Is it possible to access outer local varialbe in a PHP sub-function?
In below code, I want to access variable $l in inner function bar. Declaring $l as global $l in bar doesn't work.
function foo()
{
$l = "xyz";
function bar()
{
echo $l;
}
bar();
}
foo();
You could probably use a Closure, to do just that...
Edit : took some time to remember the syntax, but here's what it would look like :
function foo()
{
$l = "xyz";
$bar = function () use ($l)
{
var_dump($l);
};
$bar();
}
foo();
And, running the script, you'd get :
$ php temp.php
string(3) "xyz"
A couple of note :
You must put a ; after the function's declaration !
You could use the variable by reference, with a & before it's name : use (& $l)
For more informations, as a reference, you can take a look at this page in the manual : Anonymous functions
You must use the use keyword.
$bar = function() use(&$l) {
};
$bar();
In the very very old PHP 5.2 and earlier this didn't work. The syntax you've got isn't a closure, but a definition of a global function.
function foo() { function bar() { } }
works the same as:
function foo() { include "file_with_function_bar.php"; }
If you execute function foo twice, PHP will complain that you've tried to re-define a (global) function bar.
You can read default value by:
function(){
return preg_match(
"yourVar = \d+"
, str_file_get_contents(functionFile)
, arrayToPutFieldsValue
);
}
If You would use two functons in the same time - it's like someone's using a spoon and You want to take a food from that spoon - You'll waste a food or some of You will starv.
Anyway - You would have to set a pointer somehow in a hard way.
It's impossible to get any field from other function or class without calling it to life.
Functions/methods are instance-like - they need to be called.
Share the common fields by accessing a global fields with synchronized functions.
function a()
{
function val1($arg=null)
{
static $a;
if ($arg !== null) $a = $arg;
else return $a;
}
function b()
{
val1('1234');
echo val1() . '<br>'; // shows: 1234
val1('my custom data');
echo val1() . '<br>'; // shows: my custom data
}
b();
}
a();
Used val1('my custom data') to set my value
Used val1() to get my value
Related
How can i set the function as a variable in php
Similar to list function
example: list($x,$y)=array(1,2); // this is okey ,but...
How do I create such a structure?
You are talking about variable function that if a variable name has parentheses appended to it, PHP will look for a function with the same name as whatever the variable evaluates to, and will attempt to execute it. Among other things, this can be used to implement callbacks, function tables, and so forth.
Here is the little example from PHP manual Variable Functions
function foo() {
echo "In foo()<br />\n";
}
function bar($arg = '')
{
echo "In bar(); argument was '$arg'.<br />\n";
}
// This is a wrapper function around echo
function echoit($string)
{
echo $string;
}
$func = 'foo';
$func(); // This calls foo()
$func = 'bar';
$func('test'); // This calls bar()
$func = 'echoit';
$func('test'); // This calls echoit()
and the other scenario is Anonymous functions, also known as closures, allow the creation of functions which have no specified name. They are most useful as the value of callback parameters, but they have many other uses.
$greet = function($name)
{
printf("Hello %s\r\n", $name);
};
$greet('World');
$greet('PHP');
<?php
$my_array = array("Dog","Cat","Horse");
list($a, $b, $c) = $my_array;
echo "I have several animals, a $a, a $b and a $c.";
?>
The list() function is used to assign values to a list of variables. Like array(), this is not really a function, but a language construct. list() is used to assign a list of variables in one operation.
I'm trying to store anonymous functions in a static array property of my class. These functions should be invoked later by their index, but calling
self::$arr['index']()
just doesn't work, while
$a = self::$arr['index'];
$a();
does!
This doesn't work:
class A {
private static $func = array('a' => '');
public function __construct() {
self::$func['a'] = create_function('$str', 'echo "$str";');
}
public function go($str) {
self::$func['a']($str); // Call the function directly
}
}
$a = new A();
$a->go("hooray"); // Outputs "Undefined variable: func"
But this does:
class A {
private static $func = array('a' => '');
public function __construct() {
self::$func['a'] = create_function('$str', 'echo "$str";');
}
public function go($str) {
$a = self::$func['a']; // Pass the function name to a variable
$a($str); // Call the function via the variable
}
}
$a = new A();
$a->go("hooray"); // Outputs "hooray"
Why?
