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：

例 19-1. 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.

例 19-2. 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;
    }
}
?>

例 19-3. 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);


}
?>

注意: 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。

例 19-4. 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。

例 19-5. 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。

例 19-6. 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

mep_eisen at web dot de
10-Aug-2007 01:06


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;

Stephen
08-Jul-2007 11:54


jcastromail at yahoo dot es stated:

****

in php 5.2.0 for classes



$obj1  =  $obj2;

is equal to

$obj1  =  &$obj2;"

****



However, that is not completely true. While both = and =& will make a variable refer to the same object as the variable being assigned to it, the explicit reference assignment (=&) will keep the two variables joined to each other, whereas the assignment reference (=) will make the assigned variable an independent pointer to the object. An example should make this clearer:



<?php



class z {

    public $var = '';

    }

    

$a = new z();

$b   =&  $a;

$c  = $a;



$a->var = null;



var_dump($a);

print '<br>';

var_dump($b);

print '<br>';

var_dump($c);

print '<br><br>';



$a = 2;



var_dump($a);

print '<br>';

var_dump($b);

print '<br>';

var_dump($c);

print '<br><br>';



?> 



This outputs:



object(z)#1 (1) { ["var"]=>  NULL }

object(z)#1 (1) { ["var"]=> NULL }

object(z)#1 (1) { ["var"]=> NULL }



int(2)

int(2)

object(z)#1 (1) { ["var"]=> NULL } 



So although all 3 variables reflect changes in the object, if you reassign one of the variables that were previously joined by reference to a different value, BOTH of those variables will adopt the new value.



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.

realbart at hotmail dot com
25-Apr-2007 05:42


Classes do not seem to be passed to functions. correctly in PHP4

I'll let you know when I find out why



<?php



class XmlNode

{

    var $name;

    var $attrs;

    var $parentNode;

    var $firstChild;

    var $nextSibling;

}



function startElement($parser, $name, $attrs)

{

    global $XmlRootNode;

    global $XmlPreviousSibling;

    global $XmlParentNode;



    if (is_null($XmlRootNode))

    {

        $XmlRootNode = new XmlNode;

        $currentNode = &$XmlRootNode;

    }

    else

    {

        if (is_null($XmlParentNode->firstChild))

        {

            $XmlParentNode->firstChild = new XmlNode;

            $currentNode = &$XmlParentNode->firstChild;

        }

        else

        {

            $XmlPreviousSibling->nextSibling = new XmlNode;

            $currentNode = &$XmlPreviousSibling->nextSibling;

        }

        $currentNode->parentNode = &$XmlParentNode;

    }

    $currentNode->name = $name;

    $currentNode->attrs = $attrs;

    $XmlPreviousSibling = &$currentNode;

    $XmlParentNode = &$currentNode;

    echo $XmlParentNode->name;

}



function endElement($parser, $name)

{

        global $XmlParentNode;

        $XmlParentNode = &$XmlParentNode->parentNode;

}



// nothing wrong with these lines

$file = "vragen.xml";

$xml_parser = xml_parser_create();

xml_set_element_handler($xml_parser, "startElement", "endElement");

if (!($fp = fopen($file, "r")))    die("could not open XML input");

while ($data = fread($fp, 4096)) if (!xml_parse($xml_parser, $data, feof($fp))) 

die(sprintf("XML error: %s at line %d", xml_error_string(xml_get_error_code($xml_parser)), xml_get_current_line_number($xml_parser)));

xml_parser_free($xml_parser);



//the xml file should now be converted to xml nodes...



echo $XmlRootNode->name; // returns the name of the root object

echo $XmlRootNode->firstChild; // should return "Object" yet returns nothing.

echo $XmlRootNode->firstChild->name; // shoud return the name of the first object but doesn't

echo $XmlRootNode->firstChild->nextSibling->name; //also returns nothing

?>

Andrew V Azarov
14-Dec-2006 08:20


Remember that $this->var and $this->$var are different

when using smth like

<?

class test

{

        var $var;

        function Setvar()

        {

                 $this->$var = "test";

        }

}



$m = new test();

$m->Setvar();

$m->var; 

?>

will result in an fatal error or empty var



do instead

<?

