I have this error shown in the browser: Notice: Undefined variable: link in C:\xampp\htdocs\xxxx\login.php on line 40

Question

Hi!
The browser shows the errors below in php after clicking login. so it fails to go to home page
It shows the following errors

Connected successfully
Notice: Undefined variable: link in C:\xampp\htdocs\social\login.php on line 40

Warning: mysqli_prepare() expects parameter 1 to be mysqli, null given in C:\xampp\htdocs\social\login.php on line 40

The code below:

<?php
// Initialize the session
session_start();

// Check if the user is already logged in, if yes then redirect him to Home page
if(isset($_SESSION["loggedin"]) && $_SESSION["loggedin"] === true){
header("location: home.php");
exit;
}

// Include config file
require_once "connect.php";

// Define variables and initialize with empty values
$username = $password = "";
$username_err = $password_err = "";

// Processing form data when form is submitted
if($_SERVER["REQUEST_METHOD"] == "POST"){

// Check if username is empty
if(empty(trim($_POST["UserName"]))){
$username_err = "Please enter username.";
} else{
$username = trim($_POST["UserName"]);
}

// Check if password is empty
if(empty(trim($_POST["Password"]))){
$password_err = "Please enter your Password.";
}
$password = trim($_POST["Password"]);
}

// Validate credentials
if(empty($username_err) && empty($password_err)){
// Prepare a select statement
$sql = "SELECT id, UserName, Password FROM members WHERE username = ?";

if($stmt = mysqli_prepare($link, $sql)){ //line 40 here
echo mysqli_error($conn);
// Bind variables to the prepared statement as parameters
mysqli_stmt_bind_param($stmt, "s", $param_username);

// Set parameters
$param_username = $username;

// Attempt to execute the prepared statement
if(mysqli_stmt_execute($stmt)){
// Store result

Answer 1

Stephen Phiri
Change this line to this at line 40

if ($stmt = $mysqli->prepare($sql))

to

if ($stmt = mysqli_prepare($link, $sql))

where $link is the variable of connection file where connection code is stored in $link

Answer 2

Notice: Undefined variable

Happens when you try to use a variable that wasn't previously defined.

Answer 3

People usually use $link or $conn as name for variables that contains a database connection object. In your code I see both are used. In `mysqli_prepare` you use $link, and in `mysqli_error` you use $conn.

You need to decide, which of these two ($link and $conn) is the actual and valid database connection object to use.

Considering the code is only partial copy, pasted as raw text, and truncated - It would rather be difficult to assist you.

I would suggest you to save the code in SoloLearn and share the code link instead of raw text like that (notice that it is also truncated due to character limits).

If your code is more than 10-15 lines it is best to save and share the link. Follow the below guide to sharing links, in case you didn't know how

https://www.sololearn.com/post/74857/?ref=app

Answer 4

I removed $link because I didn't included it but I don't know about this.

Warning: mysqli_prepare() expects exactly 2 parameters, 1 given in C:\xampp\htdocs\xxxx\login.php on line 40