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可变变量> <预定义变量
Last updated: Sun, 25 Nov 2007

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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,有些人可能漫不经心的改变一个全局变量。PHP 中全局变量在函数中使用时必须申明为全局。

global 关键字

首先,一个使用 global 的例子:

Example#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”。在函数中申明了全局变量 $a$b,任何变量的所有引用变量都会指向到全局变量。对于一个函数能够申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:

Example#2 使用 $GLOBALS 替代 global

<?php
$a 
1;
$b 2;

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

Sum();
echo 
$b;
?>

$GLOBALS 数组中,每一个变量为一个元素,键名对应变量名,值对应变量的内容。$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:

Example#3 演示超全局变量和作用域的例子

<?php
function test_global()
{
    
// 大多数的预定义变量并不 "super",它们需要用 'global' 关键字来使它们在函数的本地区域中有效。
    
global $HTTP_POST_VARS;

    echo 
$HTTP_POST_VARS['name'];

    
// Superglobals 在任何范围内都有效,它们并不需要 'global' 声明。Superglobals 是在 PHP 4.1.0 引入的。
    
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:

Example#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:

Example#5 使用静态变量的例子

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

现在,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。一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:

Example#6 静态变量与递归函数

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

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

Note: 静态变量可以按照上面的例子声明。如果在声明中用表达式的结果对其赋值会导致解析错误。

Example#7 声明静态变量

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

    
$int++;
    echo 
$int;
}
?>

全局和静态变量的引用

在 Zend 引擎 1 代,它驱动了 PHP4,对于变量的 staticglobal 定义是以 references 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;
    
$obj = &new stdclass;
}

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

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

执行以上例子会导致如下输出:

NULL
object(stdClass)(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)) {
        
// 将一个引用赋值给静态变量
        
$obj = &new stdclass;
    }
    
$obj->property++;
    return 
$obj;
}

function &
get_instance_noref() {
    static 
$obj;

    echo 
'Static object: ';
    
var_dump($obj);
    if (!isset(
$obj)) {
        
// 将一个对象赋值给静态变量
        
$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();
?>

执行以上例子会导致如下输出:

Static object: NULL
Static object: NULL

Static object: NULL
Static object: object(stdClass)(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



可变变量> <预定义变量
Last updated: Sun, 25 Nov 2007
 
add a note add a note User Contributed Notes
变量范围
moraesdno at gmail dot com
26-Oct-2009 09:17
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>";
?>
pike
28-Aug-2009 12:58
While writing php code that was served as a "view" from within a cms (in my case ezpublish), the "global" keyword seemed to have stopped working in php5. example #1 on this page simply returned "2", not "3".

However, the error was somewhere else. Since my php code was included within other code by the CMS,  the first assignments (a=1,b=2) were apparently not really in the global scope, and hence could not be referenced by the keyword "global" in php5. This worked in php4.

In my particular case, prepending
<code>
global $a,$b;
</code>
to the code (which is invalid php in a root context afaik) fixed it.

$2c,
*-pike
Stephen Dewey
12-Aug-2009 11:06
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');

?>
akam at akameng dot com
12-Jul-2009 10:39
Many Times Globality of variables will be the small issue, after long time I decided to use super globals.

Super globals exists any where:
$_SERVER, $_GET, $_POST .....

Now for example:

<?php
$foo
[] = range(0, 3);
$_POST['foo'] = $foo;
a(); //no parameters needed.
b();
$foo = $_POST['foo'];

Print_r($foo);
/* out

Array
(
    [0] => Array
        (
            [0] => 0
            [1] => 1
            [2] => 2
            [3] => 3
        )

    [1] => Array
        (
            [0] => 4
            [1] => 5
            [2] => 6
            [3] => 7
        )

    [2] => Array
        (
            [0] => 8
            [1] => 9
            [2] => 10
        )

)

*/
function a(){
   
$_POST['foo'][] = range(4, 7);
}

function
b(){
$_POST['foo'][] = range(8, 10);
}
?>
Note: the key must not be passed by the page via _POST method by the form, else the value will be over written
emartin at sigb dot net
03-Jul-2009 10:32
If you are used to include files which declare global variables, and if you now need to include these files in a function, you will see that those globals are declared in the function's scope and so they will be lost at the end of the function.

