PHP WHy this does not work?

Question

HI,

This doesnt work. It said it coudnt send the information( of course i wrote that) but it doesnt . I dont see why .

<?php include_once("scripts/global.php") ?>
<?php
$message = '';
if(isset($_POST['username'])){

    $username = $_POST['username'];
    $fname = $_POST['fname'];
    $lname = $_POST['lname'];
    $email = $_POST['email'];
    $pass1 = $_POST['pass1'];
    $pass2 = $_POST['pass2'];

    //error handeling
    if((!$username) || (!$fname) || (!$lname) || (!$email) || (!$pass1) || (!$pass2)){
        $message = 'Please insert all fields in the form below!';
    }else{
        if($pass1 != $pass2){
            $message = 'Your password fields do not match!';
        }else{ 
            //securing the data
            $username = preg_replace("#[^0-9a-z]#i","",$username);
            $fname = preg_replace("#[^0-9a-z]#i","",$fname);
            $lname = preg_replace("#[^0-9a-z]#i","",$lname);
            $pass1 = sha1($pass1);

            $email = mysqli_real_escape_string($con, $email);

            //check for dublicates
            $user_query = mysqli_query($con, "SELECT username FROM members WHERE username='$username'LIMIT 1") or die("Could not chekc username");
            $count_username = mysqli_num_rows($user_query);

            $email_query = mysqli_query($con, "SELECT email FROM members WHERE email='$email'LIMIT 1") or die("Could not chekc username");
            $count_email = mysqli_num_rows($email_query);

            if($count_username > 0){
                $message = 'Your username is allready in use';
            }else if($count_email > 0){
                $message = 'Your email is allready in use!';
            }else{
                //insert members
                $ip_address = $_SERVER['REMOTE_ADDR'];
                $query = mysqli_query($con, "INSERT INTO members (username,firstname,lastname,email,password,ip_address,sign_up_date)VALUES('$username','$fname','$lname','$email','$pass1',$ip_address',now())") or die("Could not insert your information");
                $member_id = mysqli_insert_id();
                mkdir("users/$member_id",0755);
                $message = 'You have now been registered';

            }
        }

    }
}

?>
<!DOCTYPE html>
<html>
    <head>
        <meta charset="utf-8" />
        <title>Membership website</title>
        <link rel="stylesheet" type="text/css" href="css/style.css">
    </head>
    <body>
        <div class="container center">
            <h1>Register to my site b filling the fields below!</h1>
            <p><?php echo("$message"); ?></p>
            <form action="register.php" method="post">
                <input type="text" name="username" placeholder="Username" /><br />
                <input type="text" name="fname" placeholder="Firstname" /><br />
                <input type="text" name="lname" placeholder="Lastname" /><br />
                <input type="text" name="email" placeholder="Email Address" /><br />
                <input type="password" name="pass1" placeholder="Password" /><br />
                <input type="password" name="pass2" placeholder="Validate Password" /><br />

                <input type="submit" value="Register!" />

            </form>
        </div><!--/ container -->


    </body>
</html>

connection:

<?php

$con = mysqli_connect("localhost","root","root","membership") or die("Could not connect to server!");


?>

global.php:

<?php
session_start();
include_once("connect.php");

ini_set('display_startup_errors',1);
ini_set('display_errors',1);
error_reporting(-1);

?>

Answer 1 · 2015-04-07T12:54:46Z

April 7, 2015 12:54pm

Has mysqli not got a function where you can pull out the specific error message?

I would first investigate that so you can narrow down what's wrong!

Answer 2 · 2015-04-11T08:02:03Z

April 11, 2015 8:02am

It sounds like you're not setting $con properly to me or there is a mysql syntax error?

<?php

$query = "SELECT username FROM members WHERE username='$username'LIMIT 1";

// add a space between $username' and LIMIT.
// add curly braces around variables in strings.

$query =  "SELECT username FROM members WHERE username = '{$username}' LIMIT 1";

What happens if you:

<?php

var_dump($con);
exit;

?>

inside of your post block?

<?php

if(isset($_POST['username'])){

Is it set properly?

Also, to improve the readability of your code, I would suggest this:

<?php

$user_query = mysqli_query($con, "SELECT username FROM members WHERE username='$username'LIMIT 1");

if ( ! $user_query) {
    die("Could not check username");
}

Otherwise you're going to end up with some incredibly long lines!

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Konrad Pilch

Konrad Pilch

PHP WHy this does not work?

2 Answers

thomascawthorn

thomascawthorn

Konrad Pilch

Konrad Pilch

thomascawthorn

thomascawthorn

Konrad Pilch

Konrad Pilch

thomascawthorn

thomascawthorn

Konrad Pilch

Konrad Pilch

Konrad Pilch

Konrad Pilch

Konrad Pilch

Konrad Pilch