OOP mysqli database functions - php

Question

I am a beginner in OOP in PHP and I am trying to play with my code. So first thing I wanted to do is to write one basic class with connect and insert functions for database. So my desired situation is this:
-I wanna create on class which will controll connect and insert functions. The problem is, my $connect variable isn't working from another function, so what could I do to make that possible?
You will understand more by the code provided.
<?php
class DB {
protected $dbhost = 'localhost';
protected $dbuser = 'root';
protected $dbpass = '';
protected $dbname = 'newdb';
public function connect()
{
$connect = new mysqli($this->dbhost, $this->dbuser, $this->dbpass, $this->dbname);
if($connect->error)
{
echo "Failed to connect";
}
else
{
echo "connected";
}
}
public function insert($name, $second)
{
$insert = "INSERT INTO posts (name, second) VALUES ('$name', '$second')";
if ($connect->query($insert) === TRUE)
{
echo "New record created successfully";
}
else
{
echo "Error: " . $sql . "<br>" . mysqli_connect_error();
}
}
}
require_once 'classes/DB.php';
$db = new DB;
if(isset($_POST['submit']))
{
$name = $_POST['name'];
$second = $_POST['second'];
$db->insert($name, $second);
}
?>
<form method="POST">
Add smth<input type="text" name="name"><br>
Also omg<input type="text" name="second"><br>
<input type="submit" name="submit">
</form>

Answer 1

Add $connect as a property of your class, so you may reuse it everywhere within the class using $this->connect:
class DB
{
protected $dbhost = 'localhost';
protected $dbuser = 'root';
protected $dbpass = '';
protected $dbname = 'newdb';
protected $connect;
public function connect()
{
$this->connect = new mysqli($this->dbhost, $this->dbuser, $this->dbpass, $this->dbname);
if ($this->connect->error)
{
echo "Failed to connect";
}
else
{
echo "connected";
}
}
public function insert($name, $second)
{
$insert = "INSERT INTO posts (name, second) VALUES ('$name', '$second')";
if ($this->connect->query($insert) === TRUE)
{
echo "New record created successfully";
}
else
{
echo "Error: " . $sql . "<br>" . mysqli_connect_error();
}
}
}

Answer 2

I have rewritten your class for a bit, so you can insert your database settings in the constructor. So it really is a object, and you can use multiple databases. What you did wrong in your class was using the variable $connect without declaring it into a variable usable in the whole class.
The class
<?php
class DB {
protected $dbhost;
protected $dbuser;
protected $dbpass;
protected $dbname;
protected $connection;
public function __construct($dbhost, $dbuser, $dbpass, $dbname)
{
$this->dbhost = $dbhost;
$this->dbuser = $dbuser;
$this->dbpass = $dbpass;
$this->dbname = $dbname;
$connection = new mysqli($this->dbhost, $this->dbuser, $this->dbpass, $this->dbname);
if($connection->error)
die('Could not connect with the database!');
$this->connection = $connection;
}
public function connect()
{
$this->__construct($this->dbhost, $this->dbuser, $this->dbpass, $this->dbname);
}
public function insert($name, $second)
{
$insert = "INSERT INTO posts (name, second) VALUES ('$name', '$second')";
if ($this->connection->query($insert) === TRUE)
{
echo "New record created successfully";
}
else
{
echo "Error: " . $sql . "<br>" . mysqli_connect_error();
}
}
public function getConnection()
{
return $this->connection;
}
}
Usage
$db = new DB('localhost', 'root', '', 'newdb');
$db->insert('name', 'second');
I hope this helps you in your journal through OOP, sorry for my bad English.