I'm using PHP Version 5.4.3
this is the behavior of php's parser
$functionName['a'] = "hello";
self::$functionName['a']();
calls
self::hello();
... the very sad thing is that in php you can't do this:
(self::$functionName['a'])(); // doesn't work in php :(
as you can do in javascript, for example
what you can do is... use a temporary variable like you said
$a = self::$func['a'];
$a($parameter);
or
call_user_func(self::$func['a'], $parameter);
hope this helps...
in latest phps these features were added
$a['sss'] = function(){ echo 'bla'; };
$a['sss']();
class Bla
{
private $a;
function test()
{
$this->a['sss'] = function(){ echo 'bla2'; };
$this->a['sss']();
}
}
$c = new Bla();
$c->test();
and they work properly... so for some reason, this syntax doesn't work only when using the scope resolution operator ( ClassName:: self:: etc)
Well, in php you simply can not do that, it is a php feature. But you can use call_user_func or its relatives:
return call_user_func(self::$func['$a'], $str);
This is a consequence of how the PHP parser currently works. Since the function call () is evaluated before the static operator ::, you end up with the parser attempting to reference the local variable $func instead, and then giving you the error about $func being undefined (which it is, since there is no variable named $func in the method).
As you've discovered, you can solve this by doing two separate statements.
This is probably not the wisest question, but is it possible to return a variable from a function without redeclaring it?
function test(){
$a = 1;
return $a;
}
Would it somehow be possible to access $a with just calling the function? test();
not $a = test();
Is that the only way? I want to make $a available after calling the function.
You have to put the variable into the global scope:
http://codepad.org/PzKxWGbU
function test()
{
$GLOBALS['a'] = 'Hello World!';
}
test();
echo $a;
But it would be better to use a return value or references:
function test(&$ref)
{
$ref = 'Hello World!';
}
$var = '';
test($var);
I assume you want to change some context when you call the function.
This is possible. For example, all global variables share the same context across your application. If a function set's a global variable, its available in the global scope:
function change_global_variable($name, $value)
{
$GLOBALS[$name] = $value;
}
change_global_variable('a', 1);
echo $a;
However, whenever you do something like this, take care that you're destroying the modularity of your code. For example, if you have 50 functions like this and 80 variables, and then you need to change something, you need to change many, many places. And even harder, you don't remember which places you need to change and how they belong to each other.
It's normally better to make the context an object and pass it to the function that changes it, so it's more clear what the function does:
class Context extends ArrayObject {}
$context = new Context;
function change_context(Context $context, $name, $value)
{
$context[$name] = $value;
}
change_context($context, 'b', 2);
echo $context['b'], "\n";
This example demonstrates something that's called dependency injection. Instead of modifying the only one (global) context that exists in PHP, you are telling the function what it should modify, instead that it modifies hard-encoded global variables
There is a mix form of it, using the global variable scope to contain the context, so you spare to pass it as an additional parameter, however I only show it here for completeness, because it's not helping to retain modularity that well (however it's better than using more than one global variable):
class Context extends ArrayObject {}
$context = new Context;
function change_global_context($name, $value)
{
global $context;
$context[$name] = $value;
}
change_global_context('c', 3);
echo $context['c'], "\n";
Rule of thumb: Avoid global variables. Consider them of being very expensive and you don't want your code to be expensive.
No, it is not due to the scoping.
An option would be to declare $a before and then making it a global. It will then be available inside test:
$a = '';
function test(){
global $a;
$a = 1;
return $a;
}
echo $a; // 1
But may I ask why you want to do this? Are you just playing around or do you have a use case? You're most probably doing it wrong if you need this.
Maybe if you just want to show the variable you can use
function test(){
$a = 1;
echo $a;
return $a;
}
in this way if you want you can save the variable $a in a variable if not you can just show it calling the test function.
Hope it helps.
Nicola.
I was looking at the source for Drupal 7, and I found some things I hadn't seen before. I did some initial looking in the php manual, but it didn't explain these examples.
