PHP Include variables into a class

Question

look, i have a class that connects to a mysql database, but i have another file that contents the username, host, password and table names, If i don't include it and write them, it works fine, but the problem begins when I include it, it returns me "Undefined variable". Thank you, here's my class:

<?php
include '../config/conexiongeneral.php';
class DbConnector {
var $theQuery;
var $link;
public function DbConnector(){
        $host = $elnombredelhost;
        $db = $labasededatos;
        $user = $elnombredelusuario;
        $pass = $lacontasena;

        $this->link = mysql_connect($host, $user, $pass);
        mysql_select_db($db);
        register_shutdown_function(array(&$this, 'close'));
}
    function query($query) {

        $this->theQuery = $query;
        return mysql_query($query, $this->link);

    }
    function fetchArray($result) {

        return mysql_fetch_array($result);

    }
    function close() {

        mysql_close($this->link);

    }
}
?>

Well, i forget that i call this function with:

<?php
include 'dbConnector.php';
$connector = new DbConnector();

$username = trim(strtolower($_POST['username']));
$username = mysql_real_escape_string($username);
$query = "SELECT usuario FROM $latablatres WHERE usuario = '$username' LIMIT 1";
$result = $connector->query($query);
$num = mysql_num_rows($result);

echo $num;
mysql_close();
?>

"but the problem begins when I include it" But, what IS the Problem? — samayo
– samayo, Commented Jul 15, 2013 at 0:17
@Darkness Please try to include as much useful information as you can: the name of your "undefined variable", the number of the line with the error, etc.). — Racso
– Racso, Commented Jul 15, 2013 at 0:19
Yes, $elnombredelhost, $labasededatos, $elnombredelusuario, $lacontasena. — Darkness
– Darkness, Commented Jul 15, 2013 at 0:20

SharkofMirkwood · Accepted Answer · 2013-07-15 00:19:48Z

4

The problem is that you're including the file outside of the class so it's in the wrong scope (the class doesn't know about those variables).

Try this:

public function DbConnector(){
    include '../config/conexiongeneral.php';
    $host = $elnombredelhost;
    $db = $labasededatos;
    $user = $elnombredelusuario;
    $pass = $lacontasena;

    $this->link = mysql_connect($host, $user, $pass);
    mysql_select_db($db);
    register_shutdown_function(array(&$this, 'close'));
}

answered Jul 15, 2013 at 0:19

SharkofMirkwood

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4 Comments

SharkofMirkwood Over a year ago

What's inside your conesiongeneral.php file?

Darkness Over a year ago

<?php $elnombredelhost = "host"; $elnombredelusuario = "user"; $labasededatos = "database"; $latablauno = "table"; $latablados = "table"; $latablatres = "teble"; $lacontasena = "pass"; ?>

Darkness Over a year ago

Oh... forget it, You had all the reason, you was right, the problem was <? ?> originally In conexiongeneral.php i wrote <? ?> and correctly is <?php ?> I don't why... but anyway...

Darkness Over a year ago

Ahahaha Vale, muchas gracias! :D

ponysmith · Accepted Answer · 2013-07-15 00:21:46Z

1

Use the global keyword to make the variables available inside the function. E.g.:

public function DbConnector(){
    global $elnombredelhost;
    $host = $elnombredelhost;
    // ... rest of your code
}

Alternatively, you could declare the connection parameters a constants in the included file and use the constants in the connection script:

In conexiongeneral.php

define('DB_HOST', 'your_db_host');

In your connection class

public function DbConnector(){
    $host = DB_HOST;
    // ... rest of your code
}

answered Jul 15, 2013 at 0:21

ponysmith

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1 Comment

Yang Over a year ago

+1 because global solves current problem. We aren't talking about how bad Global state is

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