Php在构造函数中创建对象[关闭]

Question

I have two php classes, in two files.

....db.php ....sql.php

db contains:

class DataBase {

private $link;
private $host;
private $username;
private $password;
private $database;

public function __construct() {
    print'<br>in constructor DB!<br>';
    $this->host     = 'xxxx';
    $this->username = 'xxxx';
    $this->password = 'xxxx';
    $this->database = 'xxxx';
    $this->connect();
}

public function connect() {
    $this->link = mysql_connect($this->host,$this->username,$this->password);
    if (!$this->link)   die('Could not connect: ' . mysql_error());
    mysql_select_db($this->database, $this->link);
}

public function getConnection() {
    return $this->link;
}

}

sql.php contains :

include 'db.php';

class SQL {

    public $dataBase;
    public $con;

    public function __construct() {
        $dataBase = new DataBase; 
        print 'database -> '.$dataBase; 
        $con = $dataBase->getConnection(); 
    }

    public function login($email, $pass) {
        $sql = "SELECT * FROM clienti WHERE email='".$email."' and pass= md5('".$pass."')";
        $result = mysql_query($sql, $con) or die(mysql_error());
        $row = mysql_fetch_assoc($result);
        if(!$row) return 'Email or password is incorrect.';
        return '';
    }
}

in login.php

<?php
include 'include/sql.php';

$email  = $_GET['email'];
$pass= $_GET['pass'];

$error  = '';

if($email == '' || $pass== '') {
    $error = 'Email or password is empty.';
}
else {
    $sql = new SQL(); //if I print inside constructor, $con variable, is something like #3 resources
    echo $sql->con;  // but here is empty   ???
    $sql->login($email, $pass);
    $sql->close();
}   

What I am doing wrong?
The includes are ok, I tested them. But why variables of sql Object are empty??? thank you.

Answer 1

$con is a property of the class not a variable, you refer to it by using keyword $this:

$con = $dataBase->getConnection(); //WRONG

change to:

$this->con = $dataBase->getConnection(); //CORRECT