After validation name doesnt post to database???

[mikebyrne]

Once my code runs through the validation process ok and posts all the data to the database apart from the name field. I cant figure out why though??

 

My code is

<?php
require_once("adminconnect.php");

$tbl_name="adminusers";


$name = $_POST['name'];
$address = $_POST['address'];
$address1 = $_POST['address1'];
$address2 = $_POST['address2'];
$address3 = $_POST['address3'];
$address4 = $_POST['address4'];
$county = $_POST['county'];
$zip = $_POST['zip'];
$telephone = $_POST['telephone'];
$email = $_POST['email'];
$password =$_POST['password'];
$username = $_POST['username'];
$num =$_POST ['num'];


if($_POST["action"] == "signup"){

        $valid=1;

                if ($_POST['name']=="") {
			echo 'got no name<br>';
                $valid=0;
                $style_name = "background-color:#FF5959";
                $error_name = "Your name seems to be mising?<br>";
                                }

			if ($address == "" || strlen($address) < 2) {
    			echo 'got no address1<br>';//added by me to denote failure in this statement 
  		 		$valid=0;
   				$style_address = "background-color:#FF5959";
  				$error_address = "There is a problem with the address field?<br>";
							}


			if ($address1 == "" || strlen($address1) < 2) {
    			echo 'got no address1<br>';//added by me to denote failure in this statement 
  		 		$valid=0;
   				$style_address = "background-color:#FF5959";
  				$error_address1 = "There is a problem with the address field?<br>";
							}

			if ($address2 == "" || strlen($address2) < 2) {
    			echo 'got no address2<br>';//added by me to denote failure in this statement 
  		 		$valid=0;
   				$style_address = "background-color:#FF5959";
  				$error_address2 = "There is a problem with the address field?<br>";
							}

			if ($address3 == "" || strlen($address3) < 2) {
    			echo 'got no address3<br>';//added by me to denote failure in this statement 
  		 		$valid=0;
   				$style_address = "background-color:#FF5959";
  				$error_address3 = "There is a problem with the address field?<br>";
							}

			if ($address4 == "" || strlen($address4) < 2) {
    			echo 'got no address4<br>';//added by me to denote failure in this statement 
  		 		$valid=0;
   				$style_address = "background-color:#FF5959";
  				$error_address4 = "There is a problem with the address field?<br>";
							}
			if ($county == "" || strlen($county)<2) {
			echo 'got no county<br>';
                $valid=0;
                $style_county = "background-color:#FF5959";
                $error_county = "The County field is blank?<br>";
        						} 
			if ($zip == "" || strlen($zip)<2) {
			echo 'got no zip<br>';
                $valid=0;
                $style_zip = "background-color:#FF5959";
                $error_zip = "Theres a problem with the zip code?<br>";
        						}   
			if (!eregi("^[0-9]+",$telephone)) {
			echo 'got no phone<br>';
                $valid=0;
                $style_telephone = "background-color:#FF5959";
                $error_telephone = "Theres a problem with the telephone number?<br>";
        						}   
			  if (!eregi("^[A-Za-z0-9.-]+",$email)) {
			echo 'got no mail';
                $valid=0;
                $style_email = "background-color:#FF5959";
                $error_email = "Theres a problem with the email address?<br>";
       						 	}
        
        		if ($password == "" || strlen($password)<7) {
			echo 'got no password';
                $valid=0;
                $style_password = "background-color:#FF5959";
                $error_password = "Theres a problems with your password?<br>";
       							 }      


$user = mysql_real_escape_string(htmlspecialchars($_POST['username']));


$sql = "SELECT name FROM adminusers WHERE username ='$user'";
echo ".$user.";
$result = mysql_query($sql) or die("Error in SQL: ".mysql_error());
$row = mysql_fetch_array($result);
$count = mysql_num_rows($result);