$this->var = "test";

?>

then u may access it like <? $m->var; ?>

Andrew V Azarov
14-Dec-2006 08:18


Remember that $this->var and $this->$var are different

when using smth like



class test

{

        var $var;

        function Setvar()

        {

                 $this->$var;

        }

}



$m = new test();

$m->Setvar()

$m->var; will result in an fatal error or empty var

jcastromail at yahoo dot es
29-Nov-2006 03:39


in php 5.2.0 for classes



$obj1  =  $obj2;

is equal to

$obj1  =  &$obj2;



creating a reference of the object.



<?

class test {

    public $var;

}



$instance=new test();

$assigned=new test();

$reference=new test();

$copy=new test();



$instance->var='var of instance';

$assigned->var='var of asigned';

$reference->var='var of reference';

$copy->var='var of copy';



$assigned  =  $instance;

$reference  =& $instance;



$reference->var='var of reference';



echo $instance->var."<br>";

echo $assigned->var."<br>";

echo $reference->var."<br>";

echo $copy->var."<br>";

?>



it's return 



var of reference

var of reference

var of reference

var of copy

thisbizness at gmail dot com
15-Nov-2006 09:16


The best alternatives I think one has right now to nest a class are:



1- create the nested class in an external file and 'include' that file inside a nesting class function. I prefer to use a static function for this.



2- put the nested class definition in a string inside a nesting class function and use the 'eval' keyword to define the class.



<?php

class MyParent{



    public function parentFunction( ){

    }

    

    public static function nestedClasses(){

        eval('

            class MyChild{

              function __construct(){

                  echo "this is a nested class";

                }

                public function childFunction( )

                {

                

                }

            }

        ');      

    }



}



MyParent::nestedClasses();

$y = new MyChild();



?>

nicholas at nicholaswilliams dot info
13-Nov-2006 11:09


Note that you cannot nest classes (as of <5.1.2; higher versions are not available yet in my distribution). So, the following



<?php



error_reporting( E_ALL | E_STRICT );



class MyParent

{

    public function parentFunction( )

    {

        

    }

    

    class MyChild

    {

        public function childFunction( )

        {

        

        }

    }

}



$x = new MyParent();



?>



Generates the error "Parse error: syntax error, unexpected T_CLASS, expecting T_FUNCTION in [filename] on line 12". This is contrary to typical OO programming (specifically Java). Hopefully they'll change it sometime soon.



I know in part this is a feature request, but I'm also putting it here in case anyone's looking for information about it: simply, you can't yet do it.



Nicholas

Dan Dascalescu
27-Oct-2006 06:00


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()                                                                                                           

    { 

...

phpprogrammer at artspad dot net
12-Aug-2006 11:29


"Once again, I reiterate, there's no need to overwrite variables and open this can of worms: just make a new variable and be done with it. As long as you don't overwrite variables, it doesn't matter whether or not you use references or assignments (unless you're planning for PHP4 compatibility). "



Always suspect advice that disregards nuanced functionality available to you.



The difference between getting a reference versus a copy is subtle but powerful. It really only became highly relevant to me when I was working in Flash with XML. The nodes from the xml structure were returned as references. So getting a node and changing its value didn't just change the node variable I had, it updated the node within the xml document. Had I only gotten a copy I would have had to have routines for removing and inserting the node in the overall structure.

 

The moral of this story is that there are times when overwriting variables in a reference is actually very practical and a sign of evovled coding.



Also, there's the old trick of passing an object reference to a function as a parameter so that you can change its properties and not have to return anything. This is particularly useful if you pass more than one object reference to the function, since a function can only ever return one value.



There is a caveat in using references in that so long as one variable within scope is holding a copy of the reference the object will remain in memory.



