As I am learning how PHP OOP works, I stumbled upon the following confusion.
class Foo {
static function baz() {
echo 'works';
}
}
# 1
$a = 'Foo';
$a::baz();
# 2
Foo::baz();
PHP manual states that since 5.3.0, it is possible to reference a static class via a string (http://php.net/manual/en/language.oop5.static.php)
What I don't understand is, what is the different between #1 and #2? Aren't they technically the same since both are calling a static function without initiating a class? Where can #1 be applied in practical scenario?
The difference is in a version. Before PHP5.3 you cannot use static methods with variable.
Also, second one 'hide' using of a class. Your IDE could not find class usage.
Also you could use object to call static methods.
class Foo {
static function baz() {
echo 'works';
}
}
eval(
'$a = "Foo";
$a::baz();');
eval(
'$a = new Foo();
$a::baz();');
eval(
'Foo::baz();');
http://3v4l.org/WOK44
Related
what is the best way (memory safe) to use call_user_func from self instance eg.:
call_user_func(array($this, 'method'));
or by reference, eg.:
call_user_func(array(&$this, 'method'));
PS. i know the best way is $this->method but for framework design pattern reason (Wordpress) i do use call_user_func
If I am aware, it doesn't really matter, because PHP will do the same thing in both cases: will just pass reference to the same object (the object will be not cloned). So for simplicity, I would go without &
Some test:
class test
{
function foo() {
call_user_func([$this, 'bar']);
call_user_func([&$this, 'bar']);
}
private function bar() {
echo spl_object_hash($this), PHP_EOL;
}
}
and outputs:
php > $obj = new test();
php > $obj->foo();
0000000025077e4a0000000075174bd2
0000000025077e4a0000000075174bd2
I use eval() in my current project like this:
if (class_exists($class_name)) //$class_name depends on user input
eval($class_name.'::MyStaticMethod()');
eval() is executed if and only if class with the name $class_name exists so it's kinda safe, but I still don't think that this is the best solution.
Can I do the same what code above does without eval()?
I have recently answered this question. The last part of my answer perfectly answers this question and is much more useful for future readers than answers provided here. That's why I am answering my own question.
PHP has features that gives possibility to avoid using eval in most cases:
PHP is very dynamic language. It has ability to do following stuff with strings:
Define and/or get variable (supported from PHP 4.3). For example:
$variableName = 'MyVariable';
// Create new variable with the name defined in variable $variableName
${$variableName} = 'MyValue';
//Outputs: string(7) "MyValue"
var_dump($MyVariable);
//Outputs: string(7) "MyValue"
var_dump(${'MyVariable'});
Demo
Call function (supported from PHP 4.3). For example:
// Create function with the name defined in variable $functionName
function MyFunction($argument) {
return 'Argument passed is: '.$argument;
}
$functionName = 'MyFunction';
// Outputs:
// string(48) "Argument passed is: Calling MyFunction directly."
var_dump(MyFunction('Calling MyFunction directly.'));
// Outputs:
// string(51) "Argument passed is: Calling MyFunction with string."
var_dump($functionName('Calling MyFunction with string.'));
Demo
Create instance of class (supported from PHP 5.0). For example:
class MyClass {
public function __construct() {
echo 'Constructing MyClass'."\n";
}
}
$className = 'MyClass';
$objFromString = new $className();
// Outputs: object(MyClass)#1 (0) {}
var_dump($objFromString);
Demo
Call static method (supported from PHP 5.0). For example:
class MyClass {
public static function staticMethod() {
return 'MyClass::staticMethod called';
}
}
$staticMethodName = 'staticMethod';
// Outputs: string(28) "MyClass::staticMethod called"
var_dump(MyClass::$staticMethodName());
Demo
And from PHP 5.3 class name can also be defined by string. Example:
class MyClass {
public static function staticMethod() {
return 'MyClass::staticMethod called';
}
}
$className = 'MyClass';
$staticMethodName = 'staticMethod';
var_dump($className::$staticMethodName());
var_dump($className::staticMethod());
Demo
Call instance method of object (supported from PHP 5.0). For example:
class MyClass {
public function instanceMethod() {
return 'MyClass::instanceMethod called';
}
}
$methodName = 'instanceMethod';
$obj = new MyClass();
// Outputs: string(30) "MyClass::instanceMethod called"
var_dump($obj->$methodName());
Demo
Access static and instance properties of object (supported from PHP 5.0). For example:
class MyClass {
public static $myStaticProperty;
public $myInstanceProperty;
}
$staticPropertyName = 'myStaticProperty';
$instancePropertyName = 'myInstanceProperty';
MyClass::${$staticPropertyName} = 'my static value';
$obj = new MyClass();
$obj->{$instancePropertyName} = 'my instance value';
var_dump(MyClass::${$staticPropertyName});
var_dump($obj->{$instancePropertyName});
Demo
PHP has two functions: call_user_func and call_user_func_array for dynamic function/method calls. Both are perfectly documented so I won't go in details here.