$name = $row['name'];


			if ($count > 0) { // username should only exist once.
			$valid=0;
			$style_username = "background-color:#FF5959";
			$error_username = "Error! The username " . $user . " already exists in the database.";
							}

			if ($password == "" || strlen($password)<7) {
                $valid=0;
                $style_password = "background-color:#FF5959";
                $error_password = "Theres a problems with your password?<br>";

        						}
if ($valid==1) {       
                $sql="INSERT INTO $tbl_name(name, address, address1, address2, address3, address4, county, zip, telephone, email, username, password, usertype)VALUES('$name', '$address', '$address1', '$address2','$address3', '$address4','$county' ,'$zip', '$telephone', '$email', '$username', '$password' , 2)";
$result=mysql_query($sql)or die(mysql_error()."<p>With Query<br>$sql");
}
?>

 

The sql for the database is

 

CREATE TABLE `adminusers` (

  `name` varchar(255) collate latin1_general_ci default NULL,

  `address` varchar(255) collate latin1_general_ci default NULL,

  `address1` varchar(255) collate latin1_general_ci default NULL,

  `address2` varchar(255) collate latin1_general_ci default NULL,

  `address3` varchar(255) collate latin1_general_ci default NULL,

  `address4` varchar(255) collate latin1_general_ci default NULL,

  `county` varchar(255) collate latin1_general_ci default NULL,

  `zip` varchar(255) collate latin1_general_ci default NULL,

  `telephone` decimal(10,0) default NULL,

  `email` varchar(255) collate latin1_general_ci default NULL,

  `username` varchar(255) collate latin1_general_ci default NULL,

  `password` varchar(255) collate latin1_general_ci default NULL,

  `usertype` decimal(10,0) default NULL

) ENGINE=MyISAM DEFAULT CHARSET=latin1 COLLATE=latin1_general_ci;

 

 

[revraz]

echo the query and see how it looks

[mikebyrne]

if ($valid==1) {       
                $sql="INSERT INTO $tbl_name(name, address, address1, address2, address3, address4, county, zip, telephone, email, username, password, usertype)VALUES('$name', '$address', '$address1', '$address2','$address3', '$address4','$county' ,'$zip', '$telephone', '$email', '$username', '$password' , 2)";
echo $result;

$result=mysql_query($sql)or die(mysql_error()."<p>With Query<br>$sql");
}

 

The echo seems to be Resource id #4

[revraz]

echo $sql

[mikebyrne]

INSERT INTO adminusers(name, address, address1, address2, address3, address4, county, zip, telephone, email, username, password, usertype)VALUES('', 'kjsfksjdisjdsklj', 'aixasixuxiau', 'iduwiuewieu','iudisudisuci', 'isucdiuvdifuiof','idudiufdifueiof' ,'ocdicodicsodisoi', '92839483984390', 'oskfodjfoefkodvop', 'bisgosowo', 'okcodfodfosi' , 2)

 

seems to be leaving name out

[revraz]

You use $name twice.

 

Once here at the top

 

$name = $_POST['name'];

 

Then again here

 

$name = $row['name'];

 

Looks like you are overwriting it.

[mikebyrne]

Yeah I see what you mean

 

$name = $row['name']; is used as part of this validation

 

$user = mysql_real_escape_string(htmlspecialchars($_POST['username']));


$sql = "SELECT name FROM adminusers WHERE username ='$user'";
echo ".$user.";
$result = mysql_query($sql) or die("Error in SQL: ".mysql_error());
$row = mysql_fetch_array($result);
$count = mysql_num_rows($result);

$name = $row['name'];


			if ($count > 0) { // username should only exist once.
			$valid=0;
			$style_username = "background-color:#FF5959";
			$error_username = "Error! The username " . $user . " already exists in the database.";
							}

 

Would you recommend I change the varible names? If so, how would you word it?

[mikebyrne]

I've changed it to

 

$test = $row['name'];

 

and that seems to have done the trick