变量范围

变量的范围即它定义的上下文背景(也就是它的生效范围)。大部分的 PHP 变量只有一个单独的范围。这个单独的范围跨度同样包含了 include 和 require 引入的文件。例如:

<?php
$a 
1;
include 
'b.inc';
?>

这里变量 $a 将会在包含文件 b.inc 中生效。但是,在用户自定义函数中,一个局部函数范围将被引入。任何用于函数内部的变量按缺省情况将被限制在局部函数范围内。例如:

<?php
$a 
1/* global scope */

function Test()
{
    echo 
$a/* reference to local scope variable */
}

Test();
?>

这个脚本不会有任何输出,因为 echo 语句引用了一个局部版本的变量 $a,而且在这个范围内,它并没有被赋值。你可能注意到 PHP 的全局变量和 C 语言有一点点不同,在 C 语言中,全局变量在函数中自动生效,除非被局部变量覆盖。这可能引起一些问题,有些人可能不小心就改变了一个全局变量。PHP 中全局变量在函数中使用时必须声明为 global。

global 关键字

首先,一个使用 global 的例子:

示例 #1 使用 global

<?php
$a 
1;
$b 2;

function 
Sum()
{
    global 
$a$b;

    
$b $a $b;
}

Sum();
echo 
$b;
?>

以上脚本的输出将是“3”。在函数中声明了全局变量 $a$b 之后,对任一变量的所有引用都会指向其全局版本。对于一个函数能够声明的全局变量的最大个数,PHP 没有限制。

在全局范围内访问变量的第二个办法,是用特殊的 PHP 自定义 $GLOBALS 数组。前面的例子可以写成:

示例 #2 使用 $GLOBALS 替代 global

<?php
$a 
1;
$b 2;

function 
Sum()
{
    
$GLOBALS['b'] = $GLOBALS['a'] + $GLOBALS['b'];
}

Sum();
echo 
$b;
?>

$GLOBALS 是一个关联数组,每一个变量为一个元素,键名对应变量名,值对应变量的内容。$GLOBALS 之所以在全局范围内存在,是因为 $GLOBALS 是一个超全局变量。以下范例显示了超全局变量的用处:

示例 #3 演示超全局变量和作用域的例子

<?php
function test_superglobal()
{
    echo 
$_POST['name'];
}
?>

使用静态变量

变量范围的另一个重要特性是静态变量(static variable)。静态变量仅在局部函数域中存在,但当程序执行离开此作用域时,其值并不丢失。看看下面的例子:

示例 #4 演示需要静态变量的例子

<?php
function Test()
{
    
$a 0;
    echo 
$a;
    
$a++;
}
?>

本函数没什么用处,因为每次调用时都会将 $a 的值设为 0 并输出 0。将变量加一的 $a++ 没有作用,因为一旦退出本函数则变量 $a 就不存在了。要写一个不会丢失本次计数值的计数函数,要将变量 $a 定义为静态的:

示例 #5 使用静态变量的例子

<?php
function test()
{
    static 
$a 0;
    echo 
$a;
    
$a++;
}
?>

现在,变量 $a 仅在第一次调用 test() 函数时被初始化,之后每次调用 test() 函数都会输出 $a 的值并加一。

静态变量也提供了一种处理递归函数的方法。递归函数是一种调用自己的函数。写递归函数时要小心,因为可能会无穷递归下去。必须确保有充分的方法来中止递归。以下这个简单的函数递归计数到 10,使用静态变量 $count 来判断何时停止:

示例 #6 静态变量与递归函数

<?php
function test()
{
    static 
$count 0;

    
$count++;
    echo 
$count;
    if (
$count 10) {
        
test();
    }
    
$count--;
}
?>

Static variables can be assigned values which are the, result of constant expressions, but dynamic expressions, such as function calls, will cause a parse error.

示例 #7 声明静态变量

<?php
function foo(){
    static 
$int 0;          // correct
    
static $int 1+2;        // correct 
    
static $int sqrt(121);  // wrong  (as it is a function)

    
$int++;
    echo 
$int;
}
?>

静态声明是在编译时解析的。

全局和静态变量的引用

对于变量的 staticglobal 定义是以引用的方式实现的。例如,在一个函数域内部用 global 语句导入的一个真正的全局变量实际上是建立了一个到全局变量的引用。这有可能导致预料之外的行为,如以下例子所演示的:

<?php
function test_global_ref() {
    global 
$obj;
    
$new = new stdclass;
    
$obj = &$new;
}

function 
test_global_noref() {
    global 
$obj;
    
$new = new stdclass;
    
$obj $new;
}

test_global_ref();
var_dump($obj);
test_global_noref();
var_dump($obj);
?>

以上例程会输出:

