Trying to write a function that displays data in an SQL database - php

Question

Trying to write a function that displays the id, and filepath of all public content in an SQL database. Here's what I have so far, but it failed to run.
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "pricosha";
// Create connection
$connection = mysql_connect($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
//Added two extra variables, username and content_name
$query = "SELECT id, username, file_path, content_name FROM Content";
$result = mysql_query($query);
//Loop through the results of the query
while($row = mysql_fetch_array($result)){
echo "ID: " . $row["id"]. " Username: " . $row["username"]. " File Path: " . $row["file_path"]. " Content Name: " . $row["content_name"]. "<br>";
}
$conn->close();
?>

Answer 1

Your variable for create the connection to mysql is $connection, while you use $conn for this :
$conn->connect_error
and this :
$conn->close();
Now try to change $conn to $connection or just change this :
$connection = mysql_connect($servername, $username, $password, $dbname);
to this :
$conn = mysql_connect($servername, $username, $password, $dbname);
So far now, the function mysql_connect, mysql_query, and others have been deprecated. You need to change it to mysqli I think. Try this.
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "pricosha";
$connection = new mysqli($servername, $username, $password, $dbname) or die(mysqli_errno());
$query = "SELECT id, username, file_path, content_name FROM Content";
$result = mysqli_query($connection, $query);
while($row = mysqli_fetch_assoc($result)){
echo "ID: " . $row["id"]. " Username: " . $row["username"]. " File Path: " . $row["file_path"]. " Content Name: " . $row["content_name"]. "<br>";
}
?>

Answer 2

Try this
$conn = mysql_connect($servername, $username, $password, $dbname);