What does the keyword static do to a variable inside a function?
function module_load_all($bootstrap = FALSE) {
static $has_run = FALSE
It makes the function remember the value of the given variable ($has_run in your example) between multiple calls.
You could use this for different purposes, for example:
function doStuff() {
static $cache = null;
if ($cache === null) {
$cache = '%heavy database stuff or something%';
}
// code using $cache
}
In this example, the if would only be executed once. Even if multiple calls to doStuff would occur.
Seems like nobody mentioned so far, that static variables inside different instances of the same class remain their state. So be careful when writing OOP code.
Consider this:
class Foo
{
public function call()
{
static $test = 0;
$test++;
echo $test . PHP_EOL;
}
}
$a = new Foo();
$a->call(); // 1
$a->call(); // 2
$a->call(); // 3
$b = new Foo();
$b->call(); // 4
$b->call(); // 5
If you want a static variable to remember its state only for current class instance, you'd better stick to a class property, like this:
class Bar
{
private $test = 0;
public function call()
{
$this->test++;
echo $this->test . PHP_EOL;
}
}
$a = new Bar();
$a->call(); // 1
$a->call(); // 2
$a->call(); // 3
$b = new Bar();
$b->call(); // 1
$b->call(); // 2
Given the following example:
function a($s){
static $v = 10;
echo $v;
$v = $s;
}
First call of
a(20);
will output 10, then $v to be 20. The variable $v is not garbage collected after the function ends, as it is a static (non-dynamic) variable. The variable will stay within its scope until the script totally ends.
Therefore, the following call of
a(15);
will then output 20, and then set $v to be 15.
Static works the same way as it does in a class. The variable is shared across all instances of a function. In your particular example, once the function is run, $has_run is set to TRUE. All future runs of the function will have $has_run = TRUE. This is particularly useful in recursive functions (as an alternative to passing the count).
A static variable exists only in a
local function scope, but it does not
lose its value when program execution
leaves this scope.
See http://php.net/manual/en/language.variables.scope.php
To expand on the answer of Yang
If you extend a class with static variables, the individual extended classes will hold their "own" referenced static that's shared between instances.
<?php
class base {
function calc() {
static $foo = 0;
$foo++;
return $foo;
}
}
class one extends base {
function e() {
echo "one:".$this->calc().PHP_EOL;
}
}
class two extends base {
function p() {
echo "two:".$this->calc().PHP_EOL;
}
}
$x = new one();
$y = new two();
$x_repeat = new one();
$x->e();
$y->p();
$x->e();
$x_repeat->e();
$x->e();
$x_repeat->e();
$y->p();
outputs:
one:1
two:1
one:2
one:3 <-- x_repeat
one:4
one:5 <-- x_repeat
two:2
http://ideone.com/W4W5Qv
static variable in a function means that no matter how many times you call the function, there's only 1 variable.
<?php
class Foo{
protected static $test = 'Foo';
function yourstatic(){
static $test = 0;
$test++;
echo $test . "\n";
}
function bar(){
$test = 0;
$test++;
echo $test . "\n";
}
}
$f = new Foo();
$f->yourstatic(); // 1
$f->yourstatic(); // 2
$f->yourstatic(); // 3
$f->bar(); // 1
$f->bar(); // 1
$f->bar(); // 1
?>
Inside a function, static means that the variable will retain its value each time the function is called during the life of the page load.
Therefore in the example you've given, if you call a function twice, if it set $has_run to true, then the function would be able to know that it had previously been called because $has_run would still be equal to true when the function starts the second time.
The usage of the static keyword in this context is explained in the PHP manual here: http://php.net/manual/en/language.variables.scope.php
In PHP, I am able to use a normal function as a variable without problem, but I haven't figured out how to use a static method. Am I just missing the right syntax, or is this not possible?