NULL
object(stdClass)#1 (0) {
}

类似的行为也适用于 static 语句。引用并不是静态地存储的:

<?php
function &get_instance_ref() {
    static 
$obj;

    echo 
'Static object: ';
    
var_dump($obj);
    if (!isset(
$obj)) {
        
$new = new stdclass;
        
// 将一个引用赋值给静态变量
        
$obj = &$new;
    }
    if (!isset(
$obj->property)) {
        
$obj->property 1;
    } else {
        
$obj->property++;
    }
    return 
$obj;
}

function &
get_instance_noref() {
    static 
$obj;

    echo 
'Static object: ';
    
var_dump($obj);
    if (!isset(
$obj)) {
        
$new = new stdclass;
        
// 将一个对象赋值给静态变量
        
$obj $new;
    }
    if (!isset(
$obj->property)) {
        
$obj->property 1;
    } else {
        
$obj->property++;
    }
    return 
$obj;
}

$obj1 get_instance_ref();
$still_obj1 get_instance_ref();
echo 
"\n";
$obj2 get_instance_noref();
$still_obj2 get_instance_noref();
?>

以上例程会输出:

Static object: NULL
Static object: NULL

Static object: NULL
Static object: object(stdClass)#3 (1) {
  ["property"]=>
  int(1)
}

上例演示了当把一个引用赋值给一个静态变量时,第二次调用 &get_instance_ref() 函数时其值并没有被记住

add a note add a note

User Contributed Notes 41 notes

up
153
warhog at warhog dot net
18 years ago
Some interesting behavior (tested with PHP5), using the static-scope-keyword inside of class-methods.

<?php

class sample_class
{
  public function
func_having_static_var($x = NULL)
  {
    static
$var = 0;
    if (
$x === NULL)
    { return
$var; }
   
$var = $x;
  }
}

$a = new sample_class();
$b = new sample_class();

echo
$a->func_having_static_var()."\n";
echo
$b->func_having_static_var()."\n";
// this will output (as expected):
//  0
//  0

$a->func_having_static_var(3);

echo
$a->func_having_static_var()."\n";
echo
$b->func_having_static_var()."\n";
// this will output:
//  3
//  3
// maybe you expected:
//  3
//  0

?>

One could expect "3 0" to be outputted, as you might think that $a->func_having_static_var(3); only alters the value of the static $var of the function "in" $a - but as the name says, these are class-methods. Having an object is just a collection of properties, the functions remain at the class. So if you declare a variable as static inside a function, it's static for the whole class and all of its instances, not for each object.

Maybe it's senseless to post that.. cause if you want to have the behaviour that I expected, you can simply use a variable of the object itself:

<?php
class sample_class
{ protected $var = 0;
  function
func($x = NULL)
  {
$this->var = $x; }
}
?>

I believe that all normal-thinking people would never even try to make this work with the static-keyword, for those who try (like me), this note maybe helpfull.
up
128
dodothedreamer at gmail dot com
12 years ago
Note that unlike Java and C++, variables declared inside blocks such as loops or if's, will also be recognized and accessible outside of the block, so:
<?php
for($j=0; $j<3; $j++)
{
     if(
$j == 1)
       
$a = 4;
}
echo
$a;
?>

Would print 4.
up
72
HOSSEIN doesn&#39;t want spam at TAKI.IR
13 years ago
Please note for using global variable in child functions:

This won't work correctly...

<?php
function foo(){
   
$f_a = 'a';
   
    function
bar(){
        global
$f_a;
        echo
'"f_a" in BAR is: ' . $f_a . '<br />'// doesn't work, var is empty!
   
}
   
   
bar();
    echo
'"f_a" in FOO is: ' . $f_a . '<br />';
}
?>

This will...

<?php
function foo(){
    global
$f_a;   // <- Notice to this
   
$f_a = 'a';
   
    function
bar(){
        global
$f_a;
        echo
'"f_a" in BAR is: ' . $f_a . '<br />'// work!, var is 'a'
   
}
   
   
bar();
    echo
'"f_a" in FOO is: ' . $f_a . '<br />';
}
?>
up
31
Michael Bailey (jinxidoru at byu dot net)
19 years ago
Static variables do not hold through inheritance.  Let class A have a function Z with a static variable.  Let class B extend class A in which function Z is not overwritten.  Two static variables will be created, one for class A and one for class B.

Look at this example:

<?php
class A {
    function
Z() {
        static
$count = 0;       
       
printf("%s: %d\n", get_class($this), ++$count);
    }
}

class
B extends A {}

$a = new A();
$b = new B();
$a->Z();
$a->Z();
$b->Z();
$a->Z();
?>

This code returns:

A: 1
A: 2
B: 1
A: 3

As you can see, class A and B are using different static variables even though the same function was being used.
up
26
andrew at planetubh dot com
15 years ago
Took me longer than I expected to figure this out, and thought others might find it useful.

