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PHP - Display Number Of Days Between A Date And Today's Date
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 Similar TutorialsI'm looking for a simple little code to display today's date, month, day, year and countdown to 365 days. Can anyone please help. 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)); (continuing from topic title) So if I set a date of July 7 2011 into my script, hard coded in, I would like the current date to be checked against the hard coded date, and return true if the current date is within a week leading up to the hard coded date. How could I go about doing this easily? I've been researching dates in php but I can't seem to work out the best way to achieve what I'm after. Cheers Denno Hi guys, I'm putting together a small event system where I want the user to add his own date and time into a textfield (I'll probably make this a series of drop-downs/a date picker later). This is then stored as a timestamp - "0000-00-00 00:00:00" which displays fine until I try to echo it out as a UK date in this format - jS F Y, which just gives today's date but not the inputted date. Here's the code I have right now: Code: [Select] $result = mysql_query("SELECT * FROM stuff.events ORDER BY eventdate ASC"); echo "<br />"; echo mysql_result($result, $i, 'eventvenue'); echo ", "; $dt = new DateTime($eventdate); echo $dt->format("jS F Y"); In my mysql table eventdate is set up as follows: field - eventdate type - timestamp length/values - blank default - current_timestamp collation - blank attributes - on update CURRENT_TIMESTAMP null - blank auto_increment - blank Any help as to why this could be happening would be much appreciated, thanks. I want to see if a date is more than 10 days overdue. if ($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 ! 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 Is there a way of getting today's date (in European format - day-month-year) into the body of an email sent via phpmailer? Many thanks. is it possible to do something like Code: [Select] $today = date("Y-m-d"); $result = mysql_query("SELECT * FROM staff where date = '.$today.' "); also, is it normal for the first entry in the database not to be displayed? i have 6 entries in a table and only 2-6 are shown. when i changed the id for 1 to 7, it only displayed 3-7. still shows all records in the database... any idea how i go about this? Code: [Select] $date= date('y-m-d'); $query=mysql_query("SELECT * FROM listing WHERE date >= $date") or die (mysql_error()); Hellow, i need help please, writing code and it doesn't work. please help...
Here it is
WHERE start_date BETWEEN 'start_date".strtotime('-3 day')."' AND 'start_date'";
without this code everithing works fine
Thank you
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>"; 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 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! Hi... Good day! I have table that has a field from_date and to_date. Now I just want to know if how can I display as table format the dates between from_date to_date. Like this from_date: 2011-12-16 to_date: 2011-12-31 I want to display it: 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 // table format. Thank you 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 Hi I am trying to add a field to a database that is 4 days from the date the record is added, but it is not adding a value Code: [Select] $end_date=strtotime("+ 4 days"); $add_vehicle_sql=mysql_query("INSERT INTO `tbl_auction_lot`(`cust_id`,`reserve`,`make`,`model`,`spec`,`fuel`,`doors`,`mot_date`,`fns`,`fos`,`rns`,`ros`,`condition`,`reg_no`,`service_history`,`sale_type`,`status`,`keepers`,`gearbox`,`emissions`,`colour`,`date_first_reg`,`date_manufacture`,`bhp`,`engine_size`,`end_date`) VALUES ('$seller_id','$reserve','$make','$model','$body_style','$fuel_type','$no_of_doors','$mot','$fns','$fos','$rns','$ros','$vehicle_condition','$vrm','$service_history','auction','$status','$prev_keepers','$gearbox','$emissions','$colour','$date_reg','$date_man','$bhp','$engine_size','$end_date')") or die(mysql_error()); What am I doing wrong and what is there a better way to achieve the desired result. |
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