Even if everything above is not enough PHP 5 comes with great Reflection API. Unfortunately, documentation has few examples but reflection is quite large topic to cover here. Basically, It's not a big deal to use reflection after reading how it works.
I'd suggest call_user_func.
An alternative to call_user_func() would be calling it like this:
$class_and_method = 'Class::MyStaticMethod()';
$class_and_method();
yes:
call_user_func(array($class_name, 'MyStaticMethod'));
As of PHP 5.3+,
$class_name::MyStaticMethod();
Adisory:
userinput + eval = security hole;
Also eval is an expensive operation requiring parsing the string into an actionable format (parse tree, abstract syntax tree, etc.) and executing the new found logic.
You don't want to eval every little tidbit of code. Use eval if you have something for it to chew on or rather put that logic somewhere where it's reusable and parametrized such as a function.
Also as of php 5.4
$method = array('class_name', 'method_name');
$method(); // calls class_name::method_name()
I have found different information regarding static variables in PHP but nothing that actually explains what it is and how it really works.
I have read that when used within a class that a static property cannot be used by any object instantiated by that class and that a static method can be used by an object instantiated by the class?
However, I have been trying to research what a static variable does within a function that is not in a class. Also, does a static variable within a function work somewhat like closure in javascript or am I totally off in this assumption?
I have read that when used within a class that a static property cannot be used by any object instantiated by that class
It depends on what you mean by that. eg:
class Foo {
static $my_var = 'Foo';
}
$x = new Foo();
echo $x::$my_var; // works fine
echo $x->my_var; // doesn't work - Notice: Undefined property: Foo::$my_var
and that a static method can be used by an object instantiated by the class???
Yes, an instantiated object belonging to the class can access a static method.
The keyword static in the context of classes behave somewhat like static class variables in other languages. A member (method or variable) declared static is associated with the class and rather than an instance of that class. Thus, you can access it without an instance of the class (eg: in the example above, I could use Foo::$my_var)
However, I have been trying to research what a static variable does within a function that is not in a class.
Also, does a static variable within a function work somewhat like closure in javascript or am I totally off in this assumption.
Outside of classes (ie: in functions), a static variable is a variable that doesn't lose its value when the function exits. So in sense, yes, they work like closures in JavaScript.
But unlike JS closures, there's only one value for the variable that's maintained across different invocations of the same function. From the PHP manual's example:
function test()
{
static $a = 0;
echo $a;
$a++;
}
test(); // prints 0
test(); // prints 1
test(); // prints 2
Reference: static keyword (in classes), (in functions)
static has two uses in PHP:
First, and most commonly, it can be used to define 'class' variables/functions (as opposed to instance variables/functions), that can be accessed without instantiating a class:
class A {
public static $var = 'val'; // $var is static (in class context)
public $other_var = 'other_val'; // non-static
}
echo A::$var; // val
echo A::$other_var // doesn't work (fatal error, undefined static variable)
$a = new A;
echo $a->var // won't work (strict standards)
echo $a->other_var // other_val
Secondly, it can be used to maintain state between function calls:
function a() {
static $i = 0;
$j = 0;
return array($i++, $j++);
}
print_r(a()); // array(0, 0)
print_r(a()); // array(1, 0)
print_r(a()); // array(2, 0)
//...