I created a function (safeinclude), which I use to include files; it does processing before the file is actually included (determine full path, check it exists, etc).

Problem: Because the include was occurring inside the function, all of the variables inside the included file were inheriting the variable scope of the function; since the included files may or may not require global variables that are declared else where, it creates a problem.

Most places (including here) seem to address this issue by something such as:
<?php
//declare this before include
global $myVar;
//or declare this inside the include file
$nowglobal = $GLOBALS['myVar'];
?>

But, to make this work in this situation (where a standard PHP file is included within a function, being called from another PHP script; where it is important to have access to whatever global variables there may be)... it is not practical to employ the above method for EVERY variable in every PHP file being included by 'safeinclude', nor is it practical to staticly name every possible variable in the "global $this" approach. (namely because the code is modulized, and 'safeinclude' is meant to be generic)

My solution: Thus, to make all my global variables available to the files included with my safeinclude function, I had to add the following code to my safeinclude function (before variables are used or file is included)

<?php
foreach ($GLOBALS as $key => $val) { global $$key; }
?>

Thus, complete code looks something like the following (very basic model):

<?php
function safeinclude($filename)
{
   
//This line takes all the global variables, and sets their scope within the function:
   
foreach ($GLOBALS as $key => $val) { global $$key; }
   
/* Pre-Processing here: validate filename input, determine full path
        of file, check that file exists, etc. This is obviously not
        necessary, but steps I found useful. */
   
if ($exists==true) { include("$file"); }
    return
$exists;
}
?>

In the above, 'exists' & 'file' are determined in the pre-processing. File is the full server path to the file, and exists is set to true if the file exists. This basic model can be expanded of course.  In my own, I added additional optional parameters so that I can call safeinclude to see if a file exists without actually including it (to take advantage of my path/etc preprocessing, verses just calling the file exists function).

Pretty simple approach that I could not find anywhere online; only other approach I could find was using PHP's eval().
up
9
zweibieren at yahoo dot com
8 years ago
Take to heart this hard-won rule:
        Declare AT THE TOP any variable that is to be global.
        Both at the top of the FILE
        AND at the top of any FUNCTION where it appears.

Why AT THE TOP? So it is sure to be declared before use. Otherwise a non-global version of the variable will be created and your code will fail.

Why at the top of a FUNCTION? Because otherwise the function will refer only to its local version of the variable and your code will fail.

Why at the top of the FILE? Because someday--a day that you cannot now imagine--you will want to "include" the file. And when you do, instances of the variable outside functions will not go in the global scope and your code will fail. (When the "include" is inside a calling function, variables in the included file go into the scope of the calling function.)

Example file where variable $x is used outside and inside functions:
    |<!DOCTYPE html ...>
    |<html xmlns ...>
    |    <?php global $x; ?>
    |<head>
    |    Some html headers
    |    <?php
   
|        $x = 1;
    |        function
bump_x() {
    |            global
$x;
    |           
$x += 1;
    |        }
    |   
?>
    |</head>
    |<body>
    |    More html
    |    <?php echo $x; bump_x(); ?>
    |    Yet more html.
    |</body>
</html>
up
18
larax at o2 dot pl
18 years ago
About more complex situation using global variables..

Let's say we have two files:
a.php
<?php
   
function a() {
        include(
"b.php");
    }
   
a();
?>

b.php
<?php
    $b
= "something";
    function
b() {
        global
$b;
       
$b = "something new";
    }
   
b();
    echo
$b;
?>

You could expect that this script will return "something new" but no, it will return "something". To make it working properly, you must add global keyword in $b definition, in above example it will be:

global $b;
$b = "something";
up
4
shaman_master at list dot ru
4 years ago
Cycle not create inner scope:
<?php
foreach (range(1, 3) as $step) {
    echo
sprintf('%d: %s  ', $step, isset($a) ? 'yes' : 'no');
    if (! isset(
$a)) {
       
$a = 1;
    }
}
// 1: no 2: yes 3: yes
?>
up
12
ddarjany at yahoo dot com
15 years ago
Note that if you declare a variable in a function, then set it as global in that function, its value will not be retained outside of that function.  This was tripping me up for a while so I thought it would be worth noting.

<?PHP

foo
();
echo
$a; // echoes nothing

bar();
echo
$b; //echoes "b";

function foo() {
 
$a = "a";
  global
$a;
}

function
bar() {
  global
$b;
 
$b = "b";
}

?>
up
12
php at keith tyler dot com
13 years ago
Sometimes a variable available in global scope is not accessible via the 'global' keyword or the $GLOBALS superglobal array. I have not been able to replicate it in original code, but it occurs when a script is run under PHPUnit.

