PHP - Check Db If A Phone Number Was Added To Db Less Than 31 Days Ago?

I know this should be easy.... However, to validate a 7-digit phone number, I wrote this:

Code: [Select]
<?php
if(!empty($_POST['mobile_number'])) {
          if(!is_numeric($_POST['mobile_number']) || strlen($_POST['mobile_number']) != 7) {
            $errors[] = 'Please enter a 7-digit number without spaces or dashes for your <strong>Mobile</strong>.';
          } else {
            $_SESSION['mobile_number'] = $_POST['mobile_number'];
          }
      } else {
        $_SESSION['mobile_number'] = '';
      }
... but now I realise that numeric/integer validation won't work because where I am - in Ireland - mobile numbers can start with an initial zero, too. So, the following mobile numbers are possible:
0123456 0224466
How can I ensure that:
each digit in the mobile number is numeric an initial zero is possible there are exactly 7 digits
I suppose I loop through each digit, ensuring that that digit is numeric and break out of the loop if a non-numeric digit is encountered..?
Any suggestions?

TIA

Hi,

 I want to make the phone number in the below code as clickable. Below is the code php and mysql code I used.

 

    <?php  
    $connect = mysqli_connect('localhost', 'root', '1234', 'contact');  
    $query ='SELECT * FROM tbl_employee ORDER BY ID DESC';  
    $result = mysqli_query($connect, $query); 
    ?>

   

<table id="employee_data" class="table table-striped table-bordered">
                <thead>  
                <tr>  
                <td>Name</td>  
                <td>Location</td>  
                <td>Department</td>  
                <td>Phone Number</td>  
                </tr>  
                </thead>  
                          <?php  
                          while($row = mysqli_fetch_array($result))  
                          {  
                               echo '  
                               <tr>  
                                    <td>'.$row["name"].'</td>  
                                    <td>'.$row["location"].'</td>    
                                    <td>'.$row["department"].'</td>  
                                    <td>'.$row['phone_number'].'</td>
                  
                               </tr>  
                               ';  
                          }  
                          ?>  
</table>

 

Hi All,

I guess there must be someway to do this with regex, but I'm not too hot on regex.

What I want to do is check a text area for the presence of a phone number, and if it is there remove it. Ideally I don't want to remove all number as there might be a need for them to use a number, for example... I am 25 years old would be fine, but entering their number wouldn't be.

Is there anyway of tacking this with regex, or is it going to be tricky?

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 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 cyberRobot
fixed typo

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));

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,

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

Hello,

I'm working on an error form, and I want to check if the user inserted a number into the name field, and if so return an error.

I thought this might of worked, but it doesn't:

Code: [Select]
if(is_int($_POST['name'])){
     $arrErrors['name'] = 'Invalid name, remove numbers.';
}
Many thanks

Hi Guys just need some advice really

I would like to know if there is a way I can check if a number is in a times table

for example

if(3 is in 4 times table){ do bla bla }
the result for above will be false as 4 times table does not contain 3

if(12 is in 4 times table){ do bla bla }
the above would be true

Thanks guys

Hi,

I'm totally new to PHP and I'm afraid I don't understand a lot yet. However, I need to make this: when you fill in a number in a form, it should be displayed whether the number is odd or even. I tried the following code, which is obviously not working and I have no idea on how to fix this.

Code: [Select]
<?php 
$var = $_POST['var'];

for ($i = 0; $i < $var; $i++) {
  if ($i % 2 == 0) {
    print("even");
  } else {
   print("odd");
  }
}
?>

Hello,
I'm currently creating a little script, and want to assign a random number to a row in the MySQL table as soon as it will be inserted. But that number may not have been used before. How will I do this?

I tried if else thingies, but then the code would be very long (just in case it randomly picks 10 times after each other a number that exists).

$randomnumber = rand(10000000, 99999999);
if(mysql_num_rows(mysql_query("SELECT * FROM table WHERE randomnumber='".$randomnumber."')) == 1) {
  $randomnumber = rand(10000000, 99999999);
  if(mysql_num_rows(mysql_query("SELECT * FROM table WHERE randomnumber='".$randomnumber."')) == 1) {
    $randomnumber = rand(10000000, 99999999);
  }else {
   // Insert into database
  }
}else {

// Insert into database
}


I hope you understand my problem.

Regards,

I was wondering if anyone knows a good way to check for an even number of each character in a string.  For example if the string is "abbacc" then it will return true as there are two of each character.  However if the string is "ababc" it will return false as there is only 1 c.  Thanks for any help

Hi,

How can I check if a variable only has these characters:
- number (0-9)
- decimal point (.)
- positive (+)
- negative (-)
- dollar ($)

I tried is_numeric, but it doesn't like negative values...

I need to add the variable value to another variable. I have to accept numbers (including negative and decimals)...

ex:
0.1 = TRUE
-.1 = TRUE
12 = TRUE
1000.00 = TRUE
12a = FALSE



thanks