Php Mysqli内部错误

我一直在尝试一切超过4小时，并将错误固定到此代码段。你能告诉我这里可能有什么错吗？Php Mysqli内部错误

<?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(); 
?>

来源

2015-05-04 Mohamed Ashmawy

'tbl_restaurants'删除。引用和尝试。不要删除第一个和最后一个报价 –

@anantkumarsingh它们不是引号。 –

从哪里获得变量'$ db_servername' ...？可能是这些变量的范围是原因。 –

我认为$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); 
    //... 
}

一个更好的设计是，给数据库连接的参数到型号：

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

来源

2015-05-04 06:26:31

OMG！谢谢你，兄弟！ –

您假定数据库连接的详细信息，但没有从任何地方获取。

数据库未连接。

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

通过它明确为：

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

或将它们设置为类变量，并设置它通过构造函数：

类模型{

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

，并让他们为：

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

来源

2015-05-04 06:24:13 Pupil

你能进一步解释一下吗？ *编辑：刚刚尝试了你说的，仍然没有工作。 –

$ db_servername，$ db_username，$ db_password，$ db_name，$ db_port未在类中定义。试试这个：

<?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();

来源

2015-05-04 06:26:11

假设数据库详细信息是问题，您可以将它们全局化，也可以不使用变量，而是使用database.php中的定义

而不是$ db_servername =“localhost”;等等。你可以这样做：

define('DB_SERVERNAME','localhost');

然后，而不是在新的mysqli使用$ db_servername（功能，你会怎么做：

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

来源

2015-05-04 06:27:56

试试这个

<?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); 
?>

来源

2015-05-04 07:35:34

Php Mysqli内部错误

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