PHPUnit provides a variable "$filename" that reflects the name of the file loaded on its command line. This is available in global scope, but not in object scope. For example, the following phpUnit script (call it GlobalScope.php):

<?php
print "Global scope FILENAME [$filename]\n";
class
MyTestClass extends PHPUnit_Framework_TestCase {
  function
testMyTest() {
    global
$filename;
    print
"Method scope global FILENAME [$filename]\n";
    print
"Method scope GLOBALS[FILENAME] [".$GLOBALS["filename"]."]\n";
  }
}
?>

If you run this script via "phpunit GlobalScope.php", you will get:

Global scope FILENAME [/home/ktyler/GlobalScope.php]
PHPUnit 3.4.5 by Sebastian Bergmann.

Method scope global FILENAME []
Method scope GLOBALS[FILENAME] []
.

You have to -- strange as it seems -- do the following:

<?php
$GLOBALS
["filename"]=$filename;
print
"Global scope FILENAME [$filename]\n";
class
MyTestClass extends PHPUnit_Framework_TestCase {
  function
testMyTest() {
    global
$filename;
    print
"Method scope global FILENAME [$filename]\n";
    print
"Method scope GLOBALS[FILENAME] [".$GLOBALS["filename"]."]\n";
  }
}
?>

By doing this, both "global" and $GLOBALS work!

I don't know what it is that PHPUnit does (I know it uses Reflection) that causes a globally available variable to be implicitly unavailable via "global" or $GLOBALS. But there it is.
up
10
dexen dot devries at gmail dot com
7 years ago
If you have a static variable in a method of a class, all DIRECT instances of that class share that one static variable.

However if you create a derived class, all DIRECT instances of that derived class will share one, but DISTINCT, copy of that static variable in method.

To put it the other way around, a static variable in a method is bound to a class (not to instance). Each subclass has own copy of that variable, to be shared among its instances.

To put it yet another way around, when you create a derived class, it 'seems  to' create a copy of methods from the base class, and thusly create copy of the static variables in those methods.

Tested with PHP 7.0.16.

<?php

require 'libs.php';
require
'setup.php';

class
Base {
    function
test($delta = 0) {
        static
$v = 0;
       
$v += $delta;
        return
$v;
    }
}

class
Derived extends Base {}

$base1 = new Base();
$base2 = new Base();
$derived1 = new Derived();
$derived2 = new Derived();

$base1->test(3);
$base2->test(4);
$derived1->test(5);
$derived2->test(6);

var_dump([ $base1->test(), $base2->test(), $derived1->test(), $derived2->test() ]);

# => array(4) { [0]=> int(7) [1]=> int(7) [2]=> int(11) [3]=> int(11) }

# $base1 and $base2 share one copy of static variable $v
# derived1 and $derived2 share another copy of static variable $v
up
8
danno at wpi dot edu
22 years ago
WARNING!  If you create a local variable in a function and then within that function assign it to a global variable by reference the object will be destroyed when the function exits and the global var will contain NOTHING!  This main sound obvious but it can be quite tricky you have a large script (like a phpgtk-based gui app ;-) ).

example:

<?php
function foo ()
{
   global
$testvar;

  
$localvar = new Object ();
  
$testvar = &$localvar;
}

foo ();
print_r ($testvar);   // produces NOTHING!!!!
?>

hope this helps someone before they lose all their hair
up
11
Anonymous
11 years ago
It will be obvious for most of you: changing value of a static in one instance changes value in all instances.

<?php

   
class example {
        public static
$s = 'unchanged';
       
        public function
set() {
           
$this::$s = 'changed';
        }
    }

   
$o = new example;
   
$p = new example;

   
$o->set();

    print
"$o static: {$o::$i}\n$p static: {$p::$i}";

?>

Output will be:

$o static: changed
$p static: changed
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6
pogregoire##live.fr
7 years ago
writing : global $var; is exactely the samething that writing : $var =& $GLOBALS['var'];
It creates a reference on $GLOBALS['var'];

<?php
$var
=1;
function
teste_global(){
    global
$var;
    for (
$var=0; $var<5; $var++){

    }
}

teste_global();
var_dump($var);// return : int(5).
?>
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6
Randolpho
20 years ago
More on static variables:

A static variable does not retain it's value after the script's execution. Don't count on it being available from one page request to the next; you'll have to use a database for that.

