Why not use a database function class in php [duplicate]

Question

I adopted code given in this example: PHP database connection class

But i changed the code with the link above and it is as follows:

class Database {
    public function connect() {
        define('DB_HOST', 'localhost');
        define('DB_NAME', 'university');
        define('DB_USER', 'root');
        define('DB_PASSWORD', 'root');
        $db = mysqli_connect(DB_HOST,DB_USER,DB_PASSWORD,DB_NAME);
    }
}
function Information(){
    Database::connect();
    $query = "SELECT * FROM student WHERE id "; $result = mysqli_query($db, $query);
        while($row = mysqli_fetch_array($result)) {
            echo $row['name'];
        }
}
Information();

It gives error:

Notice: Undefined variable: db in /Applications/XAMPP/file.php on line 12.

What could be the reason and what am I doing wrong?

$db belongs in the connect() function scope. Why not treat $db as a class property instead and access it via $class->db? — treyBake
– treyBake, Commented Jan 8, 2020 at 11:56
WHERE id ?? think it would also be useful to show errors. To get errors out of PHP even in a LIVE environment add these 4 lines to the top of any MYSQLI_ based script you want to debug ini_set('display_errors', 1); ini_set('log_errors',1); error_reporting(E_ALL); mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);. This will force any MYSQLI_ errors to generate an Exception that you can see on the browser as well as normal PHP errors. — RiggsFolly
– RiggsFolly, Commented Jan 8, 2020 at 12:04

delboy1978uk · Accepted Answer · 2020-01-08 12:03:01Z

0

Make a property that all methods can share:

<?php

class Database 
{
    private $db;

    public function __construct() 
    {
        define('DB_HOST', 'localhost');
        define('DB_NAME', 'university');
        define('DB_USER', 'root');
        define('DB_PASSWORD', 'root');
        $this->db = mysqli_connect(DB_HOST,DB_USER,DB_PASSWORD,DB_NAME);
    }

    public function query($sql)
    {
        return mysqli_query($this->db, $query);
    }
}

And then refactor your function

function Information()
{
    $db = new Database();
    $sql = "SELECT * FROM student WHERE id "; 
    $result = $db->query($sql);
    while($row = mysqli_fetch_array($result)) {
         echo $row['name'];
    }
}

Information();

answered Jan 8, 2020 at 12:03

delboy1978uk

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Comments

Danyal Sandeelo · Accepted Answer · 2020-01-08 12:13:50Z

-1

The variable db is unknown to the function on this line:

Since $dbis a class property, just create it as a variable in the class

 private $db;

and set it in the connect function as mentioned by delboy1978uk like this:

$this->db = mysqli_connect(DB_HOST,DB_USER,DB_PASSWORD,DB_NAME);

edited Jan 8, 2020 at 12:13

answered Jan 8, 2020 at 11:59

Danyal Sandeelo

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2 Comments

delboy1978uk Over a year ago

Your solution will not work as you call connect() statically. $this won't wiork

Danyal Sandeelo Over a year ago

@delboy1978uk updated it

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