Why does my Register form still not input data into my MySQL tables

Question

I'm trying to link a register page with my login page to allow new users to register an account to use on my application. I've linked the form up to my tbl_Users table on MySQL in which holds all the information that the users would input into this form. I've properly set everything up using queries and such and the form displays properly at the very least. However when I click submit, the page just refreshed with the fields now empty again and no new data within my table on the database. Where am I going wrong? (Extra-note: I'm still in the process of coding in the safety code to prevent sql-injections)

ConnectorCode.php

<?php

 $conn = mysqli_connect("localhost", "b4014107", "Win1", "b4014107_db2") or die (mysqli_connect_error());

 ?>

Register.php

<?
error_reporting(E_ALL);
ini_set('display_errors', 1);
session_start();
include('ConnectorCode.php');

if(isset($_POST['submit'])) { 
$FName = $_POST['First_Name'];
$LName = $_POST['Last_Name'];
$Email = $_POST['Email'];
$UName = $_POST['User_Name'];
$Password  = $_POST['Password']; 

$FName = mysqli_real_escape_string($conn, $FName);
$LName = mysqli_real_escape_string($conn, $LName);
$Email = mysqli_real_escape_string($conn, $Email);
$UName = mysqli_real_escape_string($conn, $UName);
$Password = mysqli_real_escape_string($conn, $Password);

$sql = "SELECT Email FROM tbl_Users WHERE Email='$Email'";
$result = mysqli_query($conn, $sql);
$row = mysqli_fetch_array($result, MYSQLI_ASSOC);

if(mysqli_num_rows($result) == 1)
{
    echo "Sorry, the email you are trying to enter already exists";
}
else
{
    $query = mysqli_query($conn, "INSERT INTO tbl_Users(First_Name, Last_Name, Email, User_Name, Password) VALUES ('$FName', '$LName', '$Email', '$UName', '$Password')");
if($query)
{
 echo "Thank you for registering";
}
header('Location: Index.php');
}

}
?>

<!DOCTYPE HTML>
<head>
<title>Register</title>
</head>
<body>

<h1> Register Page </h1>

<p> Please fill in the form to register <p>

<form method="post" action="">
<fieldset>
First Name: <br />
<input name="First_Name" type="text" class="input" size="25" required /> <br /> <br />
 Last Name:  <br />
 <input name="Last_Name" type="text" class="input" size="25" required /> <br /> <br />
Email: <br />
<input name="Email" type="email" class="input" size="25" required /> <br /> <br />
Username: <br />
<input name="User_Name" type="text" class="input" size"25" required /> <br /> <br />
Password: <br />
<input name="Password" type="password" class="input" size="25" required /> <br /> <br/>
<input type="submit" name="submit" value="Register!" />
</fieldset>
</form>
</body>
</html>

first of `session_start();` need to be on top. second form code requird? — Alive to die - Anant, Apr 01 '16 at 14:44
can you add the code of your form, and please post the error it shows you (if there is any) — Unex, Apr 01 '16 at 14:45
The form code is there, it's at the bottom of the code sheet. — Henry Green, Apr 01 '16 at 14:46
possible duplicate of [PHP: Notice: Undefined variable and Notice: Undefined index](http://stackoverflow.com/questions/4261133/php-notice-undefined-variable-and-notice-undefined-index) — Funk Forty Niner, Apr 01 '16 at 14:47
Oh god I'm such an idiot! Let me check if that'll make it work — Henry Green, Apr 01 '16 at 14:48
one more suggestion is never ever check submit button value directly check posted values. — Alive to die - Anant, Apr 01 '16 at 14:49
I've changed the code to suit your recommendations but I'm still getting the same issue. I've updated the code sheet on to show what I've edited to. — Henry Green, Apr 01 '16 at 14:50
Like I said earlier... this `if(isser($_POST['Submit'])) {...}` will never fire up. Undefined index notices is what you failed to tell us what you were getting with error reporting. — Funk Forty Niner, Apr 01 '16 at 14:51
Man @Fred-ii- telling you again and again problem in this `isset($_POST['submit != Submit'])`. change it to `isset($_POST['submit'])` — Alive to die - Anant, Apr 01 '16 at 14:53
That's the thing though. I had error reporting set up and I was getting no errors being displayed. — Henry Green, Apr 01 '16 at 14:53
So instead of `isset($_POST['submit'])` I'd want it to check the values directly. Something like `isset($_POST['VALUES'])`? — Henry Green, Apr 01 '16 at 14:58
@Fred-ii- Even then, I'm still getting the same issue. The page just refreshes and the database doesn't have any updated rows in that table. — Henry Green, Apr 01 '16 at 15:16
Ooop... my bad. I thought you used what Anant said and what I suggested earlier. `if(isset($_POST['VALUES'])) {` that needs to read as `if(isset($_POST['submit'])) {` to match the input's `name="submit"` — Funk Forty Niner, Apr 01 '16 at 15:21
@Fred-ii- I just updated the code on the post, so is it like that? — Henry Green, Apr 01 '16 at 15:24
@Fred-ii- I'm still getting no updated rows in the table, even with the `if(isset($_POST['submit'])) {` line — Henry Green, Apr 01 '16 at 15:26
@KyleE4K Even with the `exit;` it doesn't return to the index.php — Henry Green, Apr 01 '16 at 15:29
`$sql = "SELECT Email FROM tbl_Users WHERE Email='$Email'";` ??? You're querying the same thing you're returning ??? this is just going to return what `$email` holds — Jaquarh, Apr 01 '16 at 15:31
@KyleE4K On my understanding it was set to check if the email was already part of the table, meaning it cannot be reused. So it's checking if the email being submitted matches the email in any of the field of the email column and would then tell the user to try a different email. — Henry Green, Apr 01 '16 at 15:34
This is a really bad infrastructure, I'll re-write a working verison... — Jaquarh, Apr 01 '16 at 15:36