Second, here's a good pattern to use for declaring a static variable based on some complex logic:

<?php
 
function buildStaticVariable()
  {
     
$foo = null;
     
// some complex expression or set of
      // expressions/statements to build
      // the return variable.
     
return $foo;
  }

  function
functionWhichUsesStaticVar()
  {
      static
$foo = null;
      if(
$foo === null) $foo = buildStaticVariable();
     
// the rest of your code goes here.
 
}
?>

Using such a pattern allows you to separate the code that creates your default static variable value from the function that uses it. Easier to maintain code is good. :)
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4
kouber at php dot net
19 years ago
If you need all your global variables available in a function, you can use this:

<?php
function foo() {
 
extract($GLOBALS);
 
// here you have all global variables

}
?>
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5
Jonathan Kenigson
10 years ago
Just a note about static properties declared at class level:

class Test_Class {
  static $a = 0;
  public function ReturnVar(){
    return $this->a;
  }
  }
  $b = new Test_Class();
  echo $b->ReturnVar();

Will not output "0"  because $a is declared static. Changing "static" to "public" or "private" will produce the output "0".
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3
gried at NOSPAM dot nsys dot by
8 years ago
In fact all variables represent pointers that hold address of memory area with data that was assigned to this variable. When you assign some variable value by reference you in fact write address of source variable to recepient variable. Same happens when you declare some variable as global in function, it receives same address as global variable outside of function. If you consider forementioned explanation it's obvious that mixing usage of same variable declared with keyword global and via superglobal array at the same time is very bad idea. In some cases they can point to different memory areas, giving you headache. Consider code below:

<?php

error_reporting
(E_ALL);

$GLOB = 0;

function
test_references() {
    global
$GLOB; // get reference to global variable using keyword global, at this point local variable $GLOB points to same address as global variable $GLOB
   
$test = 1; // declare some local var
   
$GLOBALS['GLOB'] = &$test; // make global variable reference to this local variable using superglobal array, at this point global variable $GLOB points to new memory address, same as local variable $test

   
$GLOB = 2; // set new value to global variable via earlier set local representation, write to old address

   
echo "Value of global variable (via local representation set by keyword global): $GLOB <hr>";
   
// check global variable via local representation => 2 (OK, got value that was just written to it, cause old address was used to get value)

   
echo "Value of global variable (via superglobal array GLOBALS): $GLOBALS[GLOB] <hr>";
   
// check global variable using superglobal array => 1 (got value of local variable $test, new address was used)
   
   
echo "Value ol local variable \$test: $test <hr>";
   
// check local variable that was linked with global using superglobal array => 1 (its value was not affected)
   
   
global $GLOB; // update reference to global variable using keyword global, at this point we update address that held in local variable $GLOB and it gets same address as local variable $test
   
echo "Value of global variable (via updated local representation set by keyword global): $GLOB <hr>";
   
// check global variable via local representation => 1 (also value of local variable $test, new address was used)
}

test_references();
echo
"Value of global variable outside of function: $GLOB <hr>";
// check global variable outside function => 1 (equal to value of local variable $test from function, global variable also points to new address)
?>
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4
eduardo dot ferron at zeion dot net
13 years ago
There're times when global variables comes in handy, like universal read only resources you just need to create once in your application and share to the rest of your scripts. But it may become quite hard to track with "variables".
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9
Stephen Dewey
14 years ago
For nested functions:

This is probably obvious to most people, but global always refers to the variable in the global (top level) variable of that name, not just a variable in a higher-level scope. So this will not work:
<?php

// $var1 is not declared in the global scope

function a($var1){

    function
b(){
        global
$var1;
        echo
$var1; // there is no var1 in the global scope so nothing to echo
   
   
}

   
b();
}

a('hello');

?>
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2
info AT SyPlex DOT net
19 years ago
Some times you need to access the same static in more than one function. There is an easy way to solve this problem:

<?php
 
// We need a way to get a reference of our static
 
function &getStatic() {
    static
$staticVar;
    return
$staticVar;
  }

 
// Now we can access the static in any method by using it's reference
 
function fooCount() {
   
$ref2static = & getStatic();
    echo
$ref2static++;
  }

 
fooCount(); // 0
 
fooCount(); // 1
 
fooCount(); // 2
?>
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3
moraesdno at gmail dot com
14 years ago
Use the superglobal array $GLOBALS is faster than the global keyword. See:

<?php
//Using the keyword global
$a=1;
$b=2;
function
sum() {
    global
$a, $b;
   
$a += $b;
}

$t = microtime(true);
for(
$i=0; $i<1000; $i++) {
    
sum();
}
echo
microtime(true)-$t;
echo
" -- ".$a."<br>";

//Using the superglobal array
$a=1;
$b=2;
function
sum2() {
   
$GLOBALS['a'] += $GLOBALS['b'];
}

 
$t = microtime(true);
for(
$i=0; $i<1000; $i++) {
    
sum2();
}
echo
microtime(true)-$t;
echo
" -- ".$a."<br>";
?>
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3
jakub dot lopuszanski at nasza-klasa dot pl
13 years ago
If you use __autoload function to load classes' definitons, beware that "static local variables are resolved at compile time" (whatever it really means) and the order in which autoloads occur may impact the semantic.

