將數據庫對象從一種方法傳遞到另一種方法的正確方法是什麼

Question

我需要幫助理解將在一種方法中創建的對象（MySQL連接）傳遞給同一類中的另一種方法的正確方法。

我想創建一個數據庫類，使用構造函數和所有其他方法加載連接，使用此連接執行各自的功能。

在嘗試這樣做，我不斷收到此錯誤：

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

我的代碼：

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"]); 
      } 

     } 

    } 
} 

任何幫助，將不勝感激。謝謝大家。

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>";  
} 

而且將更好地爲protected。

Answer 2

你不是在任何地方設置$this->mysqli，你需要在你的構造中做$this->mysqli = $mysqli;而不是return $this->mysqli。

Answer 3

你實際上並沒有創建$ mysqli作爲你的類的一個屬性。

您需要：

class database 
{ 

public $mysqli; 

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>"; 
    return $this->mysqli; 

} 

或

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>"; 
    $this->mysqli = $mysqli; 
    return $this->mysqli; 

} 

Answer 4

正如其他人已經提到的，你應該這樣設置的mysqli：

$this->mysqli = new mysqli($database_server,$database_user, $database_pass, $database_name); 

我建議不要在死構造函數，那不是數據庫類決定，而是數據庫類的用戶。改變這樣的：

class database 
{ 
    private $mysqli; 
    public $error = false; 

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

     if ($mysqli->connect_error) 
      $this->error = 'Connect Error (' . $mysqli->connect_errno . ') '. $mysqli->connect_error); 

    } 

在構建這樣$db = new Database(...);您可以檢查錯誤屬性，然後決定做什麼數據庫。