PHP - Calculate Absent Days

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 !

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

Hi there,

for every case we have "gooo" as commen in all
how this could be one code using anything like elseif or switch whatever...

if($refere && (stripos($r, $refere) === false)){ 
echo "x1"
}else{
echo"goooo"
}

And

if($limited && ($line[hits] >= $limited)){ 
echo "x2"
}else{
echo"goooo"
}

And

if($pword && (isset($_POST['password']) && $_POST['password'] == $pword) ){
echo"goooo"
}else{
echo"x3"
}

And

if ($line[capt] == 1 && ($_SESSION["security_code"]) ){
echo"goooo"
}else{
echo"x4"
}


thanks in advance

This topic has been moved to PHP Regex.

http://www.phpfreaks.com/forums/index.php?topic=320501.0

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.

Hey there,

Thanks for taking the time to read my thread.

My issue is if I'm given a time stamp in PHP how could I calculate the number of days until that time stamp?.

Thanks for your time.

Hello. I want to generate the initial of the day of all the days of a chosen month. Meaning, if I choose this month it will give me:

TFSSMTWTFSSMTWTFSSMTWTFSSMTWTF

Is that clear? Can anybody help me? thanks!

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.

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 have db table that records the days (Sunday, Monday...) when the employee login to the system. Say that the employee logged in on Monday then Logged in on Wednesday so this means he was absent on Tuesday. I calculated the number of days to get the answer 2 (between Wednesday and Monday before logging in on Wednesday) how to get the name of the missing day "Tuesday"?

code:

$curdate = date("Y-m-d");
$currday = date('l');
$lastdate = $row['indate'];
$lastdayin = $row['lastdayin'];
					
if ($lastdate !=Null)
{
						
	$misseddates = strtotime($curdate) - strtotime($lastdate);
	$misseddates = $misseddates / 86400;
	echo $misseddates;
						
	$misseddays = strtotime($lastdayin) - strtotime($currday);
	$misseddays = $misseddays / 86400;
	echo $misseddays;
}

I tried to get the name of day in the last step but it only calculates 5 in numbers how to get the name. I want the answer to be "Tuesday" as of the example.

Thanks.

Hello,

I am trying to put together a mysql query that will return the number of visitors for four days ago, what i am trying to do is plot the last seven days visitors on a graph in the format of day seven, day six, etc and need to find away to get a count for each of those days.  At the moment i am thinking about running various queries with each returning the results for a specific day.  The code below is supposed to get the count of visitors four days ago.  Not between now and four days ago but just for the 24 hour period which covers day 4.  The current code just returns a value of 0.   Any help would be appreciated.

Code: [Select]
$result = mysql_query("SELECT ip FROM ip_stats WHERE  date= date_sub(NOW(), interval 4 DAY)");
$num_rows = mysql_num_rows($result);
echo "$num_rows";

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 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.

please could some one help me i need to limit the query by timestamp for the last 7 days here is my code on the php side

Code: [Select]
<?php
 
             $db = @mysql_connect("localhost", "8conv", "*****") or die("Connection Error: " . mysql_error());
             mysql_select_db("tempmonitor") or die("Error connecting to db.");
             $sql = "SELECT *, UNIX_TIMESTAMP(datetime) AS datetime FROM 8conv";
             $result = mysql_query($sql);
             $data = array();
             while ($row = mysql_fetch_array($result)) {
                $temp1[] = array($row['datetime'] * 1000, (int)$row['temp1']);;

             }
            
 $temp1 = json_encode($temp1);