For example if you have:
<?php
class Singleton{
  static public function
get_instance(){
     static
$instance = null;
     if(
$instance === null){
       
$instance = new static();
     }
     return
$instance;
  }
}
?>

and two separate files A.php and B.php:
class A extends Singleton{}
class B extends A{}

then depending on the order in which you access those two classes, and consequently, the order in which __autoload includes them, you can get strange results of calling B::get_instance() and A::get_instance().

It seems that static local variables are alocated in as many copies as there are classes that inherit a method at the time of inclusion of parsing Singleton.
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2
simon dot barotte at gmail dot com
7 years ago
To be vigilant, unlike Java or C++, variables declared inside blocks such as loops (for, while,...) or if's, will also be recognized and accessible outside of the block, the only valid block is the BLOCK function so:

<?php
for($j=0; $j<5; $j++)
{
     if(
$j == 1){
       
$a = 6;
     }
}

echo
$a;
?>

Would print 6.
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2
Anonymous
16 years ago
I was pondering a little something regarding caching classes within a function in order to prevent the need to initiate them multiple times and not clutter the caching function's class properties with more values.

I came here because I remembered something about references being lost. So I made a test to see if I could pull what I wanted to off anyway. Here's and example of how to get around the references lost issue. I hope it is helpful to someone else!

<?php
class test1{}
class
test2{}
class
test3{}

function
cache( $class )
{
    static
$loaders = array();
   
   
$loaders[ $class ] = new $class();

   
var_dump( $loaders );
}
print
'<pre>';
cache( 'test1' );
cache( 'test2' );
cache( 'test3' );

?>
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1
jameslee at cs dot nmt dot edu
18 years ago
It should be noted that a static variable inside a method is static across all instances of that class, i.e., all objects of that class share the same static variable.  For example the code:

<?php
class test {
    function
z() {
        static
$n = 0;
       
$n++;
        return
$n;
    }
}

$a =& new test();
$b =& new test();
print
$a->z();  // prints 1, as it should
print $b->z();  // prints 2 because $a and $b have the same $n
?>

somewhat unexpectedly prints:
1
2
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0
2505036090 at qq dot com
2 years ago
About assignment of object:
```php
<?php
function aaa() {
    global
$obj;
   
$cl = new stdclass;
   
$obj = &$cl; //  assign By Ref
   
$obj->name = 'jack';
   
$obj = NULL; // the change of $obj leads to the change of $cl
   
var_dump($cl); // output: NULL
}
aaa();
var_dump($obj); // output: NULL

function bbb() {
    global
$obj;
   
$cl = new stdclass;
   
$obj = $cl; // general assignment of object
   
$obj->name = 'jack';
   
$obj = NULL; // the value of $obj changes to NULL, but the value of $cl  is still original;
   
var_dump($cl); //output:  object(stdClass)#2 (1) { ["name"]=> string(4) "jack" }
}
bbb();
var_dump($obj); // output: NULL
?>
```
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1
alan
17 years ago
Using the global keyword inside a function to define a variable is essentially the same as passing the variable by reference as a parameter:

<?php
somefunction
(){
   global
$var;
}
?>

is the same as:

<?php
somefunction
(& $a) {

}
?>

The advantage to using the keyword is if you have a long list of variables  needed by the function - you dont have to pass them every time you call the function.
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0
atesin () 6m4i1 ! c0m
3 years ago
i hadn't found info on this so i had to make some tests...

inside a function, when globalling a variable that doesn't exist in the parent scope, it will set to null... i mean...

<?php

function f()
{
  global
$g;
 
var_dump($g); // prints 'NULL' with no log message, means if $g is unset globally then here IS SET to NULL (*)
 
var_dump($l); // prints 'NULL' and logs the 'undefined variable l' notice in this line
 
  // (*) notice until the variable $g IS SET to null here, isset($g) will still return false
  //     because for isset() the var has to be set and not null (see doc for details)...
  //     maybe the only way to find if a variable is actually unset is throught the log warnings
}

var_dump($g); // prints 'NULL' and logs the 'undefined variable g' in this line
var_dump($g); // same as above but in this line, means the variable is still unset
f();

?>

.
knowing this could be useful in a case like mine... consider something like this...

