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Hello. I was doing a login form that when the login is successful, it redirect in a welcome page. The problem is: in the welcome page, there is a email space, but it give me this error: Notice: Undefined index: email in C:\xampp\htdocs\RegistrationLogin\header1.php on line 8.

Here is the code:
<?php
include_once('link.php');
session_start();
Line 8 $email = $_SESSION['email'];
?>

<nav class="navbar navbar-default">
	<div class="container-fluid">
		<div class="navbar-header">
			<a href="#" class="navbar-brand">Registration Login</a>
		</div>
		<div class="dropdown navbar-right" id="right">
  <button class="btn btn-primary dropdown-toggle" type="button" data-toggle="dropdown"><?php echo $email;?>
  <span class="caret"></span></button>
  <ul class="dropdown-menu">
  	<li><a href="account.php">Account</a></li>
    <li><a href="logout.php">Logout</a></li>
  </ul>
</div>
	</div>
</nav>


My welcome page gives another error Notice: Undefined index: id in C:\xampp\htdocs\RegistrationLogin\welcome.php on line 10

Here is the code:
<?php
include_once('link.php');
include_once('header1.php');
require_once('connection.php');

Line 10 $id = $_SESSION['id'];
$nome = $username = $email = $gender = '';
$sql = "SELECT * FROM registrationdb WHERE ID='$id'";
$result = mysqli_query($conn, $sql);
if(mysqli_num_rows($result) == 0)
{
	while($row = mysqli_fetch_assoc($result))
	{
		$nome = $row["Nome"];
		$username = $row["Username"];
		$email = $row["Email"];
		$gender = $row["Gender"];
	}
}

?>
<div class="jumbotron">
	<center>
		<h1>Welcome <?php echo $nome." ".$username; ?></h1>
	</center>
</div>


What I have tried:

I tried to change the name of the variables
Posted
Updated 5-Jul-19 12:04pm
v3

1 solution

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Solution 1

Quote:
Notice: Undefined index: email in C:\xampp\htdocs\RegistrationLogin\header1.php on line 8.

This means that email does not exist in $_SESSION.
Use the debugger to check which variable/index exist or not, we can't do it for you with the little information we have.

Your code do not behave the way you expect, or you don't understand why !

There is an almost universal solution: Run your code on debugger step by step, inspect variables.
The debugger is here to show you what your code is doing and your task is to compare with what it should do.
There is no magic in the debugger, it don't know what your code is supposed to do, it don't find bugs, it just help you to by showing you what is going on. When the code don't do what is expected, you are close to a bug.
To see what your code is doing: Just set a breakpoint and see your code performing, the debugger allow you to execute lines 1 by 1 and to inspect variables as it execute.

Debugger - Wikipedia, the free encyclopedia[^]

Mastering Debugging in Visual Studio 2010 - A Beginner's Guide[^]
Basic Debugging with Visual Studio 2010 - YouTube[^]

phpdbg | php debugger[^]
Debugging techniques for PHP programmers[^]

The debugger is here to only show you what your code is doing and your task is to compare with what it should do.

-----
$sql = "SELECT * FROM registrationdb WHERE ID='$id'";

Not necessary a solution to your question, but another problem you have.
Never build an SQL query by concatenating strings. Sooner or later, you will do it with user inputs, and this opens door to a vulnerability named "SQL injection", it is dangerous for your database and error prone.
A single quote in a name and your program crash. If a user input a name like "Brian O'Conner" can crash your app, it is an SQL injection vulnerability, and the crash is the least of the problems, a malicious user input and it is promoted to SQL commands with all credentials.
SQL injection - Wikipedia[^]
SQL Injection[^]
SQL Injection Attacks by Example[^]
PHP: SQL Injection - Manual[^]
SQL Injection Prevention Cheat Sheet - OWASP[^]
How can I explain SQL injection without technical jargon? - Information Security Stack Exchange[^]
   

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