Prepared statement fetch row returns nothing - php

Question

I want to fetch this row and save it into $notescheck, but when I try to do this the $notescheck is empty when I want to echo and there are no errors. With non-prepared statements it works fine.
Code:
if($user_ok == true) {
$sql = "SELECT notescheck FROM users WHERE username=? LIMIT 1";
$stmt = $conn->prepare($sql);
$stmt->bind_param("s",$log_username);
$stmt->execute();
$row = $stmt->fetch();
$notescheck = $row[0];
$stmt->close();
}
With non-prepared statement it would look like this:
if($user_ok == true) {
$sql = "SELECT notescheck FROM users WHERE username='$log_username' LIMIT 1";
$query = mysqli_query($conn, $sql);
$row = mysqli_fetch_row($query);
$notescheck = $row[0];
mysqli_close($conn);
}

Answer 1

This isn't how fetch() works with prepared statements, you're not fetching an array like you think you are. You also need to bind the result of the select into variables, then use those to display. If there are multiple records, you'd use a while($stmt->fetch){ echo $notescheck };
if($user_ok == true) {
$sql = "SELECT notescheck FROM users WHERE username=? LIMIT 1";
$stmt = $conn->prepare($sql);
$stmt->bind_param("s",$log_username);
$stmt->execute();
$stmt->bind_result($notescheck);
$stmt->fetch();
$stmt->close();
}
echo $notescheck;
You should check into reading this:
http://php.net/manual/en/mysqli-stmt.fetch.php
Multiple records matching username=x would look like this:
if($user_ok == true) {
$sql = "SELECT notescheck FROM users WHERE username=? LIMIT 1";
$stmt = $conn->prepare($sql);
$stmt->bind_param("s",$log_username);
$stmt->execute();
$stmt->bind_result($notescheck);
$stmt->store_result()
while($stmt->fetch()){
echo $notescheck;
}
$stmt->close();
}