<?php

//--- account.php ---

$user = 'john doe';

//--- main.php ---

function welcome()
{
  global
$user;
  echo
"welcome $user"; // if $user is not set in parent scope it just prints "welcome "

  // the right way afaik
 
if ( isset($user) )
    echo
"welcome $user";
}

if (
login_ok() )
 
incude 'account.php';

welcome();

?>
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0
Duke dot Bouvier at NOPSAM dot Gmail dot com
3 years ago
The recursion example has some excess code:

"$count--" is superfluous.
"$count++" can be built into the loop (but be sure to use pre-increment not a post-incremenet

Thus:

<?php
function test()
{
    static
$count = 1;

    echo
$count." ";

    if (++
$count < 10) {
       
test();
    }
}
?>

Though with a static, all you are really doing is creating a loop but with lots of function calls pushed onto the stack.

A great things about recursive functions is that you don't actually have to have any static data.  You simply supply an argument to pass the value (and can use a default value for the argument for your initialisation if desired.)

<?php
function test($count = 1)
{
    echo
$count." ";

    if (
$count < 10) {
       
test(++$count);
    }

}
test()
?>
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0
jtunaley at gmail dot com
3 years ago
It's worth noting that block statements without a control structure also don't affect variable scope.

<?php
{
   
$a = 4;
}
echo
$a; // Outputs "4"
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1
jake dot tunaley at berkeleyit dot com
5 years ago
Beware of using $this in anonymous functions assigned to a static variable.

<?php
class Foo {
    public function
bar() {
        static
$anonymous = null;
        if (
$anonymous === null) {
           
// Expression is not allowed as static initializer workaround
           
$anonymous = function () {
                return
$this;
            };
        }
        return
$anonymous();
    }
}

$a = new Foo();
$b = new Foo();
var_dump($a->bar() === $a); // True
var_dump($b->bar() === $a); // Also true
?>

In a static anonymous function, $this will be the value of whatever object instance that method was called on first.

To get the behaviour you're probably expecting, you need to pass the $this context into the function.

<?php
class Foo {
    public function
bar() {
        static
$anonymous = null;
        if (
$anonymous === null) {
           
// Expression is not allowed as static initializer workaround
           
$anonymous = function (self $thisObj) {
                return
$thisObj;
            };
        }
        return
$anonymous($this);
    }
}

$a = new Foo();
$b = new Foo();
var_dump($a->bar() === $a); // True
var_dump($b->bar() === $a); // False
?>
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0
ketansheladiya1
4 years ago
As per PHP standards
The new operator returns a reference automatically, so assigning the result of new by reference is not allowed as of PHP 7.0.0, results in an E_DEPRECATED message as of PHP 5.3.0, and an E_STRICT message in earlier versions.
https://www.php.net/manual/en/language.operators.assignment.php

so you can use following code to check

<?php
function &get_instance_ref() {
    static
$obj;

    echo
'Static object: ';
   
var_dump($obj);
    if (!isset(
$obj)) {
       
// Assign a reference to the static variable
       
$obj = new stdclass;
    }
   
$obj->property++;
    return
$obj;
}

function &
get_instance_noref() {
    static
$obj;

    echo
'Static object: ';
   
var_dump($obj);
    if (!isset(
$obj)) {
       
// Assign the object to the static variable
       
$obj = new stdclass;
    }
   
$obj->property++;
    return
$obj;
}

$obj1 = get_instance_ref();
$still_obj1 = get_instance_ref();
echo
"\n";
$obj2 = get_instance_noref();
$still_obj2 = get_instance_noref();
?>
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-1
iphstich at gmail dot com
4 years ago
It seems that static variables defined within a class method or static function are stored separately for the class and each extended class of it.

It's sort of like, every time you extend a class, functions with static variables are re-writen, and the static variable re-created. Even if the function is defined as final!

<?php

class A {
    function
x () { static $x = 0; return ++$x; }
    static function
y () { static $y = 0; return ++$y; }
    final static function
z () { static $z = 0; return ++$z; }
}

class
B extends A { }
class
C extends B { }

// As methods, A and B have different values for x
$a = new A();
echo
$a->x(); // 1
$b = new B();
echo
$b->x(); // 1
echo $b->x(); // 2

// And different instances of C will have the same values for x
$c1 = new C();
$c2 = new C();
$c3 = new C();
echo
$c1->x(); // 1
echo $c2->x(); // 2
echo $c3->x(); // 3

// The same behavior applies to static functions
echo A::y(); // 1
echo B::y(); // 1
echo B::y(); // 2
echo C::y(); // 1
echo C::y(); // 2
echo C::y(); // 3

// The same behavior still applies to FINAL static functions
echo A::z(); // 1
echo B::z(); // 1
echo B::z(); // 2
echo C::z(); // 1
echo C::z(); // 2
echo C::z(); // 3
?>

Tested with version 7.0.13
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0
Hayley Watson
6 years ago
static variables are implicitly initialised to NULL if no explicit initialisation is made.