score 0 · Answer 1 · answered Apr 01 '16 at 15:22

0

You do not meet the condition isset($_POST['VALUES']) because you don't have field with name="VALUES".

Change

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

to

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

answered Apr 01 '16 at 15:22

Akairis

407
2
11

I've changed it but I'm still getting no new rows in my table on the database. – Henry Green Apr 01 '16 at 15:27
No! That's what's so confusing, no error messages are displayed on the top of the screen, all that happens is the field refresh and no new data is entered into the table. – Henry Green Apr 01 '16 at 15:35

Jaquarh · Answer 2 · 2016-04-01T15:52:22.553

0

You had a lot missing...

<?php

error_reporting(E_ALL);
ini_set('display_errors', 1);
session_start();
require_once 'ConnectorCode.php';

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

    $FName = mysqli_real_escape_string($conn, $_POST['First_Name']);
    $LName = mysqli_real_escape_string($conn, $_POST['Last_Name']);
    $Email = mysqli_real_escape_string($conn, $_POST['Email']);
    $UName = mysqli_real_escape_string($conn, $_POST['User_Name']);
    $Password = mysqli_real_escape_string($conn, $_POST['Password']); // why did you need to repeat this all twice?

    $sql = "SELECT * FROM tbl_Users WHERE Email='$Email'"; // ? didn't understand why you was asking for the Email using the Email...
    $result = $conn->query($sql);
    if(count($result) == 0)
    {
         $insert_sql = "INSERT INTO tbl_Users (First_Name,Last_Name,Email,User_Name,Password) VALUES ('$FName','$LName','$Email','$UName','$Password')";
         if($conn->query($insert_sql)) // You forgot to query this
         {
             echo "Thank you for registering";
             header('Location: index.php'); // lowercase i
             exit; // you forgot this
         }
    }
    else
    {
        echo "Sorry, that email already exists!";
    }
    $conn->close(); // you forgot this

}

?>

Hope this helps...

edited Apr 01 '16 at 15:52

answered Apr 01 '16 at 15:42

Jaquarh

6,493
7
34
86

Thank you a lot! But now everytime I load up the php file This pops up at the top of the page `query($query)) { echo "Thank you for registering"; header('Location: index.php'); exit; } } else { echo "Sorry, that email already exists!"; } $conn->close(); } ?> ` – Henry Green Apr 01 '16 at 15:51
Try that now, edited. Anyway, I'm finishing work now so that should fix your issue. Good luck. – Jaquarh Apr 01 '16 at 15:51
Now it's showing this `query($sql); if(count($result) == 0) { $insert_sql = "INSERT INTO tbl_Users (First_Name,Last_Name,Email,User_Name,Password) VALUES ('$FName','$LName','$Email','$UName','$Password')"; if($conn->query($insert_sql)) { echo "Thank you for registering"; header('Location: index.php'); exit; } } else { echo "Sorry, that email already exists!"; } $conn->close(); } ?> ` – Henry Green Apr 01 '16 at 15:54
It's the -> syntax I think, it doesn't seem to like it. – Henry Green Apr 01 '16 at 15:55
What PHP version are you running? It may be that $result returns a boolean rather than the query data, just use the depreciated `mysqli_query` rather than the `->` and it should all work. Good luck – Jaquarh Apr 01 '16 at 15:57
It's forming an error properly now, it's showing the echo message `"Sorry, that email already exists!"` to any entry now/ – Henry Green Apr 01 '16 at 15:59
By the way how can I convert this line into the mysqli because I'm not sure how to `if($conn->query($insert_sql)) ` – Henry Green Apr 01 '16 at 16:03
`if(mysqli_query($conn, $insert_sql))` – Jaquarh Apr 04 '16 at 08:05

Why does my Register form still not input data into my MySQL tables

2 Answers2