PHP - Anyway To Make Count(*) Return 0?

Is it possible to count how many returns there are on a if statement?

Where you can see
Code: [Select]
echo 'yes';
it will echo this twice according to my condition, all I want is the the value 2 that it has returned twice... ?

Code: [Select]
              while($row = mysql_fetch_array($bookings))
              {           
                  if(time() > $row['returnby'] && $row['returned']==0)
                  {
                      $books_overdue = count($row['returned']);                 
                  }
                  if(time() > $row['returnby'] && $row['returned']==1)
                  {
                      $returned_ontime = count($row['returned']);                 
                  }

                  if(time() <= $row['returnby'])
                  {
                    echo 'yes';                     
                  }               
              }

thanks for help.

Hi guys,

I need your help. I am trying to insert the rows in the mysql database as I input the values in the url bar which it would be like this:

Code: [Select]
www.mysite.com/testupdate.php?user=tester&pass=test&user1=tester&email=me@shitmail.com&ip=myisp

However i have got a error which i don't know how to fix it.


Error: Column count doesn't match value count at row 1


<?php
session_start();
    define('DB_HOST', 'localhost');
    define('DB_USER', 'mydbusername');
    define('DB_PASSWORD', 'mydbpassword');
    define('DB_DATABASE', 'mydbname');

$errmsg_arr = array();
    $errflag = false;

    $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);
    if(!$link) {

die('Failed to connect to server: ' . mysql_error());
    }

    $db = mysql_select_db(DB_DATABASE);
    if(!$db) {

die("Unable to select database");
    }

   function clean($var){

return mysql_real_escape_string(strip_tags($var));
    }
  $username = clean($_GET['user']);
    $password = clean($_GET['pass']);
    $adduser = clean($_GET['user1']);
    $email = clean($_GET['email']);
    $IP = clean($_GET['ip']);
    if($username == '') {
  $errmsg_arr[] = 'username is missing';
  $errflag = true;
    }
    if($password == '') {
  $errmsg_arr[] = 'PASSWORD is missing';
  $errflag = true;
    }
    if($errflag) {
  $_SESSION['ERRMSG_ARR'] = $errmsg_arr;
  echo implode('<br />',$errmsg_arr);
   }
   else {
   $sql = "INSERT INTO `members` (`username`,`email`,`IP`) VALUES ('$adduser','$email','$IP')";

if (!mysql_query($sql,$link))
  {
  die('Error: ' . mysql_error());
  }
echo "The information have been updated.";
  }
?>


Here's the name of the columns i have got in my database:

Code: [Select]
username
email
IP


I have input the correct columns names, so I can't correct the problem I am getting.

Please can you help?

This topic has been moved to MySQL Help.

http://www.phpfreaks.com/forums/index.php?topic=331562.0

Why am I getting this error when there are 3 Fields with 3 values?

Column count doesn't match value count at row 1

Code: [Select]
$sql5="INSERT INTO participants (participant_name, team_no, sport)
VALUES ('".implode("','$_POST[team]'),('",$_POST['participant_name'])."','$_POST[team]','$sport')";