<?php
function foo()
{
  static
$v;
  echo
gettype($v);
}

foo();
?>

will echo NULL without complaining that $v is undefined.

In short: "static $v;" is equivalent to "static $v = null;".
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0
Semyon Naitur
6 years ago
function f(){
    global $a; // global $a is declared, local reference is created
    $a = 'a';  // global $a is set
    unset($a); // local reference is unset, global $a remains set
    $a = 'b';  // local $a is declared and set
}
f();
echo $a; // prints 'a'

function f(){
    global $a; // global $a is declared, local reference is created
    $a = 'a';  // global $a is set
    unset($a); // local reference $a is unset, global var $a remains set
    global $a; // local reference is created again
    $a .= 'b';
}
f();
echo $a; // prints 'ab'
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0
nino dot skopac at gmail dot com
8 years ago
Interesting behavior in PHP 5.6.12 and PHP 7 RC3:

<?php
class Foo {
    public function
Bar() {
        static
$var = 0;
       
        return ++
$var;
    }
}

$Foo_instance = new Foo;

print
$Foo_instance->Bar(); // prints 1
print PHP_EOL;

unset(
$Foo_instance);

$Foo_instance2 = new Foo;

print
$Foo_instance2->Bar(); // prints 2
print PHP_EOL;
?>

How can a 2 be printed, since we unseted the whole instance before?

Consider a similar example:

<?php
class Foo {
    public static
$var = 0;
   
    public static function
Bar() {
        return ++
self::$var;
    }
}

$Foo_instance = new Foo;

print
$Foo_instance->Bar(); // prints 1
print PHP_EOL;

unset(
$Foo_instance);

$Foo_instance2 = new Foo;

print
$Foo_instance2->Bar(); // prints 2
print PHP_EOL;
?>

No idea why is this happening.
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0
Ray.Paseur often uses Gmail
10 years ago
Variable "Visibility" in PHP Object Oriented Programming is documented here:
http://php.net/manual/en/language.oop5.visibility.php
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0
pedro at worcel dot com
13 years ago
Another way of working with a large ammount of global variables could be the following.

<?php

$var
= "3";
$smarty = new Smarty();

function
headers_set_404() {
extract($globals);

echo
$var . "<br />";
print_r($smarty);

return;

}

?>

Regards,
Droope
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-1
nullhility at gmail dot com
15 years ago
Like functions, if you declare a variable in a class, then set it as global in that class, its value will not be retained outside of that class either.

<?php
class global_reference
{
    public
$val;
   
    public function
__construct () {
        global
$var;
       
$this->val = $var;
    }
   
    public function
dump_it ()
    {
       
debug_zval_dump($this->val);
    }
   
    public function
type_cast ()
    {
       
$this->val = (int) $this->val;
    }
}
$var = "x";
$obj = new global_reference();
$obj->dump_it();
$obj->type_cast();
echo
"after change ";
$obj->dump_it();
echo
"original $var\n";
?>

The work-around is of course changing the assignment in the constructor to a reference assignment as such:

<?php
   
//....
       
$this->val = &var;
   
//....
?>

If the global you're setting is an object then no reference is necessary because of the way PHP deals with objects. If you don't want to reference to the same object however you can use the clone keyword.

<?php
//...
   
global $Obj;
   
$this->obj_copy = clone $Obj;
//...
?>

[EDIT BY danbrown AT php DOT net:  Merged all thoughts and notes by this author into a single note.]
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-1
Ganlv
7 years ago
<?php
$var
= 1;
function
foo() {
   
$var = &$GLOBALS['var'];
   
var_dump($var);
}
function
bar() {
    global
$var; // they are the same.
   
var_dump($var);
}
foo();
bar();
var_dump($var);
?>

In a function, 'global $var;' is to declare a local variant, and the local $var has the same reference to the global $var.

<?php
$var
= 1;
function
foo() {
    global
$var;
    unset(
$var);               // unset local $a, the global $a is still there.
   
var_dump($var);            // Undefined variable: var
   
var_dump($GLOBALS['var']); // this is ok.
}
foo();
var_dump($var);                // this is ok.
?>

<?php
$var
= 1;
function
bar() {
    global
$var;
    unset(
$GLOBALS['var']);    // unset global $a, the local $a is still here.
   
var_dump($var);            // this is ok.
   
var_dump($GLOBALS['var']); // Undefined index: var
}
foo();
var_dump($var);                // Undefined variable: var
?>

'unset($var);' is like 'var = NULL;'(var is a pointer) in the C language, instead of 'free(var);'
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