Can't echo out something after a MySqli query? - php

Question

Today, I have a question regarding PHP.
I know to be fact based on my setup that this function is being called. However, only the first requested 'echo' is being displayed (I used "cats", long story...).
I know that a lot of times this is something simple like a missed end quote, but my PHP doesn't throw me any errors. Anyting I'm doing wrong? I can't get anything after my code.
Before I do this:
$connection = new mysqli('localhost', 'root', 'root', 'VV_root') or die("There was an error connecting; " . mysqli_connect_error);
Then this is my function.
function knock_on() {
echo("cats");
// It is asssumed that from here, all information is to be kept as secure as possible in the transactions
// Get all the ingredients
$username = $_POST['username'];
$password = $_POST['password'];
// Now, we have all the ingredients
// Mix password !IMPORTANT!
// Now, run the query to see if we get any matches.
$query = $connection->query("select * from user where username = " . $username . " & password = " . $password) or die("Something went wrong");
echo("cats2");
}

Answer 1

You are running into scope issues. The $connection variable is not available inside your knock_on() function. Pass the $connection as a parameter to your function definition.
Like this..
function knock_on($connection) { //Forget the global keyword thing.

Answer 2

Firs of all turn on error displaying and then you have an error in your query:
$query = $connection->query("select * from user where username = " . $username . " & password = " . $password) or die("Something went wrong");
Here you need to use AND operator not an & as you did:
username = " . $username . " AND password = " . $password
Abnd the second : $connection variable is not available inside your knock_on() function as
Shankar Damodaran mentioned.

Answer 3

Note this code is not tested.
$connection = new mysqli('localhost', 'root', 'root', 'VV_root') or die("There was an error connecting; " . mysqli_connect_error);
function knock_on() {
global $connection;
echo("cats");
// It is asssumed that from here, all information is to be kept as secure as possible in the transactions
// Get all the ingredients
$username = $_POST['username'];
$password = $_POST['password'];
// Now, we have all the ingredients
// Mix password !IMPORTANT!
// Now, run the query to see if we get any matches.
$query = $connection->query("select * from user where username = " . $username . " & password = " . $password) or die("Something went wrong");
if($query) {
return true;
} else {
return false;
}
}

Answer 4

There were several things that I had to fix, so here's my new working code
$connection = new mysqli("localhost", "root", "root", "VV_root");
The actual function:
function knock_on() {
echo("cats");
// It is asssumed that from here, all information is to be kept as secure as possible in the transactions
// Get all the ingredients
$username = $_POST['username'];
$password = $_POST['password'];
// Now, we have all the ingredients
// Mix password !IMPORTANT!
// Now, run the query to see if we get any matches.
if ($connection) {
$query = $connection->query("select * from user where username = '" . $username . "' and password = '" . $password . "'") or die("Something went wrong");
}
else {
echo($mysqli->connect_error);
}
echo("cats2");
}