如何使用另一個文件中的PHP類

Question

我有兩個文件dbconnect.php和config.php文件

dbconnect.php

<?php 
class connect{ 
    public function __construct(){ 
     $config = require_once __DIR__ . 'config.php'; 
    } 

    private $dbhost = $config['host']; 
    private $dbuser = $config['username']; 
    private $dbpass = $config['pass']; 
    private $dbname = $config['dbname']; 

    public function startConn(){ 
     $this->DBcon = null; 
     try{ 
      $this->DBcon = new PDO("mysql:host=".$this->dbhost.";dbname=".$this->dbname, $this->dbuser, $this->dbpass); 
     }catch(Exception $e){ 
      echo "error connecting:"; 
     } 
     return $this->DBcon; 
    } 
} 

？>

的config.php

內

<?php 

    /** 
    * Contains all configurations 
    * 
    */ 
    return [ 
    'dbname' => 'user', 
    'pass' => '@user.intern1', 
    'username' => 'user1', 
    'host' => 'localhost', 
    ]; 
?> 

在我的dbconnect.php文件中;
如何從我的config.php中包含變量到類中連接

如果我按照以下方式執行操作：
它大叫我，並給我致命錯誤：
「解析錯誤：語法錯誤，意外的'$ config'（T_VARIABLE）在C：\ xampp \ htdocs \ hngfun \ profile \ adeojoemmanuel \ php-handler \ dbconfig.php第8" 行

Answer 1

我正在猜測這裏。但是我可以清楚地看到你在構造函數中將$ config設置爲局部變量。這意味着一旦你離開構造函數，它就不可用。

<?php 
class connect{ 
    public function __construct(){ 
     $config = require_once __DIR__ . 'config.php'; 
     $this->dbhost = $config['host']; 
     $this->dbuser = $config['username']; 
     $this->dbpass = $config['pass']; 
     $this->dbname = $config['dbname']; 
    } 

    private $dbhost ; 
    private $dbuser ; 
    private $dbpass ; 
    private $dbname ; 

    public function startConn(){ 
     $this->DBcon = null; 
     try{ 
      $this->DBcon = new PDO("mysql:host=".$this->dbhost.";dbname=".$this->dbname, $this->dbuser, $this->dbpass); 
     }catch(Exception $e){ 
      echo "error connecting:"; 
     } 
     return $this->DBcon; 
    } 
} 

Answer 2

聲明的屬性等private $dbhost不能分配依賴於運行時數據從PHP Docs引述的值，如$config['host'];

：

This declaration may include an initialization, but this initialization must be a constant value--that is, it must be able to be evaluated at compile time and must not depend on run-time information in order to be evaluated.

的溶液，分配VA梅毒在構造函數：

class connect{ 
    public function __construct(){ 
     $config = require_once __DIR__ . 'config.php'; 

     private $this->dbhost = $config['host']; 
     private $this->dbuser = $config['username']; 
     private $this->dbpass = $config['pass']; 
     private $this->dbname = $config['dbname']; 
    } 

    private $dbhost; 
    private $dbuser; 
    private $dbpass; 
    private $dbname; 

    public function startConn(){ 
     $this->DBcon = null; 
     try{ 
      $this->DBcon = new PDO("mysql:host=".$this->dbhost.";dbname=".$this->dbname, $this->dbuser, $this->dbpass); 
     }catch(Exception $e){ 
      echo "error connecting:"; 
     } 
     return $this->DBcon; 
    } 
} 

Answer 3

您需要內部構造設置：

private $dbhost; 
private $dbuser; 
private $dbpass; 
private $dbname; 

public function __construct(){ 
    $config = require_once __DIR__ . 'config.php'; 
    $this->dbhost = $config['host']; 
    $this->dbuser = $config['username']; 
    $this->dbpass = $config['pass']; 
    $this->dbname = $config['dbname']; 

} 

Answer 4

第一個問題是，你不能從你的config.php文件中返回任何東西。你只能在函數內返回一個結果。實現這一點的一種方法是將數組聲明爲全局變量，然後在需要該配置數組的所有其他php文件中使用它。

<?php 
/** 
* Contains all configurations 
* 
*/ 
$dbconfig = array(
    'dbname' => 'user', 
    'pass' => '@user.intern1', 
    'username' => 'user1', 
    'host' => 'localhost', 
); 
?> 

<?php 
require_once __DIR__ . 'config.php'; 

class connect{ 
    public function __construct(){ 

    } 

    private $dbhost = $dbconfig['host']; 
    private $dbuser = $dbconfig['username']; 
    private $dbpass = $dbconfig['pass']; 
    private $dbname = $dbconfig['dbname']; 

    public function startConn(){ 
     $this->DBcon = null; 
     try{ 
      $this->DBcon = new PDO("mysql:host=".$this->dbhost.";dbname=".$this->dbname, $this->dbuser, $this->dbpass); 
     }catch(Exception $e){ 
      echo "error connecting:"; 
     } 
     return $this->DBcon; 
    } 
} 
?>