You may use something like this to solve this problem:

main_file.php :
<?php

//Some innocent variables which exist before the problem
$a = 42;
$b = 33;
$c = 56;

function
some_function() {
   
//Some variables that we don't want out of the function
   
$saucisse = "saucisse";
   
$jambon = "jambon";
   
   
//Let's include another file
   
$evalt = "require_once 'anothertest_include.php';";
   
$before_eval_vars = get_defined_vars();
    eval(
$evalt);

   
//Let's extract the variables that were defined AFTER the call to 'eval'
   
$function_variable_names = array("function_variable_names" => 0, "before_eval_vars" => 0, "created" => 0);
   
//We can generate a list of the newly created variables by substracting the list of the variables of the function and the list of the variables which existed before the call to the list of current variables at this point
   
$created = array_diff_key(get_defined_vars(), $GLOBALS, $function_variable_names, $before_eval_vars);
   
//Now we globalize them
   
foreach ($created as $created_name => $on_sen_fiche)
        global $
$created_name;
   
//And we affect them
   
extract($created);
   
}

some_function();
print_r(get_defined_vars());

?>

included_file.php :
<?php

//Some variables that we want in the global scope of main_file.php
$included_var_one = 123;
$included_var_two = 465;
$included_var_three = 789;

?>
Leigh Harrison
27-Mar-2009 02:31
External variables in a function

I needed to access dynamically-created variables from an included file within a helper function. Because the list of $path_* variables I needed to access from the other file is itself dynamic, I didn't want to have to declare all possible variables within the function, and I was concerned at the overhead of declaring =all= members of $GLOBALS[] as global. However the following code worked for me:

<?php
 
function makePath($root, $atom) {
   
$pos = strrpos($atom, '/');
    if (
$pos === false) {
      global ${
'path_'.$atom}; 
     
$path = ${'path_'.$atom};
    }
    else {
      global ${
'path_'.substr($atom, 0, $pos)};
     
$path = ${'path_'.substr($atom, 0, $pos)};
    }
    if (
$path)
      return (
$pos === false)
        ?
$root.$path
       
: $root.$path.substr($atom, $pos + 1);
    else
      return
NULL;
  }
?>

Regards,

::Leigh
http://www.else.co.nz/
andrew at planetubh dot com
04-Feb-2009 04:16
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().
nullhility at gmail dot com
29-Jan-2009 11:17
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.]
ddarjany at yahoo dot com
20-Aug-2008 11:15
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";
}

?>
lgrk
28-May-2008 09:41
Useful function:
<?php
function cycle($a, $b, $i=0) {
    static
$switches = array();
    if (isset(
$switches[$i])) $switches[$i] = !$switches[$i]; else !$switches[$i] = true;
    return (
$switches[$i])?$a:$b;
}
?>

Exeample

<?php
for ($i = 1; $i<3; $i++) {
    echo
$i.cycle('a', 'b').PHP_EOL;
    for (
$j = 1; $j<5; $j++) {
        echo
' '.$j.cycle('a', 'b', 1).PHP_EOL;
        for (
$k = 1; $k<3; $k++) {
            echo
'  '.$k.cycle('c', 'd', 2).PHP_EOL;
        }
    }
}
/**
Output:
1a
 1a
  1c
  2d
 2b
  1c
  2d
 3a
  1c
  2d
 4b
  1c
  2d
2b
 1a
  1c
  2d
 2b
  1c
  2d
 3a
  1c
  2d
 4b
  1c
  2d
*/

?>
Thomas
04-Mar-2008 06:06
It might be worth noting in the article that you shouldn't define magic values at global level and use "global" to access them in a function - like I did in the past few years.

Use define() instead.
Anonymous
02-Mar-2008 03:10
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' );

?>
SID TRIVEDI
27-Oct-2007 05:46
<?php
/*
VARIABLE SCOPE : GLOBAL V/S STATIC

If variable $count is defined global as under, instead of static, it does not work well as desired in repeated function calls.

$count = 1; //if not defined STATIC, in each function call, it starts countig from one to 25.
global $count;

which gives folowing output:
0123456789101112131415161718192021222324
Total 24 numbers are printed.
So far 26 function call(s) made.

26272829303132333435363738394041424344454647484950
Total 50 numbers are printed.
So far 52 function call(s) made.
*/

function print_1to50()
{
//    $count = 1;
//    global $count;
   
static $count=1; // Initial assigment of One to $count, static declarion holds the last(previous) value of variable $count in each next function calls.
       
$limit = $count+24;
        while(
$count<=$limit)
        {
        echo
"$count";
       
$count=$count+1;
        }
       
$num_count= $count-1;
        echo
"<br>\n". "Total $num_count numbers are printed.<br>";

        return;
// return statement without parenthesis()or arguments denotes end of a function rather than returning any values to subsequent function call(s).
} // end of while loop

$count=0;
print_1to50();
$count=$count+1;
print
"So far $count function call(s) made.<br><br>";

print_1to50();
$count=$count+1;
print
"So far $count function call(s) made.<br>";
/*
Which gives following output:
12345678910111213141516171819202122232425
Now I have printed 25 numbers.
I have made 1 function call(s).
26272829303132333435363738394041424344454647484950
Now I have printed 50 numbers.
I have made 2 function call(s).
*/

?>
daniel at nohair dot com
09-Sep-2007 04:01
Ah, nested functions.  Thanks for your notes below, search on  page for "nested functions" folks.   This is how this seems to work.