Please take the time to really learn virtue of references versus copies and decide for yourself whether its a tool that can help you.

edwardzyang at thewritingpot dot com
13-Apr-2006 07:47


I read the first few comments about references versus assignment on objects, and I still couldn't get it. I'm going to talk about it in terms of practical usage, not the tomfoolery occuring behind the scenes. No talk of memory pointers and the like.



<?php



// prepend this to all examples



class Mushroom

{

    public $size;

    public function __construct($size) {

        $this->size = $size;

    }

}



$mushroom = new Mushroom(1);

$assigned_shroom = $mushroom;

$reference_shroom =& $mushroom;



?>



Assignments and references, when it comes down to calling functions and modifying member variables, don't make ANY difference at all.



<?php



$mushroom->size = 5;



// all shrooms are size 5



?>



The only deviation of behavior is when you assign totally new contents to a variable. Usually, this never happens unless you're doing some strange reflection or injection.



<?php



$mushroom = 5;



// mushroom and reference_shroom are the integer 5, assigned_shroom is still a mushroom



?>



<?php



$reference_shroom = 5;



// same effect as example above



?>



<?php



$assigned_shroom = 5;



// only assigned_shroom is an integer



?>



The same would apply for any object type.



Once again, I reiterate, there's no need to overwrite variables and open this can of worms: just make a new variable and be done with it. As long as you don't overwrite variables, it doesn't matter whether or not you use references or assignments (unless you're planning for PHP4 compatibility).

ben AT chemica.co.uk
01-Feb-2006 01:40


PHP classes helpfully allow you to define member variables on the fly. On top of that, dynamic variables mean you can create them by name from strings.



This means you can ape the 'Ruby On Rails' view system - or replace heavyweight templating systems like Smarty - with very little code:



view.class.php:



<?



class View

{

    public function AddVariables($aVariables){

        foreach($aVariables as $name => $value){

            $this->AddVariable($name, $value);

        }

    }

    

    public function AddVariable($name, $value){

        $this->$name = $value; // <--- Note the extra $

    }

    

    public function Render($template){

       include($template . ".php");  

// You could put validation in here, so it exits cleanly if the template is missing.

    }

    

    protected function AddParagraph($strText){

        echo("<p>" . $strText . "</p>");

    }

    

    protected function AddOptions($aOptions, $strSelected){

        foreach($aOptions as $value => $text){

            echo "<option value='" . $value . "'";

            if($value==$strSelected){echo " selected";}

                echo ">" . $text . "</option>\n";

        }

    }

}



// The code below would normally be in a calling page/class.



$v = new View();

$v->AddVariables(array("peach"=>"fruit", "carrot"=>"vegetable"));

$v->AddVariables(array("options" => array("1"=>"Jan", "2"=>"Feb", "3"=>"March")));

$v->AddVariable("bodytext", "Some content goes here");



$v->Render("test");



?>



test.php:



<html>

<body>

<p>A peach is a <? echo $this->peach; ?>.</p>

<p>A carrot is a <? echo $this->carrot; ?>.<p>



<form action="#" method="GET">

<select id="test">

<? $this->AddOptions($this->options, 2); ?>

</select>

</form>

<? $this->AddParagraph($this->bodytext); ?>

</body>

</html>



This means you can unleash the awesome power of PHP on your templates, allowing all kinds of cool tricks, while ensuring you're not mixing up your display and business code. You only have access to the variables you pass in there yourself. Oh, and the library of markup rendering member functions you'll create. In PHP.



You can re-use templates - just use the same variables with a different page. 



You can swap in different mark-up languages programatically, so you could make versions that spit out HTML, XHTML, WML, and XML for SOAP interfaces.



It's entirely extensible. Either write in more PHP member functions to automate view-based tasks, or derive new classes and extend the functionality that way. (Perfect for your proprietary XML language helper functions - they really don't belong in the same class that's generating XHTML.)



The best part is that it's all PHP, so you don't have to learn a new language (as you would do with Smarty - or if you wanted to switch over to Ruby on Rails.)