Note that declaring a variable static within a function works the same regardless of whether or not the function is defined in a class, all that matters is where the variable is declared (class member or in a function).
A static variable in a function is initialized only in the first call of that function in its running script.
At first i will explain what will happen if static variable is not used
<?php
function somename() {
$var = 1;
echo $var . "<br />";
$var++;
}
somename();
somename();
somename();
?>
If you run the above code the output you gets will be 1 1 1 . Since everytime you called that function variable assigns to 1 and then prints it.
Now lets see what if static variable is used
<?php
function somename() {
static $var = 1;
echo $var . "<br />";
$var++;
}
somename();
somename();
somename();
?>
Now if you run this code snippet the output will be 1 2 3.
Note: Static keeps its value and stick around everytime the function is called. It will not lose its value when the function is called.
class Student {
static $total_student = 0;
static function add_student(){
return Student::$total_student++;
}
}
First: for the add_student function, the best practice is to use static not public.
Second: in the add_student function, we are using Student::$total_student,not use $this->total_student. This is big different from normal variable.
Third:static variable are shared throughout the inheritance tree.
take below code to see what is the result:
class One {
static $foo ;
}
class Two extends One{}
class Three extends One{}
One::$foo = 1;
Two::$foo = 2;
Three::$foo = 3;
echo One::$foo;
echo Two::$foo;
echo Three::$foo;`
I'm getting an error that I think is because I've made some kind of mistake in a refactoring, but I can't find documentation on when $this is bound, and my error could be explained by it being bound statically.
Extra points (I can't actually give you extra points) for links to excellent documentation about this kind of thing in php.
[Edit]
The error that I'm getting is telling me that Subclass::$var doesn't exist when I do, for example, echo $this->var in a superclass. The $var exists in the subclass, though.
$this becomes available after you've called the constructor. Logically you can't use $this in a static function.
Aside from calling $this in a static function there isn't a whole lot that can go wrong timing wise as there is simply no way in PHP.
What exactly is the error you're getting? Code would useful too.
This works in PHP:
class A {
public function foo() {
echo $this->bar;
}
}
class B extends A {
public $bar = 1;
}
$b = new B;
$b->foo(); // 1
It works because of the dynamic scope resolution that PHP has (i.e.: scope is resolved at runtime as opposed to compile time). However, I'd recommend against it, because it is really a particularity of the language for one. For second, failing to declare $bar in a subclass would result in an error. I think that a class should only reference members that it is aware of.
The same code, say in C++:
class A {
public:
void foo() {
std::cout << bar;
}
};
class B : public A {
public:
int bar;
B() {
bar = 1;
}
};
...would give you a compile error (In A::foo(): 'bar' was not declared in this scope).
Yes, $this is bound dynamically, as is evidenced by the fact that the output of the following is "foo":
<?php
class Base
{
public function ecc(){
echo $this->subvar;
}
}
class Sub extends Base
{
public $subvar;
public function __construct(){
$this->subvar = 'foo';
$this->ecc();
}
}
new Sub();
?>
I am trying to access a static method, but using a variable as the class name. Is this possible? I seem to be having issues with it. I want to be able to do something like this:
class foo {
public static function bar() {
echo 'test';
}
}
$variable_class_name = 'foo';
$variable_class_name::bar();
And I want to be able to do similar using static variables as well.
That syntax is only supported in PHP 5.3 and later. Previous versions don't understand that syntax, hence your parse error (T_PAAMAYIM_NEKUDOTAYIM refers to the :: operator).
In previous versions you can try call_user_func(), passing it an array containing the class name and its method name:
$variable_class_name = 'foo';
call_user_func(array($variable_class_name, 'bar'));
You can use reflection for PHP 5.1 and above:
class foo {
public static $bar = 'foobar';
}
$class = 'foo';
$reflector = new ReflectionClass($class);
echo $reflector->getStaticPropertyValue('bar');
> foobar