PHP - Php $row['date'] + 30 Days

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));

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!

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 684425

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

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

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

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

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 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.

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

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!

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...

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.