爲什麼我不能訪問這個全局變量？

Question

我當然試過了$GLOBALS，仍然沒有去。我的語法是否正確。我的理解是$DB_USER在全球範圍內。

<?php 
    $DB_USER = 'foo'; 
    class Database 
    { 
     // this does not work 
     private $DB_USER =    $GLOBALS['DB_USER']; 
     private $DB_PASS =    'foob'; 
     private $DB_DRIVER =   'foob_foob'; 
     // ... 

Answer 1

您從類方法，這意味着你實際上是調用內部調用$DB_USER從變量本地範圍（在課堂內）。爲了解決這個問題，只是告訴你正在通過增加global $DB_USER你的方法，它將被用於內部尋找的全局變量PHP（或使用您的構造函數將其添加到類範圍）：

class Database 
{ 
    private $DB_USER =    ''; 
    private $DB_PASS =    'foob'; 
    private $DB_DRIVER =   'foob_foob'; 
    // snip 


    // Method 1: Add the variable to the class scope with the constructor   
    public function __construct() 
    { 
     global $DB_USER; 
     $this->DB_USER = $DB_USER; 
    } 


    // Method 2: tell PHP that you want the global variable in your methods 
    public function foo() 
    { 
     $global $DB_USER; 
     ... 
    } 

隨着方法1，您現在可以撥打$this->DB_USER而不是$GLOBALS['DB_USER']。

使用方法2，您將global $DB_USER添加到每個，然後只使用$DB_USER。

欲瞭解更多信息，請參閱http://php.net/manual/en/language.variables.scope.php

Answer 2

使用構造函數初始化varibale：

public function __constructor(){ 
    global $YOUR_VARIABLE; 
    $DB_USER = $YOUR_VARIABLE; 
} 

Answer 3

所以，你宣佈你variables第一，這將輸出初始化它們在__construct方法

$DB_USER = 'foo'; 
class Database 
{ 
    private $DB_USER, 
    $DB_PASS, 
    $DB_DRIVER; 
    public function __construct(){ 
     $this->DB_USER = $GLOBALS['DB_USER']; 
     $this->DB_PASS = 'foob'; 
     $this->DB_DRIVER = 'foob_foob'; 
     print_r($this); 
    } 
} 

new Database(); 

Database Object 
(
    [DB_USER:Database:private] => foo 
    [DB_PASS:Database:private] => foob 
    [DB_DRIVER:Database:private] => foob_foob 
) 

DEMO

編輯

使用依賴注入：

$DB_USER = 'foo'; 
class Database 
{ 
    private $DB_USER, 
    $DB_PASS, 
    $DB_DRIVER; 
    public function __construct($dbUser){ 
     $this->DB_USER = $dbUser; 
     $this->DB_PASS = 'foob'; 
     $this->DB_DRIVER = 'foob_foob'; 
     print_r($this); 
    } 
} 

new Database($DB_USER); 

DEMO_2