php include_once not work in class

Question

I'm working to convert a project to multilanguage but I have problems with require_once("../idioma/lang.br.php"); below

require_once("../idioma/lang.br.php");
require_once("../classe/class.usuario.dao.php");

Class UsuarioBLL 
{
    public function CadastraUsuario($campos)
    {
        $msg = array();
        // senha
        if(strlen($campos['senha']) >= 6 && strlen($campos['senha']) <= 30)
            array_push($msg,$user["TAMANHO_SENHA"]);

        if ($campos['senha'] != $campos['csenha'])
            array_push($msg, $user["SENHA_DIFERENTE"]);     

        $dao = new UsuarioDAO();

in

array_push($msg,$user["TAMANHO_SENHA"]);

the array variable

$user["TAMANHO_SENHA"]

is null

this is a include

<?php
//cadastro usuario
$user = array();
$user["PREENCHER_TUDO"]  = "Por favor preencha todos os campos.";
$user["TAMANHO_SENHA"]   = "Senha deve conter entre 6 a 30 digitos.";

if I do a var_dump ($ user); before class

require_once("../idioma/lang.br.php");
require_once("../classe/class.usuario.dao.php");
var_dump ($ user);
Class UsuarioBLL 
{
    public function CadastraUsuario($campos)
    {
        $msg = array();
        // senha
        if(strlen($campos['senha']) >= 6 && strlen($campos['senha']) <= 30)
            array_push($msg,$user["TAMANHO_SENHA"]);

        if ($campos['senha'] != $campos['csenha'])
            array_push($msg, $user["SENHA_DIFERENTE"]);       

        $dao = new UsuarioDAO();

the system print

array(2) { 
["PREENCHER_TUDO"]=> string(35) "Por favor preencha todos os campos." ["TAMANHO_SENHA"]=> string(39) "Senha deve conter entre 6 a 30 digitos." 
}

the require_once("../classe/class.usuario.dao.php");

 $dao = new UsuarioDAO();

its working

I did a test I create

class UserLang {
    public $user;
     function __construct() {
         $this->user["PREENCHER_TUDO"] = "xxxxxxxxxxxxxxxxxxxxxxxxxxxxx.";
    }
}

and it works now

$e = new UserLang();
echo $e->user["PREENCHER_TUDO"];

whyyy with class work and an array out of a class does not work?

I do not know what is happening

Thank you for your help

You should read this one... stackoverflow.com/questions/797380/…. Also as a note it is much more preferable to copy and paste code rather then screenshots. Also as that answer saws using global variables is super bad practice. — nerdlyist
– nerdlyist, Commented Jul 13, 2016 at 1:28
You should post your code rather than images. Don't post the whole script though, just the excerpts that are needed for use to understand the problem. — BeetleJuice
– BeetleJuice, Commented Jul 13, 2016 at 1:42

BeetleJuice · Accepted Answer · 2016-07-13 02:42:31Z

0

Your class cannot see what is outside. If you mean for the class to work on a specific user dataset, you should make it a class property. Here's the simplest way:

Class UsarioBLL{
    public $user; //needs to be set before other functions can use it
}

After you create the object, set the user

$usario = new UsarioBLL();
$usario->user = $user; //now  the class can see the user data!

After you've done this, your CastradaUsario function will see the data.

why with class work and an array out of a class does not work?

Note that your require_once("../idioma/lang.br.php") inserts the $user array above the class definition. Your problem has to do with variable scope. When you define the array outside of your class, the class members cannot see it

$user = [...];
Class UsarioBLL{
// $user is undefined here.
}

To gain visibility, you need to pass the data and store it in a class variable. The simplest way is in my original answer above. Another way is to pass data to the class constructor:

Class UsarioBLL{
    private $user;
    __construct($input_data){this->user = $input_data}
}

__construct is called when a new object is created. So create your new object like so:

$usario = new UsarioBLL($user); //will call __construct and pass $user

edited Jul 13, 2016 at 2:42

answered Jul 13, 2016 at 1:29

BeetleJuice

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4 Comments

Eduardo Over a year ago

yes I can pass the array as a parameter but because "class.usuario.dao.php" include working? below the line of $dao = new UsuarioDAO(); It does not display error should not this also presenting error?

Eduardo Over a year ago

Please see the post again

BeetleJuice Over a year ago

Thanks for cleaning up your question. I've edited my answer

Eduardo Over a year ago

Work .... Class UsuarioBLL { public function CadastraUsuario($campos) { require_once("../idioma/lang.br.php"); $msg = array(); // senha if(strlen($campos['senha']) >= 6 && strlen($campos['senha']) <= 30) array_push($msg,$user["TAMANHO_SENHA"]);

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