类与对象（PHP 5）
@info -- 20-April
This is because you requested class "b" before defining it, not because you defined class "b" before "a". It doesn't make a difference which class you define first.
基本概念
class
每个类的定义都以关键字 class 开头，后面跟着类名，可以是任何非 PHP 保留字的名字。后面跟着一对花括号，里面包含有类成员和方法的定义。伪变量 $this 可以在当一个方法在对象内部调用时使用。$this 是一个到调用对象（通常是方法所属于的对象，但也可以是另一个对象，如果该方法是从第二个对象内静态调用的话）的引用。看下面例子：
Example#1 面向对象语言中的 $this 变量
<?php
class A
{
function foo()
{
if (isset($this)) {
echo '$this is defined (';
echo get_class($this);
echo ")\n";
} else {
echo "\$this is not defined.\n";
}
}
}
class B
{
function bar()
{
A::foo();
}
}
$a = new A();
$a->foo();
A::foo();
$b = new B();
$b->bar();
B::bar();
?>
上例将输出：
$this is defined (a) $this is not defined. $this is defined (b) $this is not defined.
Example#2 简单的类定义
<?php
class SimpleClass
{
// 成员声明
public $var = 'a default value';
// 方法声明
public function displayVar() {
echo $this->var;
}
}
?>
Example#3 类成员的默认值
<?php
class SimpleClass
{
// 无效的类成员定义：
public $var1 = 'hello '.'world';
public $var2 = <<<EOD
hello world
EOD;
public $var3 = 1+2;
public $var4 = self::myStaticMethod();
public $var5 = $myVar;
// 正确的类成员定义：
public $var6 = myConstant;
public $var7 = self::classConstant;
public $var8 = array(true, false);
}
?>
Note: 除此之外还有不少用于处理类与对象的函数，详情参见 类 / 对象函数
new
要创建一个对象的实例，必须创建一个新对象并将其赋给一个变量。当创建新对象时该对象总是被赋值，除非该对象定义了构造函数并且在出错时抛出了一个异常。
Example#4 创建一个实例e
<?php
$instance = new SimpleClass();
?>
当把一个对象已经创建的实例赋给一个新变量时，新变量会访问同一个实例，就和用该对象赋值一样。此行为和给函数传递入实例时一样。可以用克隆给一个已创建的对象建立一个新实例。
Example#5 对象赋值
<?php
$assigned = $instance;
$reference =& $instance;
$instance->var = '$assigned will have this value';
$instance = null; // $instance and $reference become null
var_dump($instance);
var_dump($reference);
var_dump($assigned);
?>
上例将输出：
NULL NULL object(SimpleClass)#1 (1) { ["var"]=> string(30) "$assigned will have this value" }
extends
一个类可以在声明中用 extends 关键字继承另一个类的方法和成员。不能扩展多个类，只能继承一个基类。
被继承的方法和成员可以通过用同样的名字重新声明被覆盖，除非父类定义方法时使用了 final 关键字。可以通过 parent:: 来访问被覆盖的方法或成员。
Example#6 简单的类继承
<?php
class ExtendClass extends SimpleClass
{
// Redefine the parent method
function displayVar()
{
echo "Extending class\n";
parent::displayVar();
}
}
$extended = new ExtendClass();
$extended->displayVar();
?>
上例将输出：
Extending class a default value
add a note
User Contributed Notes基本概念
ashraf dot samhouri at hotmail dot com
24-May-2008 09:35
24-May-2008 09:35
info at youwanttoremovethisvakantiebaas dot nl
21-Apr-2008 06:40
21-Apr-2008 06:40
if you do this
<?php
$x = new b();
class b extends a {}
class a { }
?>
PHP will tell you "class b not found", because you've defined class b before a. However, the error tells you something different.... Got me a little confused :)
david dot schueler at tel-billig dot de
15-Feb-2008 10:16
15-Feb-2008 10:16
If you just want to create a new object that extends another object and you want to copy all variables from the father object, you may use this piece of code:
<?php
$father =& new father();
$father->a_var = "Hello World.";
$son = new son($event);
$son->say_hello();
class father {
public $a_var;
}
class son extends father {
public function __construct($father_class) {
foreach ($father_class as $variable=>$value) {
$this->$variable = $value;
}
}
public function say_hello() {
echo "Son says: ".$this->a_var;
}
}
?>
This outputs:
Son says: Hello World.
