"Notice: Undefined variable" in PHP. How can I fix it? [duplicate] - php

Question

This question already has answers here:
mysqli_prepare() expects parameter 1 to be mysqli
(3 answers)
Closed 1 year ago.
I have a registration form here. I am a dummy in PHP (this is PHP for an Android app). It worked, but I found that I can register with the same username and email, so I added functions to check the database for the same username and prevent that, as I am dummy, I get this error when trying to register -
Notice: Undefined variable: con in /storage/ssd1/448/5907448/public_html/Register.php on line 27
Warning: mysqli_prepare() expects parameter 1 to be mysqli, null given in /storage/ssd1/448/5907448/public_html/Register.php on line 27
Warning: mysqli_stmt_bind_param() expects parameter 1 to be mysqli_stmt, null given in /storage/ssd1/448/5907448/public_html/Register.php on line 28
Warning: mysqli_stmt_execute() expects parameter 1 to be mysqli_stmt, null given in /storage/ssd1/448/5907448/public_html/Register.php on line 29
{"success":true}
My PHP code
<?php
$response = array();
if (!isset($_POST["username"], $_POST["email"], $_POST["password"])) {
$response['success'] = false;
$response['Error'] = "No needed data";
echo json_encode($response);
exit(0);
}
ob_start();
$con = mysqli_connect("host", "username", "password", "database");
ob_end_clean();
if (!$con) {
$response['success'] = false;
$response['Error'] = "Error Connecting" . PHP_EOL;
$response['Error'] .= "Error Code: " . mysqli_connect_errno() . PHP_EOL;
$response['Error'] .= "Error: " . mysqli_connect_error() . PHP_EOL;
echo json_encode($response);
exit(0);
}
function registerUser() {
$username = $_POST["username"];
$email = $_POST["email"];
$password = $_POST["password"];
$statement = mysqli_prepare($con, "INSERT INTO user (username, email, password) VALUES (?, ?, ?)");
mysqli_stmt_bind_param($statement, "sss", $username, $email, $password);
mysqli_stmt_execute($statement);
}
function usernameAvailable() {
global $con, $username;
$statement = mysqli_prepare($con, "SELECT * FROM user WHERE username = ?");
mysqli_stmt_bind_param($statement, "s", $username);
mysqli_stmt_execute($statement);
mysqli_stmt_store_result($statement);
$count = mysqli_stmt_num_rows($statement);
mysqli_stmt_close($statement);
if ($count < 1){
return true;
}
else {
return false;
}
}
$response["success"] = false;
if (usernameAvailable()){
registerUser();
$response["success"] = true;
}
echo json_encode($response);
?>
PS: Connection data changed to default.

Answer 1

Add your variable $con to the parameters of your functions that need access to the database i.e. registerUser() and usernameAvailable()

Answer 2

You referenced a global, $con, and it’s returning undefined. Ensure that it has a value by debugging it using print_r or other alternatives. Due to it returning undefined no functions are running.
Although it is bad practise to reference a global variable, I would recommend either setting up a function that returns the $con variable or passing it.
function returnCon(){
$con = mysqli_connect("host", "username", "password", "database");
return $con
}
You can then use
$con = returnCon();