Problem with my form validations?

The code below is where my form goes to validate the information on it. I need to be sure that the form is completely filled along with unique email address and player name. Any help would be appriciated thank you

<?php

$dbserver = "xxxxxx";
$dblogin = "xxxxxx";
$dbpassword = "xxxxxxxx";
$dbname = "xxxxxxxx";

$pname=$_POST['pname'];
$pword=$_POST['pword'];
$email=$_POST['email'];
$fname=$_POST['fname'];
$lname=$_POST['lname'];
$country=$_POST['country'];
$referrals=$_POST['referrals'];        

//if no message entered and no playername entered print an error 
if (empty($pname)){
print "No Player Name was entered. <br>Please include your Requested in game Name.<br>";
}
//if no password entered send print an error 
elseif (empty($pword)){
print "No password was entered.<br>Please include a password.<br>";
}
//if no password entered send print an error 
elseif (empty($fname)){
print "No First Name was entered.<br>Please include your First Name.<br>";
}
//if no password entered send print an error 
elseif (empty($lname)){
print "No Last Name was entered.<br>Please include your Last Name.<br>";
}
//if no password entered send print an error 
elseif (empty($country)){
print "No Country was entered.<br>Please include Your Country.<br>";
}
//if no email entered send print an error 
elseif (empty($email)){
print "No email address was entered.<br>Please include your email. <br>";
}
//if the form has been completely filled out continue 
else {

    $con = mysql_connect("$dbserver","$dblogin","$dbpassword");
if (!$con)
  {
  die('Could not connect to the mySQL server please contact gangwars.mofiki.com technical support with the following information: ' . mysql_error());
  } 
  
mysql_select_db("$dbname", $con);


$result = mysql_query("SELECT pname FROM users WHERE pname = '$pname'");
$num_rows = mysql_num_rows($result);

if ($num_rows > 0){echo "Username $pname exists please press back and try again";}

$result = mysql_query("SELECT email FROM users WHERE email = '$email'");
$num_rows = mysql_num_rows($result);

if ($num_rows > 0){echo "Email Address $pname has already been used please press back and try again";}


$sql = mysql_query("INSERT INTO users (playerid, playername, password, email, firstname, lastname, country, referrals) VALUES ('','$_POST[pname]','$_POST[pword]','$_POST[email]','$_POST[fname]','$_POST[lname]','$_POST[country]','$_POST[referrals]')");

if (!mysql_query($sql,$con))
  {
  die('Error: ' . mysql_error());
  }
echo "Thank you for registering you will recieve an email at each phase of Gang Wars release";
  
mysql_close($con); 
} 
?>

Code (markup):

 

Okay I have changed:

$result = mysql_query("SELECT pname FROM users WHERE pname = '$pname'");
$num_rows = mysql_num_rows($result);

if ($num_rows > 0){echo "Username $pname exists please press back and try again";}

$result = mysql_query("SELECT email FROM users WHERE email = '$email'");
$num_rows = mysql_num_rows($result);

if ($num_rows > 0){echo "Email Address $pname has already been used please press back and try again";}

Code (markup):

to:

if(mysql_num_rows(mysql_query("SELECT pname FROM users WHERE pname = '$pname'")))
{
echo "Player Name $pname has already been used please press back and try again";
}

if(mysql_num_rows(mysql_query("SELECT pname FROM users WHERE pname = '$email'")))
{
echo "The Email Address $email has already been used please press back and try again";
}

Code (markup):

Still not working please help

 

What's not working about it?

foreach ($_POST as $name => value) {
    if (strlen($value)==0)
        $errors .= "{name} was left blank!<br />";
}
//check for existing email address and username and add to $errors
if ($errors=="") {
    //retrieve the data, add to db, output success message
} else {
    echo "<b><font color=\"red\">There were errors in your request!</font></b><br/>{$errors}";
}

Code (markup):

 

Error is:

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource on line 54

line 54 is:

$num_rows = mysql_num_rows($result);

If you have time to explain line by line the code you have posted above I would be interested in learning it (i'm a noob as php still I just know the basics)

Thank you

 

You get that error because there's something wrong with the queries or the database. Make sure the database name, table name, user, pass, and server are valid. Additionally, add ";" inside each of your queries (i.e. mysql_query("SELECT * FROM `table`;"). Additionally, at the end of each query you can do something like
$result = mysql_query("query here") or die(mysql_error());

 

Thank you so much for your time I will review all of my query's and see where i'm having the problem at least know I have an idea of where to look.

Thank you again

 

right now your just saying where pname = $pname not the value stored in $pname;


try

mysql_query("SELECT pname FROM users WHERE pname = '{$pname}'");

Code (markup):

when using double quotes " " around your query the php parser will recognize the curly brackets {} and replace it with the contents of the variable.

 

so would this be correct? or is there a better way to write it?

if(mysql_num_rows(mysql_query("SELECT pname FROM users WHERE pname = '{$pname}'")))
{
echo "Player Name $pname has already been used please press back and try again";
}

Code (markup):