(EDIT: the first suggested answer does not seem to work. I've extended my example to show the errors returned.)
function foo1($a,$b) { return $a/$b; }
class Bar
{
static function foo2($a,$b) { return $a/$b; }
public function UseReferences()
{
// WORKS FINE:
$fn = foo1;
print $fn(1,1);
// WORKS FINE:
print self::foo2(2,1);
print Bar::foo2(3,1);
// DOES NOT WORK ... error: Undefined class constant 'foo2'
//$fn = self::foo2;
//print $fn(4,1);
// DOES NOT WORK ... error: Call to undefined function self::foo2()
//$fn = 'self::foo2';
//print $fn(5,1);
// DOES NOT WORK ... error: Call to undefined function Bar::foo2()
//$fn = 'Bar::foo2';
//print $fn(5,1);
}
}
$x = new Bar();
$x->UseReferences();
(I am using PHP v5.2.6 -- does the answer change depending on version too?)
PHP handles callbacks as strings, not function pointers. The reason your first test works is because the PHP interpreter assumes foo1 as a string. If you have E_NOTICE level error enabled, you should see proof of that.
"Use of undefined constant foo1 - assumed 'foo1'"
You can't call static methods this way, unfortunately. The scope (class) is relevant so you need to use call_user_func instead.
<?php
function foo1($a,$b) { return $a/$b; }
class Bar
{
public static function foo2($a,$b) { return $a/$b; }
public function UseReferences()
{
$fn = 'foo1';
echo $fn(6,3);
$fn = array( 'self', 'foo2' );
print call_user_func( $fn, 6, 2 );
}
}
$b = new Bar;
$b->UseReferences();
In php 5.2, you can use a variable as the method name in a static call, but to use a variable as the class name, you'll have to use callbacks as described by BaileyP.
However, from php 5.3, you can use a variable as the class name in a static call. So:
class Bar
{
public static function foo2($a,$b) { return $a/$b; }
public function UseReferences()
{
$method = 'foo2';
print Bar::$method(6,2); // works in php 5.2.6
$class = 'Bar';
print $class::$method(6,2); // works in php 5.3
}
}
$b = new Bar;
$b->UseReferences();
?>
You could use the full name of static method, including the namespace.
<?php
function foo($method)
{
return $method('argument');
}
foo('YourClass::staticMethod');
foo('Namespace\YourClass::staticMethod');
The name array array('YourClass', 'staticMethod') is equal to it. But I think the string may be more clear for reading.
In PHP 5.3.0, you could also do the following:
<?php
class Foo {
static function Bar($a, $b) {
if ($a == $b)
return 0;
return ($a < $b) ? -1 : 1;
}
function RBar($a, $b) {
if ($a == $b)
return 0;
return ($a < $b) ? 1 : -1;
}
}
$vals = array(3,2,6,4,1);
$cmpFunc = array('Foo', 'Bar');
usort($vals, $cmpFunc);
// This would also work:
$fooInstance = new Foo();
$cmpFunc = array('fooInstance', 'RBar');
// Or
// $cmpFunc = array('fooInstance', 'Bar');
usort($vals, $cmpFunc);
?>
Coming from a javascript background and being spoiled by it, I just coded this:
function staticFunctionReference($name)
{
return function() use ($name)
{
$className = strstr($name, '::', true);
if (class_exists(__NAMESPACE__."\\$className")) $name = __NAMESPACE__."\\$name";
return call_user_func_array($name, func_get_args());
};
}
To use it:
$foo = staticFunctionReference('Foo::bar');
$foo('some', 'parameters');
It's a function that returns a function that calls the function you wanted to call. Sounds fancy but as you can see in practice it's piece of cake.
Works with namespaces and the returned function should work just like the static method - parameters work the same.
This seems to work for me:
<?php
class Foo{
static function Calc($x,$y){
return $x + $y;
}
public function Test(){
$z = self::Calc(3,4);
echo("z = ".$z);
}
}
$foo = new Foo();
$foo->Test();
?>
In addition to what was said you can also use PHP's reflection capabilities:
class Bar {
public static function foo($foo, $bar) {
return $foo . ' ' . $bar;
}
public function useReferences () {
$method = new ReflectionMethod($this, 'foo');
// Note NULL as the first argument for a static call
$result = $method->invoke(NULL, '123', 'xyz');
}
}
"A member or method declared with static can not be accessed with a variable that is an instance of the object and cannot be re-defined in an extending class"
(http://theserverpages.com/php/manual/en/language.oop5.static.php)