php - 将数据库对象从类中的一个方法传递到另一个方法的正确方法是什么

Question

我需要帮助来理解将在一个方法中创建的对象(MySQL 连接)传递给同一类中的另一个方法的正确方法。

我正在尝试创建一个数据库类，它使用构造函数和所有其他方法加载连接，使用这个连接来执行它们各自的功能。

在尝试这样做时，我不断收到此错误:

Fatal error: Call to a member function query() on a non-object in /database.php on line 44

我的代码:

class database
{ 
    public $mysqli;

    public function __construct($database_server, $database_name, $database_user, $database_pass)
    {
        $mysqli = new mysqli($database_server,$database_user, $database_pass, $database_name);

        if ($mysqli->connect_error)
        {
                  die('Connect Error (' . $mysqli->connect_errno . ') '. $mysqli->connect_error);
        }
        echo "Database connection established successfully.<br>";
        return $this->mysqli;

    }

    public function fetch_data()
    {

        $query = "select * from payment";
        if ($result = $this->mysqli->query($query))
        {

            // fetch associative array
            while ($row = $result->fetch_assoc())
            {
                printf ("%s (%s)\n", $row["id"], $row["status"]);
            }

        }

    }
}

如有任何帮助，我们将不胜感激。谢谢大家。

Answer 1

public function __construct($database_server, $database_name, $database_user, $database_pass)
{
    $mysqli = new mysqli($database_server,$database_user, $database_pass, $database_name);

    if ($mysqli->connect_error)
    {
              die('Connect Error (' . $mysqli->connect_errno . ') '. $mysqli->connect_error);
    }
    echo "Database connection established successfully.<br>";

    // instead of return, just set the local variable to the `mysqli` property.
    $this->mysqli = $mysqli;

}

或者

public function __construct($database_server, $database_name, $database_user, $database_pass)
{
    $this->mysqli = new mysqli($database_server,$database_user, $database_pass, $database_name);

    if ($this->mysqli->connect_error)
    {
       die('Connect Error (' . $this^>mysqli->connect_errno . ') '. $this->mysqli->connect_error);
    }
    echo "Database connection established successfully.<br>";    
}

public $mysqli; 会更好protected。