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PHP - Number Of Days Between Now And A Date In The Future
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)); Similar TutorialsI 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! I want to set a never expire date, so I figured I would set the date 99 years into the future. No dice. After trial and error, it appears you can only make it 25 years into futu date("Y-m-d",strtotime("+25 years")); I wonder if it is the absolute date, or the number 25. If it is the maximum date (i.e. nothing after 2035) then I will have a problem with this code next year! Hi, I have a job listing website which displays the closing date of applications using: $expired_date (This displays a date such as 31st December 2019) I am trying to show a countdown/number of days left until the closing date. I have put this together, but I can't get it to show the number of days.
<?php
$expired_date = get_post_meta( $post->ID, '_job_expires', true );
$hide_expiration = get_post_meta( $post->ID, '_hide_expiration', true );
if(empty($hide_expiration )) {
if(!empty($expired_date)) { ?>
<span><?php echo date_i18n( get_option( 'date_format' ), strtotime( get_post_meta( $post->ID, '_job_expires', true ) ) ) ?></span>
<?php
$datetime1 = new DateTime($expired_date);
$datetime2 = date('d');
$interval = $datetime1->diff($datetime2);
echo $interval->d;
?>
<?php }
}
?>
Can anyone help me with what I have wrong? Many thanks Hi, I am using this function to output a list of dates for every friday over the next 12months. This is working fine. Code: [Select] [php] //function function nextWeeksDay($date_begin,$nbrweek) { $nextweek=array(); for($i = 1; $i <= $nbrweek; $i++) { // 52 week in one year of course $nextweek[$i]=date('Y-m-d', strtotime('+'.$i.' week',$date_begin)); } return $nextweek; } /// end function /// example of a select date // var $date_begin=strtotime('this Friday'); $nbrweek=52; // call function $result=nextWeeksDay($date_begin,$nbrweek); // Insert //$date1=date('Y-m-d', $date_begin); for($i = 1; $i <= $nbrweek; $i++) { //$date=$result[$i]; echo $result[$i]."<br/>"; } [/php] My problem is that when I alter the code to try and get the date for every second Friday for the next year the script freezes. Code: [Select] [php] //function function nextWeeksDay($date_begin,$nbrweek) { $nextweek=array(); for($i = 2; $i <= $nbrweek; $i+2) { // 52 week in one year of course $nextweek[$i]=date('Y-m-d', strtotime('+'.$i.' week',$date_begin)); } return $nextweek; } /// end function /// example of a select date // var $date_begin=strtotime('this Friday'); $nbrweek=52; // call function $result=nextWeeksDay($date_begin,$nbrweek); // Insert //$date1=date('Y-m-d', $date_begin); for($i = 2; $i <= $nbrweek; $i+2) { //$date=$result[$i]; echo $result[$i]."<br/>"; } [/php] I really don't understand why?? Does anyone know why? Thanks Hi, I want to calculate the future month, i have used following code to calculate the date $date = date('Y-m-d',strtotime(date("Y-m-d", strtotime('2010-01-31')) . " +3 month")); It gives me 2010-03-03 But i need to get 2010-02-28 OR just month is enough like 2010-02 Thank you Nikhil I have this block of code that was written by someone years ago :-
<?php
class BoDelivery {
public function EstimatedDays($off, $standard_days, $saturday, $delay) {
$today = date("N"); // Weekday - number 1-7
$now = strtotime("now"); // Unix
$off_array = explode(":", str_replace(".", ":", $off));
$off_unix = mktime($off_array[0], $off_array[1], "00", date("n"), date("j"), date("Y"));
$sending_days_from_now = 0;
if ($now > $off_unix) {
$sending_days_from_now++;
}
$sending_day = $today + $sending_days_from_now;
switch ($sending_day) {
case 6:
$sending_days_from_now++;
$sending_days_from_now++;
break;
case 7:
$sending_days_from_now++;
break;
}
$sending_day = $today + $sending_days_from_now;
// Estimated delivery time
$delivery_days = $standard_days;
$over_weekends = 0;
// Add Delay
if ($delay) {
$delivery_days = $delivery_days + $delay;
}
if ($sending_day == 5 && !$saturday) {
$delivery_days++;
$delivery_days++;
$over_weekends++;
}
$delivery_day = $sending_day + $delivery_days;
switch ($delivery_day) {
case 6:
if (!$saturday) {
$delivery_days++;
$delivery_days++;
$over_weekends++;
}
break;
case 7:
$delivery_days++;
$over_weekends++;
break;
}
if ($over_weekends == 0 && $delivery_days > 5) {
$delivery_days++;
$delivery_days++;
}
$delivery_day = $sending_day + $delivery_days;
switch ($delivery_day) {
case 13:
$delivery_days++;
$delivery_days++;
$over_weekends++;
break;
case 14:
$delivery_days++;
$over_weekends++;
break;
}
$delivery_day = $sending_day + $delivery_days;
$delivery_days = $delivery_day - $today;
return $delivery_days;
}
}
?>
Which is supposed to calculate the number of days for delivery, taking into account weekends & with a delay variable that we can set. It isn't working as expected, may never have worked(??) or maybe down to PHP upgrades. For example, if today (18th)I set the delay to 2 days the delivery estimate is the 21st (which is correct), if I change the delay to 3 OR 4 it becomes the 24th, 5 then becomes the 26th! I can't get my head around the code at all, I wonder if someone could assist in what may be going on or add comments to the code snippets so I can maybe work it out?
