I have an undefined variable in my PHP - php

Question

I am very new to PHP and have been attempting to integrate with mysqli. Apparently on line 19 of my code the variable shown is undefined but as far as I can tell I defined it.
Here is the code. I've look around but I can't really find something to isolate this.
<?php
include("connect.php");
$error = "";
if(isset($_POST['submit']))
{
$characterName = $_POST['fname'];
$email = $_POST['email'];
$password = $_POST['password'];
$passwordConfirm = $_POST['passwordConfirm'];
$image = $_FILES['image']['name'];
$tmp_image = $_FILES['image']['tmp_name'];
$imageSize = $_FILES['image']['size'];
}
if(strlen($fname) < 3)
{
$error = "Character name is too short";
}
else if(!filter_var($email, FILTER_VALIDATE_EMAIL))
{
$error = "Please enter a valid email address";
}
else if(strlen($password) < 8)
{
$error = "Password must be more than 8 characters";
}
else if($password === $passwordConfirm)
{
$error = "Password does not match";
}
else if($image = "")
{
$error = "Please upload an Avatar";
}
else
{
$error = "You have successfully registered";
}
?>
Form Code:
<form method="post" action="index.php" enctype="multipart/form-data">
<label>Character Name:</label><br />
<input type="text" name="fname" /><br /><br />
<label>Email:</label><br />
<input type="text" name="email" /> <br /><br />
<label>Password:</label><br />
<input type="password" name="password" /><br /><br />
<label>Reenter Password:</label><br />
<input type="password" name="passwordConfirm" /><br /><br />
<label>Send us an Avatar:</label><br />
<input type="file" name="image" /><br /><br />
<input type="submit" name="submit" value="submit" />
</form>

Answer 1

if(strlen($fname) < 3) {
$error = "Character name is too short";
}
Here you have the error, $fname is not defined. What you there meaning is $_POST['fname'];. Which you stored in $characterName so change it to:
if(strlen($characterName) < 3){
$error = "Character name is too short";
}
Anyway, cause you define your variables only if isset($_POST['submit']), the lines below will fail if it is not set. Here is a example how it would work.
$_POST['submit'] is only defined if you call it with post parameters (formular, ajax..), so if you directly open the php file it wont work. I added a few comments to make it clear.
<?php
include("connect.php");
$error = "";
if(isset($_POST['submit'])) {
//If this block of variable declaration failed it wouldn´t define the variables
$characterName = $_POST['fname'];
$email = $_POST['email'];
$password = $_POST['password'];
$passwordConfirm = $_POST['passwordConfirm'];
$image = $_FILES['image']['name'];
$tmp_image = $_FILES['image']['tmp_name'];
$imageSize = $_FILES['image']['size'];
//So we led Php only check the variables if a submit is provided
if(strlen($characterName) < 3) {
$error = "Character name is too short";
} else if(!filter_var($email, FILTER_VALIDATE_EMAIL)) {
$error = "Please enter a valid email address";
} else if(strlen($password) < 8) {
$error = "Password must be more than 8 characters";
} else if($password === $passwordConfirm) {
$error = "Password does not match";
} else if($image = "") {
$error = "Please upload an Avatar";
} else {
$error = "You have successfully registered";
}
} else {
//If there is no submit we land here
$error = "No data provided";
}
?>

Answer 2

if submit is not posted then only you receive undefined variables error. In order to avoid those errors. You just change your code like this
if(isset($_POST['submit']))
{
$characterName = $_POST['fname'];
$email = $_POST['email'];
$password = $_POST['password'];
$passwordConfirm = $_POST['passwordConfirm'];
$image = $_FILES['image']['name'];
$tmp_image = $_FILES['image']['tmp_name'];
$imageSize = $_FILES['image']['size'];
}
else
{
$error ="Your submit is not posted.";
exit(); //Without it would again trigger the undefined variables.
}