So you dont have to clone the entire object to get the contents of the variables from the father object.
aaron at thatone dot com
16-Dec-2007 10:46
16-Dec-2007 10:46
I was confused at first about object assignment, because it's not quite the same as normal assignment or assignment by reference. But I think I've figured out what's going on.
First, think of variables in PHP as data slots. Each one is a name that points to a data slot that can hold a value that is one of the basic data types: a number, a string, a boolean, etc. When you create a reference, you are making a second name that points at the same data slot. When you assign one variable to another, you are copying the contents of one data slot to another data slot.
Now, the trick is that object instances are not like the basic data types. They cannot be held in the data slots directly. Instead, an object's "handle" goes in the data slot. This is an identifier that points at one particular instance of an obect. So, the object handle, although not directly visible to the programmer, is one of the basic datatypes.
What makes this tricky is that when you take a variable which holds an object handle, and you assign it to another variable, that other variable gets a copy of the same object handle. This means that both variables can change the state of the same object instance. But they are not references, so if one of the variables is assigned a new value, it does not affect the other variable.
<?php
// Assignment of an object
Class Object{
public $foo="bar";
};
$objectVar = new Object();
$reference =& $objectVar;
$assignment = $objectVar
//
// $objectVar --->+---------+
// |(handle1)----+
// $reference --->+---------+ |
// |
// +---------+ |
// $assignment -->|(handle1)----+
// +---------+ |
// |
// v
// Object(1):foo="bar"
//
?>
$assignment has a different data slot from $objectVar, but its data slot holds a handle to the same object. This makes it behave in some ways like a reference. If you use the variable $objectVar to change the state of the Object instance, those changes also show up under $assignment, because it is pointing at that same Object instance.
<?php
$objectVar->foo = "qux";
print_r( $objectVar );
print_r( $reference );
print_r( $assignment );
//
// $objectVar --->+---------+
// |(handle1)----+
// $reference --->+---------+ |
// |
// +---------+ |
// $assignment -->|(handle1)----+
// +---------+ |
// |
// v
// Object(1):foo="qux"
//
?>
But it is not exactly the same as a reference. If you null out $objectVar, you replace the handle in its data slot with NULL. This means that $reference, which points at the same data slot, will also be NULL. But $assignment, which is a different data slot, will still hold its copy of the handle to the Object instance, so it will not be NULL.
<?php
$objectVar = null;
print_r($objectVar);
print_r($reference);
print_r($assignment);
//
// $objectVar --->+---------+
// | NULL |
// $reference --->+---------+
//
// +---------+
// $assignment -->|(handle1)----+
// +---------+ |
// |
// v
// Object(1):foo="qux"
?>
alan at alan-ng dot net
10-Oct-2007 12:41
10-Oct-2007 12:41
The following odd behavior happens in php version 5.1.4 (and presumably some other versions) that does not happen in php version 5.2.1 (and possibly other versions > 5.1.4).
<?php
$_SESSION['instance']=...;
$instance=new SomeClass;
?>
The second line will not only create the $instance object successfully, it will also modify the value of $_SESSION['instance']!
The workaround I arrived at, after trial and error, was to avoid using object names which match a $_SESSION array key.
This is not intended to be a bug report, since it was apparently fixed by version 5.2.1, so it's just a workaround suggestion.
mep_eisen at web dot de
10-Aug-2007 09:06
10-Aug-2007 09:06
referring to steven's post:
****
Perhaps this is because =& statements join the 2 variable names in the symbol table, whereas = statements applied to objects simply create a new independent entry in the symbol table that simply points to the same location as other entries. I don't know for sure - I don't think this behavior is documented in the PHP manual, so perhaps somebody with more knowledge of PHP's internals can clarify what is going on.
****
lets talk about
a =& b;
b = c;
PHP internally marks a to be a reference to b. If You reassign b PHP does not update a. But if you access a once more PHP looks at the current value of b (now containing c).
Both statements (a=b and a=&b) seem to do the same but they don't. However this changed for objects from PHP4 to PHP5. Where PHP4 needed this operator to avoid object cloning, PHP5 does not need it.
It is explained in chapter 21 (References Explained). It's important to understand that a becomes a reference and the following code will not modify b:
a =& b;
a =& c;
Dan Dascalescu
27-Oct-2006 02:00
27-Oct-2006 02:00
If E_STRICT is enabled, the first example will generate the following error (and a few others akin to it):
Non-static method A::foo() should not be called statically on line 26
The example should have explicitly declared the methods foo() and bar() as static:
class A
{
static function foo()
{
...