Hope someone finds it useful. If anyone makes any improvements, corrections etc. I'd love to see them. :)



You'll need a recent version of PHP5 to run it - some earlier versions gave strange scoping errors with the use of $this. I know that build date 'Jan 11 2006 16:35:21' works.

mrich at runbox dot removethis dot com
06-Dec-2005 09:45


PHP 4.3.10:



<?php



class foo {



    var $what;



    function foo($newWhat = 'foo') {

        $this->what = $newWhat;

    }



    function bar() {

        print "<p>I'm a ".$this->what."</p>\n";

    }



}



$var = 'foo';



$obj = new $var;

// prints "I'm a foo"

$obj->bar();



$obj = new $var();

// prints "I'm a foo"

$obj->bar();



$obj = new $var('bar');

// prints "I'm a bar"

$obj->bar();



?>

npugh at tacc dot utah dot edu
29-Nov-2005 10:10


Note that a function defined in the parent class may not access a private member of a child class. 



<?php 

    class A

    {

        public function foo() { echo $this->x; } 

    }

    

    class B extends A

    {

        protected $x = "class B's member";

    }

    

    class C extends A

    {

        private $x = "class C's member";

    }

    

    $b = new B(); 

    $b->foo();

    $c = new C(); 

    $c->foo();

    

    // Output: 

    // class B's member

    // Fatal error: Cannot access private property C::$x

?>



In the above example, if you don't want to duplicate function foo() in all of the child classes you would have to make each member foo() operates on protected or public.  Effectively breaking encapsulation to in order to implement polymorphism.

PHP at Rulez dot com
10-Oct-2005 05:35


Check this!!!



<?php

class foo{

  function bar() {

    return $this;

  }

  function hello() {

    echo "Hello";

  }

}

$foo = new foo();

$foo->bar()->bar()->bar()->bar()->hello();

?>



Haaa! Rulezzz!

chris dot good at NOSPAM dot geac dot com
13-Sep-2005 04:32


Note that Class names seem to be case-insensitive in php 5.

eg 

class abc

{

..

{

class def extends ABC

{

..

}



works fine.

jerry
12-Jun-2005 09:30


Note the little correction on graced's notes below:



Your explanation looks correct about how instances created and destroyed, just wanted to correct some errors in your code below. But Thank you greatly for explaining that I appreciate it.



<?php

  $obj1 = new stdclass();

  var_dump($obj1);  // object #1

  $obj1 = new stdclass();

  var_dump($obj1);  // object #2

  // no more references to object #1, so it is destroyed.

  $obj1 = new stdclass();  // object #1 (object #2 destroyed right after this expression)

  var_dump($obj1);

  $obj2 = $obj1;  // two variables now reference object #1

  $obj1 = new stdclass();  // becomes object #2

  var_dump($obj1);

  $obj1 = new stdclass(); // object #3

  var_dump($obj1);

?>



Output:



object(stdClass)#1 (0) {

}

object(stdClass)#2 (0) {

}

object(stdClass)#1 (0) {

}

object(stdClass)#2 (0) {

}

object(stdClass)#3 (0) {

}

rasmus at nospaam dot flajm dot com
05-Apr-2005 10:22


A very good (in my opinion) change in PHP5 is that objects are always passed by reference. In PHP4 they where passed by value, which cased a copy to be made for every call. (as long as you didn't use "&" of cause)



This example illustrates pass-by-reference, as the name changes in the same object, passed around two different ways:



<?php



class Mother {

  private $obj = null;

  public function setChild($o) {

    $this->obj = $o;

  }

  public function getChild() {

    return $this->obj;

  }

}



class Child {

  private $name = 'noname';

  public function setName($string) {

    $this->name = $string;

  }

  public function getName() {

    return $this->name;

  }

}



$mom = new Mother();

$frasse = new Child();

$frasse->setName('Johan');



$mom->setChild($frasse);

$frasse->setName('Frans');



print $frasse->getName() . "\n";

print $mom->getChild()->getName() . "\n";



?>



This will print:



Frans

Frans



In PHP4, this would print:



Frans

Johan

graced at monroe dot NOSPAM dot wednet dot edu
29-Mar-2005 06:46


In reponse to jerry's comment: "If you instantiate a class and assign it to a variable over and over again, php will start saving instances #1 and #2 in memory but will stop at these"...



