PHP - Calculate Number Of Days Between 2 Dates
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 Similar TutorialsHello 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? 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, 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? 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 ramiwahdanmistake in dates Hi 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 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 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)); 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 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 = $row['SUM(punch_id)']; When using the current query im getting Workers days at work = 6 it should be 5 but thats why i need some solution to group dates and then sum them. Does anybody have any ideas? Hi guys, I am new to PHP/coding and am trying to look for 1. A way of comparing the words in one static array against other dynamically created arrays (created from mysql queries) 2. Work out how many similar words there are - then assign that number to that array My static array is...$comparewithme = array with values = this is an example of an array Mysql_query("select id, words from table_example") Results from query are put into an array that is named according to id.. $result2 = array with values = this is an example of queries $result3 = array with values = this is not an example of php Comparison should give the following info Comparing $comparewithme with $result2 should generate a hit rate of 5 (similar words=this is an example of)... Comparing $comparewithme with $result3 should generate a hit rate of 4 (similar words=this is an example)... Any ideas greatly appreciated...thanks in advance 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 ! 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 All, I need to subtract dates and display the number of days left. I have a 'Start' date and an 'End' date in DATETIME format in the DB. Not quite sure where to start. A simply start - end doesn't work . Start = 2011-11-01-00:00:00 End = 2011-11-30-23:59:59 Since it is now 2011-11-27, my output should equal 3. Any help is appreciated. I'm getting the dreaded " Invalid parameter number: number of bound variables does not match number of tokens" error and I've looked at this for days. Here is what my table looks like:
| id | int(4) | NO | PRI | NULL | auto_increment | | user_id | int(4) | NO | | NULL | | | recipient | varchar(30) | NO | | NULL | | | subject | varchar(25) | YES | | NULL | | | cc_email | varchar(30) | YES | | NULL | | | reply | varchar(20) | YES | | NULL | | | location | varchar(50) | YES | | NULL | | | stationery | varchar(40) | YES | | NULL | | | ink_color | varchar(12) | YES | | NULL | | | fontchosen | varchar(30) | YES | | NULL | | | message | varchar(500) | NO | | NULL | | | attachment | varchar(40) | YES | | NULL | | | messageDate | datetime | YES | | NULL |Here are my params: $params = array( ':user_id' => $userid, ':recipient' => $this->message_vars['recipient'], ':subject' => $this->message_vars['subject'], ':cc_email' => $this->message_vars['cc_email'], ':reply' => $this->message_vars['reply'], ':location' => $this->message_vars['location'], ':stationery' => $this->message_vars['stationery'], ':ink_color' => $this->message_vars['ink_color'], ':fontchosen' => $this->message_vars['fontchosen'], ':message' => $messageInput, ':attachment' => $this->message_vars['attachment'], ':messageDate' => $date );Here is my sql: $sql = "INSERT INTO messages (user_id,recipient, subject, cc_email, reply, location,stationery, ink_color, fontchosen, message,attachment) VALUES( $userid, :recipient, :subject, :cc_email, :reply, :location, :stationery, :ink_color, :fontchosen, $messageInput, :attachment, $date);"; And lastly, here is how I am calling it: $dbh = parent::$dbh; $dbh->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_WARNING); if (empty($dbh)) return false; $stmt = $dbh->prepare($sql); $stmt->execute($params) or die(print_r($stmt->errorInfo(), true)); if (!$stmt) { print_r($dbh->errorInfo()); }I know my userid is valid and and the date is set above (I've echo'd these out to make sure). Since the id is auto_increment, I do not put that in my sql (though I've tried that too), nor in my params (tried that too). What am I missing? I feel certain it is something small, but I have spent days checking commas, semi-colons and spelling. Can anyone see what I'm doing wrong? Hi guys, I am trying to do a multidates events availability calender. The script below indicates todays date by highlighting an orange colour and also indicates the start and end date of the event highlighting grey colour on the two dates (The colour are link via css classes as shown). Code: [Select] //Today's date $todaysDate = date("d/m/Y"); $dateToCompare = $daystring . '/' . $monthstring . '/' . $year; echo "<td align='center' "; if($todaysDate == $dateToCompare){ echo "class='today'"; }else{ //Compare's the event dates $sqlcount = "select event_start,event_end from b_calender where event_start ='".$dateToCompare."' AND event_end='".$dateToCompare."'"; $noOfEvent = mysql_num_rows(mysql_query($sqlcount)); if($noOfEvent >= 1){ echo "class='event'"; } } It works ok i.e. if start date = 01/01/2012 and end date = 04/01/2012 both date will be highlighted with grey colour. However I want it to also highlight grey on the dates between the 1st and 4th to show that then anydates between the 1st and 4th are not available and this is when I'm stuck. Please guys I need help. Thanks Hello,
I have problem durring binding update query. I can't find what is causing problem.
public function Update(Entry $e) { try { $query = "update entry set string = $e->string,delimiter=$e->delimiter where entryid= $e->id"; $stmt = $this->db->mysqli->prepare($query); $stmt->bind_param('ssi',$e->string,$e->delimiter,$e->id); $stmt->close(); } catch(Exception $ex) { print 'Error: ' .$ex->getMessage(); } }When I run function update I'm getting next error:Warning: mysqli_stmt::bind_param(): Number of variables doesn't match number of parameters in prepared statement Can you help me to solve this problem ? Edited by danchi, 17 October 2014 - 10:25 AM. I need to display a number(the number is retrieved from the db) in the form input field such that only the last 4 digits is visbile, the remaining can be masked as * or X or whatever is applicable. I know the last 4 can be obtained as follows: Code: [Select] $number=substr($number,-4,4); But when i hit the submit button the form validates the input field and checks if the number is a valid number of a specific format. Therefore when I click on the submit button then I should still be able to unmask the masked numbers or do something similar that would help me validate the whole number. Code: [Select] <input type="text" name="no" value="<?php if(!empty($number)){ echo $number;} ?>"> I'm trying to figure out if today's current date and time falls between Mon & Fri, but so far, I can't get it to work. Can anybody look at my code and see what I'm doing wrong? Code: [Select] <?php $day_start = date('D h:i a', strtotime("Mon 05:30")); $day_end = date('D h:i a', strtotime("Fri 10:00")); $day_current = date('D h:i a', strtotime("+1 hours")); if (($day_current > $day_start) && ($day_current < $day_end)) { echo "yup"; } else { echo "nope"; } ?> Thanks in advance When I try to add 30 days: Code: [Select] $date = date("Y-m-d"); $date = strtotime(date("Y-m-d", strtotime($date)) . " +30 days"); echo $date; and I echo date I get 1330664400 How do I get it to echo out 3/1/2012? I know the answer lies in the strtotime but I can't figure it out. I know it's a simple problem for most of you... 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. |