php mysql prepare statement not working [duplicate] - php

Question

This question already has answers here:
Fatal error: Call to a member function prepare() on a non-object in
(2 answers)
Closed 6 years ago.
I'm currently learning php and mysql and am trying to build an authentication webpage where the user registers and is able to log in to a member protected page. The registration process works fine but for some reason I'm getting this error in my login execution script.
Fatal error: Call to a member function prepare() on a non-object in /homepages/8/d459264879/htdocs/tymbi_reg/login_exec.php on line 40
Line 40 is here if($stmt = $mysqli->prepare("SELECT * FROM member WHERE username=? AND password =?"))
I've been trying hard to find where the problem is without any success.
This is the code that I'm using in my login script
<?php
session_start();
require_once('connection.php');
$errmsg_arr = array();
$errflag = false;
$username = $_POST['username'];
$password = $_POST['password'];
if($username == '') {
$errmsg_arr[] = 'Username missing';
$errflag = true;
}
if($password == '') {
$errmsg_arr[] = 'Password missing';
$errflag = true;
}
if($errflag) {
$_SESSION['ERRMSG_ARR'] = $errmsg_arr;
session_write_close();
header("location: index.php");
exit();
} else {
if($stmt = $mysqli->prepare("SELECT * FROM member WHERE username=? AND password =?"))
{
$stmt->bind_param("ss", $username, $password);
$stmt->execute();
$stmt->store_results();
if($stmt->num_rows > 0)
{
$SESSION['username'] = $username;
header("Location: home.php");
}
else
{
$error['alert'] = "Username or password are incorrect";
}
}
}
?>
Here's my connection.php code
<?php
$con=new mysqli("test","test","test","test");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
?>
And yes, I have replaced the test values with my details

Answer 1

make sure you have $mysqli defined properly
$mysqli = new mysqli('host','user','pass','database_name');

Answer 2

You haven't defined $mysqli - your connection. That is why it's not working.
Unless it's defined in connection.php. Could you try var_dump $mysqli and see what you get?

Answer 3

$link = mysqli_connect("localhost", "user", "my_password", "world");
/* check connection */
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
this helps in reading the connection error.
or you can also use
try
{
//your connection
}catch(Exception $e)
{
echo $e->getMessage();
}

Answer 4

I would recommend going all-in on using the Mysqli object and not mixing in the procedural functions. You would have caught your error earlier if you had written the connection logic like so:
$con = new mysqli("host","user","pass","db");
if ($mysqli->errno)
{
echo "Failed to connect to MySQL: " . $mysqli->error;
}
The OP declared the connection as $con and later tried to access it as $mysqli. mysqli_connect_errno() reports that $con is ok, but $mysqli->errno will fail because $mysqli is not an object at that point, because he named the the connection $con not $mysqli