This is because after assigning the new object to the variable, there is no longer anything referencing the old object so it is destroyed.  If there WAS another reference, say you assigned another variable to the first variable before setting the first to a new class, it'd be preserved.



In reality, this really makes no difference to the programmer, just thought an explanation was in order.



<?php

  $obj1 = new stdclass(); 

  var_dump($obj1);  // object #1

  $obj1 = new stdclass();

  var_dump($obj1);  // object #2

  // no more references to object #1, so it is destroyed.

  $obj1 = new stdclass();  // object #1 

  var_dump($obj1);

  $obj2 = $obj1;   // two variables now reference object #1

  $obj1 = new stdclass();  // becomes object #3

  // no more references to object #2, so it is destroyed

  var_dump($obj1);

  $obj1 = new stdclass(); // object #2

  //etc.

?>

tigr at mail15 dot com
01-Mar-2005 12:08


Objects are not being passed by reference as varables do. Let me try to explain:



Variable passing by reference means that two variables are being binded together, so that changing one variable leads to changes in the other. In fact, there is only one variable.



Object passing by reference is a bit different. It means that not the object itself is being passed (that would lead to copying it and all evil), but only reference to the object is being passed. Now, both VARIABLES point to the same object, BUT they DO NOT point to each other. There are TWO DIFFERENT variables. This means that if you change one VARIABLE, second one would still point to the same object.



So, adding reference operator still has some sense. Here is an example:



<?php



class sampleClass {

    public $id;

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

}

$object1 = new sampleClass(1);

$object2 = $object1;

echo $object1->id; // 1

echo $object2->id; // 1



$object2 = new sampleClass(2);

echo $object1->id; // 1 So, $object1 was not changed. It still links

                   // to the same object

echo $object2->id; // 2



$object3 = &$object1; // note the reference operator

$object3 = new sampleClass(3);



echo $object1->id; // 3 This time $object one was changed since it

                   // was bound with $object3

echo $object2->id; // 2

echo $object3->id; // 3



?>

jerry
12-Jan-2005 06:41


If you instantiate a class and assign it to a variable over and over again, php will start saving instances #1 and #2 in memory but will stop at these, and will not save more than those instances(ie. #3, #4, #5, #6...). Instead it will toggle between #1 and #2.



<?php

  $obj1 = new stdclass();

  var_dump($obj1);

  $obj1 = new stdclass();

  var_dump($obj1);

  $obj1 = new stdclass();

  var_dump($obj1);

  $obj1 = new stdclass();

  var_dump($obj1);

  $obj1 = new stdclass();

  var_dump($obj1);

  $obj1 = new stdclass();

  var_dump($obj1);

  $obj1 = new stdclass();

  var_dump($obj1);

?>

Richard (php at tsg1985 dot com)
13-Dec-2004 04:36


"Actually, the new variable will access a COPY of the instance and not the SAME instance. This is shown by the example: var_dump($assigned) was not NULLified because a COPY of the instance was assigned to $assigned..."



This is actually incorrect. However, it is explained really poorly in the article.



$instance = new SimpleClass();

This creates the new SimpleClass Object location in the memory. (This spot in memory stores all of the variable and the code to execute.)Then, it points the variable $instance to that location in the memory.



$assigned =$instance;

This takes the location that $instance is pointed to and points the variable $assigned to it as well.



$reference  =& $instance;

This line of code points $refrence to the memory spot which contains the location of the SimpleClass object, not the object!



$instance->var = '$assigned will have this value';

This changes the values stored in SimpleClass Object's memory. Since $assigned points to this location as well, $assinged>var is the same value.



$instance = null;

This tells instance to point at nothing. So, it breaks its link to the SimpleClass object. $refrence points to $instance still, but since $instance points to nothing $refrence will also point to nothing.



I know this is kind of tough to understand, so I made an animated gif showing the steps.

http://www.prism.gatech.edu/~gtg624r/Code_Explenation.gif

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