Home php教程 php手册 PHP 5.0对象模型深度探索之定义一个类

PHP 5.0对象模型深度探索之定义一个类

Jun 21, 2016 am 08:59 AM
name public user

  当你声明一个类,你需要列出对象应有的所有变量和所有函数—被称为属性和方法。列表1中显示了一个类的构成. 注意在大括号({})内你只能声明变量或者函数。列表2中显示了如何在一个类中定义三个属性和两个方法。

  列表1

class Name extends Another Class
{
 Access Variable Declaration
 Access Function Declaration
}


  列表2

//定义一个跟踪用户的类
class User
{
 //属性
 public $name;
 private $password, $lastLogin;

 //方法
 public function __construct($name, $password)
 {
  $this->name = $name;
  $this->password = $password;
  $this->lastLogin = time();
  $this->accesses++;
 }

 // 获取最后访问的时间
 function getLastLogin()
 {
  return(date("M d Y", $this->lastLogin));
 }
}

//创建一个对象的实例
$user = new User("Leon", "sdf123");

//获取最后访问的时间
print($user->getLastLogin() ."
n");

//打印用户名
print("$user->name n");
?>


  当你声明属性,你不需要指明数据类型。变量可能是整型,字符串或者是另一个对象,这取决于实际情况。在声明属性时增加注释是一个好主意,标记上属性的含义和数据类型。

  当你声明一个方法,你所做的和在类外部定义一个函数是一样的。方法和属性都有各自的命名空间。这意味着你可以安全地建立一个与类外部函数同名的方法,两者不会冲突。例如,一个类中可以定义一个名为date()的方法。但是你不能将一个方法命名为PHP的关键字,如for或者while。

  类方法可能包含PHP中所谓的type hint. Type hint 是另一个传递参数给方法的类的名字. 如果你的脚本调用方法并传递一个不是类的实例的变量,PHP将产生一个”致命(fatal)错误” 。你可能没有给其它类型给出type hint,就像整型,字符串,或者布尔值. 在书写的时候, type hint是否应当包含数组类型仍存在争议。

  Type hint是测试函数参数或者运算符的实例的数据类型的捷径. 你可能总是返回这个方法. 确认你强制让一个参数必须是哪种数据类型,如整型. 列表3确保编译类只产生Widget的实例。

  列表3

//组件
class Widget
{
public $name='none';
public $created=FALSE;
}

//装配器
class Assembler
{
public function make(Widget $w)
{
print("Making $w->name
n");
$w->created=TRUE;
}
}

//建立一个组件对象
$thing = new Widget;
$thing->name = 'Gadget';

//装配组件
Assembler::make($thing);
?>


  除了传递参数的变量外,方法含有一个特殊的变量. 它代表类的个别实例. 你应当用这个来指向对象的属性和其它方法.一些面向对象的语言假设一个不合格的变量提交给本地属性,但在PHP中方法的任何变量只是在方法的一定范围内. 注意在User类的构造函数中这个变量的使用(3.1.2).

  PHP在属性和方法声明前定义一个访问限定语,如public,private和protected. 另外,你可以用”static”来标记一个成员. 你也可以在类中声明常量。以后稍后会有不同访问方式的相关讨论。

  你可以在一行中列出相同访问方式的几个属性,用逗号来分隔它们. 在3.1.2中,User类有两个private属性--$password和$lastLogin。



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