OOP PHP First Class Mysql數據庫連接

Question

我正在上一個測試課，主要是為了自學。 下面是課程：

class Connection
{   
    public $con;
    public $dbSelected;
    public $activeConnection;
    public $dataBaseName;
    function __contruct($dbUserName, $dbPassword, $server = "localhost")
    {
        $this->con = mysql_connect($server,$dbUserName,$dbPassword);
        if(!$this->$con)
        {
            $this->activeConnection = false;
        }
        else
        {
            $this->activeConnection = true;
        }
    }

    public function dbConnect($dbName, $identifyer = null)
    {   
        if ($identifyer === null)
        {
            $identifyer = $this->con;
        }
        $this->dbSelected = mysql_select_db($dbName, $identifyer);
        $this->dataBaseName = $dbName;
        if($this->dbSelected != true)
        {
            $this->connectionErrorReport(__LINE__);
        }
    }


    public function cleanData( array $submission)
    {
        unset($submission["throughTheCleaners"]);
        foreach($submission as $key => $value)
        {

            if(is_array($value))
            {
                $data[$key] = $this->cleanData($value);
            }
            else
            {
                $data[$key] = mysql_real_escape_string($value);
            }
        }
        $data["throughTheCleaners"] = true;
        return $data;
    }


    public function query($query, $dataSent)
    {
        if($dataSent["throughTheCleaners"] != true)
        {
            die("you must clean the data".__LINE__);
        }

        if($this->activeConnection == true && $this->dbSelected == true)
        {
            $result = mysql_query($query) or queryErrorReport($query, __LINE__);
            $i = 0;
            while($row = mysql_fetch_array($result))
            {
                foreach($row as $key => $value)
                {
                    $data[$i][$key] = $value;
                    $i++;
                }
            }
            return $data;           
        }
        else
        {
            $this->$connectionErrorReport(__LINE__);
        }
    }   

    public function connectionErrorReport($line = __LINE__)
    {
        $error = "There has been a connection error on line ".$line."</br>";
        if($this->activeConnection == false)
        {
            $error.= "Active Connection Error <br/>";
        }
        if($this->dbSelected  == false)
        {
            $error.= "Data Base Selection Error <br/>";
        }
        die($error.mysql_error());
    }

    public function queryErrorReport($query, $line = __LINE__)
    {
        die("There was a query error on ".$line."<br />$query<br/>".mysql_error());
    }

    function __destruct() {
        mysql_close($this->con);
    }
}

我不能一輩子弄清楚為什么它在構造函數調用期間不選擇存儲在$con變量中的mysql數據庫資源，而在其他函數中使用它。

最終，將為另一個類中的字符串提供加密功能，但我目前遇到的唯一錯誤都屬於連接類。

因此，引發錯誤的頁面如下所示：

include 'connection.php';
include 'loginClass.php';
$longinConnection = new Connection('***USERNAMEHERE***','***PASSWORD***');
$longinConnection->dbConnect("***DBTOCONNECTTO***");
echo Authoriz::encryptPassword("***USERPASSWORD***",$longinConnection);

該頁面測試了encryptPassword函數，但我什至無法使Connection類正常工作。 它引發的錯誤如下：

警告：mysql_select_db（）期望參數2為資源，在第27行的[PATH]中給定null在第31行出現連接錯誤活動連接錯誤數據庫選擇錯誤

警告：mysql_close（）期望參數1為資源，在第103行的[PATH]中給出null

在我看來，由於某種原因未設置$con變量，對我正在做錯的任何見解將不勝感激。

Answer 1

您拼寫了__construct ，使它在初始化類時沒有觸發，並且沒有初始化$this->con