OOP PHP Database results to Array - php

Question

I am new to the idea of oop php and i am trying to write an irc php.
What I'm trying to do:
I am trying to query my database, get results from my database and put it into an array inside my program.
I tried making a new function to carry out the task and called it in the __construct function.
I have shortened the code but it pretty much looks like this:
Any thoughts and ideas are much appreciated.
class IRCBot
{
public $array = array();
public $servername = "localhost";
public $username = "root";
public $password = "usbw";
public $dbname = "bot";
function __construct()
{
//create new instance of mysql connection
$conn = new mysqli($this->servername, $this->username, $this->password, $this->dbname);
if ($mysqli->connect_errno)
{
echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error;
}
echo $mysqli->host_info . "\n";
$this->database_fetch();
}
function database_fetch()
{
$query = "SELECT word FROM timeoutwords";
$result = mysqli_query($query);
while($row = mysqli_fetch_assoc($result))
{
$array[] = $row();
}
}
function main()
{
print_r($array);
}
}
$bot = new IRCBot();

Answer 1

Changes
1) Change if ($mysqli->connect_errno) to if ($conn->connect_errno)
2) Change $array[] = $row(); to $array[] = $row;
3) Add return $array; in function database_fetch()
4) Call database_fetch() function inside main() function instead of constructor.
5) Add $this->conn in mysqli_query() (Thanks #devpro for pointing out.)
Updated Code
<?php
class IRCBot
{
public $array = array();
public $servername = "localhost";
public $username = "root";
public $password = "usbw";
public $dbname = "bot";
public $conn;
function __construct()
{
//create new instance of mysql connection
$this->conn = new mysqli($this->servername, $this->username, $this->password, $this->dbname);
if ($this->conn->connect_errno)
{
echo "Failed to connect to MySQL: (" . $this->conn->connect_errno . ") " . $this->conn->connect_error;
}
}
function database_fetch()
{
$query = "SELECT word FROM timeoutwords";
$result = mysqli_query($this->conn,$query);
while($row = mysqli_fetch_assoc($result)){
$array[] = $row;
}
return $array;
}
function main()
{
$data = $this->database_fetch();
print_r($data);
}
}
Quick Start
Object Oriented Programming in PHP
Classes and Objects
Principles Of Object Oriented Programming in PHP

Answer 2

First of all you need to fix error from your constructor, you can modify as:
function __construct()
{
//create new instance of mysql connection
$this->conn = new mysqli($this->servername, $this->username, $this->password, $this->dbname);
if ($this->conn->connect_errno)
{
echo "Failed to connect to MySQL: (" . $this->conn->connect_errno . ") " . $this->conn->connect_error;
}
echo $this->conn->host_info . "\n";
}
Here, you need to replace $mysqli with $conn because your link identifier is $conn not $mysqli
No need to call database_fetch() here.
You need to use $conn as a property.
Now you need to modify database_fetch() method as:
function database_fetch()
{
$query = "SELECT word FROM timeoutwords";
$result = mysqli_query($this->conn,$query);
$array = array();
while($row = mysqli_fetch_assoc($result))
{
$array[] = $row;
}
return $array;
}
Here, you need to pass add first param in mysqli_query() which should be link identifier / database connection.
Second, you need to use return for getting result from this function.
In last, you need to modify your main() method as:
function main()
{
$data = $this->database_fetch();
print_r($data);
}
Here, you need to call database_fetch() method here and than print the data where you need.