Thanks for any help. 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? 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 cyberRobotfixed typo I want to see if a date is more than 10 days overdue. if ($row['duedate'] < "todays date plus 10 days"){ How do I do that? I put in quote sup there in "english" what I want... Hi there, i am using a form with 2 inputs which are equipped with a datepicker: Date 1 & Date 2, is it possible to calculate how many days are there from Date 1 to Date 2 (including the selected ones) ? On my form the dates are in this format: September 08, 2011 (i guess i can change that to numeric only, if that helps) Tamper data shows them getting posted like this: September+14%2C+2011 Any help / hints will be appreciated ! My query is a fixtures list for my local sports team- There seems little point including fixtures from the past as they are in a results query anyway. I'm ordering by date (see below) but how can I remove the ones already past? $query = "SELECT * FROM fix10 ORDER BY Date"; TIA Nick I have a database lets say punch_log The data in the table looks like: --------------------------------------------------------------------------- |punch_log_id | user_id | punch_id | punch_time | --------------------------------------------------------------------------- | 10010 | 21 | 1 | 2010-11-10 15:04:59| | 10011 | 21 | 2 | 2010-11-10 15:50:05| | 10010 | 21 | 1 | 2010-11-11 15:04:59| | 10011 | 21 | 2 | 2010-11-11 15:50:05| | 10010 | 21 | 1 | 2010-11-12 15:04:59| | 10011 | 21 | 2 | 2010-11-12 15:50:05| | 10010 | 21 | 1 | 2010-11-13 15:04:59| | 10011 | 21 | 2 | 2010-11-13 15:50:05| | 10010 | 21 | 1 | 2010-11-14 15:04:59| | 10011 | 21 | 2 | 2010-11-14 15:50:05| | 10010 | 21 | 1 | 2010-11-14 15:50:59| <-- this is why i need this. | 10011 | 21 | 2 | 2010-11-14 15:55:05| <-- this is why i need this. ---------------------------------------------------------------------------- Im currently using : $kust = $_POST['AKuu']; $kuni3 = $_POST['LKuu']; $valitudtootaja = $_POST['TNimi']; mysql_select_db($database, $con); $query2 = "SELECT *, SUM(punch_id) FROM punch_log WHERE user_id = '".$valitudtootaja."' AND punch_time BETWEEN '$kust' AND '$kuni3' AND punch_id ='1' "; $result2 = mysql_query($query2) or die(mysql_error()); while($row = mysql_fetch_array($result2)){ $X = $row['SUM(punch_id)']; When using the current query im getting Workers days at work = 6 it should be 5 but thats why i need some solution to group dates and then sum them. Does anybody have any ideas? hi friends could someone help with this criteria? i want to restrict entry to db if the post value number was entered into db less than 31 days ago?
if($homePhone == '') {
$qry = "SELECT * FROM table WHERE homePhone='$homePhone '";
in db i store time in datetime current timestamp like 2013-07-13 21:19:15
could someone help me out?
Edited by lovephp, 12 July 2014 - 11:36 AM. 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, This one has been driving me up the wall, so hopefully some kind person can help me. I'm trying to make a validation script. $creststartdate comes in from a form as a UK formatted date (d/m/Y), and if $creststartdate is more than a month ahead of today, then it gets rejected. here's the code I have now... Code: [Select] $today = date("m/d/y"); $onemonth = strtotime ('+1 month', strtotime($today)); $nextmonth = date ('d/m/Y', $onemonth); $csd = date("m/d/y", $creststartdate); $strcsd = strtotime($csd); $newdate = date ('d/m/Y', $strcsd); if ($newdate > $nextmonth) { $creststartdatefailed = "The Crest Start Date cannot be more than 1 month in the future."; $creststartdatevalid = "NO"; } As it stands, this version of the code means that nothing is getting rejected. Please help?? Cheers 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>"; 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 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... I have date stored in database in any of the given forms 2020-06-01, 2020-05-01 or 2019-04-01 I want to compare the old date with current date 2020-06-14 And the result should be in days. Any help please? PS: I want to do it on php side. but if its possible to do on database side (I am using myslq) please share both ways🙂 Edited June 14, 2020 by 684425Hi all, I am trying to figure out how to calculate 5 working days prior to a given date. I have done some googling but can only see examples of how to add 5 working days onto a date, such as this: Code: [Select] $holidayList = array(); $j = $i = 1; while($i <= 5) { $day = strftime("%A",strtotime("+$j day")); $tmp = strftime("%d-%m-%Y",strtotime("+$j day")); if($day != "Sunday" and $day != "Saturday" and !in_array($tmp, $holidayList)) { $i = $i + 1; $j = $j + 1; } else $j = $j + 1; } $j = $j -1; echo strftime("%A, %d-%m-%Y",strtotime("+$j day")); Does anyone know how to calculate 5 working days prior to a date? Many thanks, Greens85 |
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