Php Mysqli內部錯誤

我一直在嘗試一切超過4小時，並將錯誤固定到此代碼段。你能告訴我這裏可能有什麼錯嗎？Php Mysqli內部錯誤

<?php 
include_once 'database.php'; 

class Model { 

    public $db; 
    public $data; 
    public $data_item = array(); 


    public function output(){ 
     $this->db = new mysqli($db_servername, $db_username, $db_password, $db_name, $db_port); 
     $this->data = $this->db->query('SELECT * FROM `tbl_restaurants`'); 
     while($row = $this->data->fetch_assoc()){ 

      $this->data_item['res_id'] = $row['res_id']; 

     } 

     return $this->data_item['res_id']; 
    } 

} 
$obj = new Model; 
echo $obj->output(); 
?>

來源

2015-05-04 Mohamed Ashmawy

'tbl_restaurants'刪除。引用和嘗試。不要刪除第一個和最後一個報價 –

@anantkumarsingh它們不是引號。 –

從哪裏獲得變量'$ db_servername' ...？可能是這些變量的範圍是原因。 –

我認爲$db_servername是database.php中

需要global此：

public function output(){ 
    global $db_servername, $db_username, $db_password, $db_name, $db_port; 
    $this->db = new mysqli($db_servername, $db_username, $db_password, $db_name, $db_port); 
    //... 
}

一個更好的設計是，給數據庫連接的參數到型號：

public function __construct($db) { 
    $this->db = $db; 
}

來源

2015-05-04 06:26:31

OMG！謝謝你，兄弟！ –

您假定數據庫連接的詳細信息，但沒有從任何地方獲取。

數據庫未連接。

$this->db = new mysqli($db_servername, $db_username, $db_password, $db_name, $db_port);

通過它明確爲：

public function output($db_servername, $db_username, $db_password, $db_name, $db_port){

或將它們設置爲類變量，並設置它通過構造函數：

類模型{

public $db; 
public $data; 
public $data_item = array(); 
public $db_servername; 
public $db_username; 
public $db_password; 
public $db_name; 
public $db_port;

，並讓他們爲：

public function output(){ 
    $this->db = new mysqli($this->db_servername, $this->db_username, $this->db_password, $this->db_name, $db_port);

來源

2015-05-04 06:24:13 Pupil

你能進一步解釋一下嗎？ *編輯：剛剛嘗試了你說的，仍然沒有工作。 –

$ db_servername，$ db_username，$ db_password，$ db_name，$ db_port未在類中定義。試試這個：

<?php 
include_once 'database.php'; 

class Model { 

    public $db; 
    public $data; 
    public $data_item = array(); 


    public function output(){ 
     global $db_servername, $db_username, $db_password, $db_name, $db_port; 
     $this->db = new mysqli($db_servername, $db_username, $db_password, $db_name, $db_port); 
     $this->data = $this->db->query('SELECT * FROM `tbl_restaurants`'); 
     while($row = $this->data->fetch_assoc()){ 

      $this->data_item['res_id'] = $row['res_id']; 

     } 

     return $this->data_item['res_id']; 
    } 

} 
$obj = new Model; 
echo $obj->output();

來源

2015-05-04 06:26:11

假設數據庫詳細信息是問題，您可以將它們全局化，也可以不使用變量，而是使用database.php中的定義

而不是$ db_servername =「localhost」;等等。你可以這樣做：

define('DB_SERVERNAME','localhost');

然後，而不是在新的mysqli使用$ db_servername（功能，你會怎麼做：

$this->db = new mysqli(DB_SERVERNAME, DB_USERNAME, DB_PASSWORD, DB_NAME, DB_PORT);

來源

2015-05-04 06:27:56

試試這個

<?php 
include_once 'database.php'; 

class Model { 

    public $db; 
    public $data; 
    public $data_item = array(); 


    public function output($db_servername, $db_username, $db_password, $db_name, $db_port){ 
     $this->db = new mysqli($db_servername, $db_username, $db_password, $db_name, $db_port); 
     $this->data = $this->db->query('SELECT * FROM `tbl_restaurants`'); 
     while($row = $this->data->fetch_assoc()){ 

      $this->data_item['res_id'] = $row['res_id']; 

     } 

     return $this->data_item['res_id']; 
    } 

} 
$obj = new Model; 
echo $obj->output($db_servername, $db_username, $db_password, $db_name, $db_port); 
?>

來源

2015-05-04 07:35:34

Php Mysqli內部錯誤

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