OOP PHP Database results to Array

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();

pass $conn into $this->database_fetch(); as $this->database_fetch($conn); because mysqli_query need ist param as a link identifier but this will not work — devpro
– devpro, Commented Oct 28, 2016 at 10:58
make it property $conn as $this->conn than get this property into main method and call database_fetch() inside the main() method not in constructor — devpro
– devpro, Commented Oct 28, 2016 at 11:00
Amazing what "three little words" will do to someone's day, eh? @devpro — Funk Forty Niner
– Funk Forty Niner, Commented Oct 28, 2016 at 11:27

Nana Partykar · Accepted Answer · 2016-10-28 11:16:56Z

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

friend, u missed the very important mysqli_query($query) ist param
now looks fine friend, just one suggestion u must need to initialize array $array = array();
I think, no need as in top public $array = array(); is initialised @devpro

devpro · Accepted Answer · 2016-10-28 11:29:35Z

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.

Sometimes happen. But, that was super for me $_POST('pedert').
@NanaPartykar: hahaha, actually something like shoro shoro k din Early days
Ha Ha :D Shuru Shuru Ke Din, Wo Bhi Kya Din They.. Let's Jump To Other Question Too. @devpro

Collectives™ on Stack Overflow

OOP PHP Database results to Array

2 Answers 2

3 Comments

5 Comments

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

3 Comments

5 Comments

Related