would be extremely appreciated.
I am aware that I need to validate this data and insure  that it is a valid date. That I'm okay with. I would like some help on getting it into the database.

Alright, I have a Datetime field in my database which I'm trying to store information in. Here is my code to get my Datetime, however it's returning to me the wrong date. It's returning:
1969-12-31 19:00:00

$mysqldate = date( 'Y-m-d H:i:s', $phpdate );
$phpdate = strtotime( $mysqldate );
echo $mysqldate;


Is there something wrong with it?

I've tried several scripts to get the number of days between two dates, but none of them will work. Obviously I'm doing something wrong. One thing I know that is causing a problem is that I'm using variables instead of an actual typed in date. Which in my opinion is how most people would do it - variables not actual dates.

I've tried these:

$todaydate = date('Y/m/d');
$now = date('Y-m-d');
$dd2 = strtotime($now);
$thisyear = 2019;
$payment_day = 29;
$pay_month = 6;

if($pay_month == 1){$newpaymentmonth = 2;}
if($pay_month == 2){$newpaymentmonth = 3;}
if($pay_month == 3){$newpaymentmonth = 4;}
if($pay_month == 4){$newpaymentmonth = 5;}
if($pay_month == 5){$newpaymentmonth = 6;}
if($pay_month == 6){$newpaymentmonth = 7;}
if($pay_month == 7){$newpaymentmonth = 8;}
if($pay_month == 8){$newpaymentmonth = 9;}
if($pay_month == 9){$newpaymentmonth = 10;}
if($pay_month == 10){$newpaymentmonth = 11;}
if($pay_month == 11){$newpaymentmonth = 12;}
if($pay_month == 12){$newpaymentmonth = 1;}

$threezero = array(4, 6, 9, 11); // months with 30 days

// Deal with months that have only 28 days or 30 days, setting the payment day to accommodate months with fewer days.
if ($payment_day > 28 && $pay_month == 2)
{
    $payment_day = 28;
}
elseif ($payment_day == 31 && in_array($pay_month, $threezero))
{
    $payment_day = 30;
}
else
{
    $payment_day = $payment_day;
}

if ($newpaymentmonth == 1){$newpaymentyear = $thisyear + 1;}else{$newpaymentyear = $thisyear;}
$newpaymentdate = date($newpaymentyear.'-'.$newpaymentmonth.'-'.$payment_day);
echo date("Y-m-d", $newpaymentdate) . "<br><br>";

$ddate = new DateTime($thisyear.'-'.$newpaymentmonth.'-'.$payment_day);
$ddate->add(new DateInterval('P5D'));
echo $ddate->format('Y-m-d') . " - New month payment date with 5 day grace period added.<br><br>";

Now, how to calculate the difference in days between today's date and $ddate? I tried the below, but none worked.
 

function DateDiff($strDate1,$strDate2){
return (strtotime($strDate2) - strtotime($strDate1))/  ( 60 * 60 * 24 );  // 1 day = 60*60*24
}
echo "Date Diff = ".DateDiff($now,$ddate)."<br>";

$timeDiff = abs($now - $ddate);
$numberofdays = $timeDiff/86400;
echo "<p>$numberofdays - days between today's date and payment w/grace date.</p>";

$date1 = $now;
$date2 = $ddate;
$diff = date_diff($date1,$date2);
echo 'Days Count - '.$diff->format("%a");

$date1=$now;
$date2=$ddate;
function dateDiff($date1, $date2)
{
      $date1_ts = strtotime($date1);
      $date2_ts = strtotime($date2);
      $diff = $date1_ts - $date2_ts;
      return round($diff / 86400);
}
$dateDiff= dateDiff($date1, $date2);
printf("Difference between in two dates : " .$dateDiff. " Days ");
print "</br>";

This one returns 18112 Days. Should be days 7 days from 2019-08-05.

None of these work, so I'm doing something wrong. Any ideas?

Thanks

Edited August 9, 2019 by cyberRobot
fixed typo

Hi, I have this code :-

$datestarted = $row['datestarted'];
$datestart=date("l d/m/Y @ H:i:s",strtotime($datestarted)); 

Which is been echo'd like this :-

echo '
<tr align="center"'.$rowbackground.'>
<td>'.$datestart.'</td>							
<td align="center">
';

So for example I get this output :-Sunday 18/04/2021 @ 10:45:26

The data in the DB for the above is stored like this :-2021-04-18 10:45:26

 

What I'd like to do is also echo how many days ago this date was, all of the examples I've tried don't seem to work though?