The child function is seen at global level only after they have been seen once.  But, variables inside functions are only reachable within the functions scope.

<?php
$var1
= "This is \$var1 OUTSIDE parent function <br />";
function
parent_function() {
    echo
"Now inside parent <br />";
   
$var1 = "This is \$var1 INSIDE parent function <br />";
   
$var2 = "This is \$var2 INSIDE parent function <br />";
    function
child_function() {
        echo
"now inside child <br />";
       
//global $var1; //Calls var1 outside parent_function;
       
echo $var1; //doesn't work without global;
        // even if we comment out $var1 outside parent function.
        // global $var1 doesn't reach the one inside parent function.
       
echo $var2; //doesn't work; Can't seem to reach parent variables.
   
}
    echo
"Now calling child<br />";
   
//child_function();    //works
}

// child_function(); //causes fatal error: call to undefined function;
parent_function();  //works;
child_function(); //now works;

?>

[EDIT BY danbrown AT php DOT net: The author is referring to http://php.net/manual/en/language.variables.scope.php#52148 and http://php.net/manual/en/language.variables.scope.php#20407 ]
crack wilding
24-Aug-2007 11:04
Another way of dealing with a large number of globals is to declare a single global array and then put all your global variables into it. Like this:

<?php
$_G
= array(
   
'foo' => 'some text',
   
'bar' => 4,
   
'boo' => 'more text,
    '
far' => 'yet more text'
);
?>

Now you just declare the one global array in each function:

<?php
function blah() {
    global $_G;
    echo $_G['
foo']; // or whatever
}
?>

You can freely add to it without having to go back and add variable declarations to your functions. Kinda like using the $GLOBALS superglobal, except you don'
t have to  type so much.
mod
15-Mar-2007 03:03
Can not access to global variables from destructor, if obj is not unseted at the end:

<?php

 
class A
  
{
     function
__destruct()
      {
        global
$g_Obj;
        echo
"<br>#step 2: ";
       
var_dump($g_Obj);
      }

     function
start()
      {
        global
$g_Obj;
        echo
"<br>#step 1: ";
       
var_dump($g_Obj);
      }
   };

 
$g_Obj = new A();        // start here
 
$g_Obj->start();
 
$g_Obj = NULL;        // !!! comment line and result will changed !!!

?>

Result, if line is not commented:

#step 1: object(A)#1 (0) { }
#step 2: object(A)#1 (0) { }

Result, if line is commented:

#step 1: object(A)#1 (0) { }
#step 2: NULL
Rohan
26-Jan-2007 02:11
<?php
$a
= 20;
function
myfunction($b){
   
$a=30;   //Local Variable
   
global $a,$c; //here global $a overrides the local
   
return $c=($b+$a);
}
print
myfunction(40)+$c;
?>

The output of this function will be 120.
alan
13-Sep-2006 01:53
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.
sami doesn't want spam at no-eff-eks com
22-Jul-2006 12:18
PHP 5.1.4 doesn't seem to care about the static keyword. It doesn't let you use $this in a static method, but you can call class methods through an instance of the class using regular -> notation. You can also call instance methods as class methods through the class itself. The documentiation here is plain wrong.

<?php
class Foo {
  public static function
static_fun()
  {
    return
"This is a class method!\n";
  }
 
  public function
not_static_fun()
  {
    return
"This is an instance method!\n";
  }
}

echo
'<pre>';
echo
"From Foo:\n";
echo
Foo::static_fun();
echo
Foo::not_static_fun();
echo
"\n";

echo
"From \$foo = new Foo():\n";
$foo = new Foo();
echo
$foo->static_fun();
echo
$foo->not_static_fun();
echo
'</pre>';
?>

You'll see the following output:

From Foo:
This is a class method!
This is an instance method!

From $foo = new Foo():
This is a class method!
This is an instance method!
larax at o2 dot pl
23-Mar-2006 07:38
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";
franp at free dot fr
11-Feb-2006 08:25
If you want to access a table row using $GLOBALS, you must do it outside string delimiters or using curl braces :

<?php
$siteParams
["siteName"] = "myweb";

function
foo() {
$table = $GLOBALS["siteParams"]["siteName"]."articles"// OK
echo $table; // output  "mywebarticles"
$table = "{$GLOBALS["siteParams"]["siteName"]}articles"; // OK
echo $table; // output  "mywebarticles"
$table = "$GLOBALS[siteParams][siteName]articles";       // Not OK
echo $table; // output  "Array[siteName]article"

$result = mysql_query("UPDATE $table ...");
}
?>

Or use global :

<?php
function foo() {
global
$siteParams;
$table = "$siteParams[siteName]articles";         // OK
echo $table; // output  "mywebarticles"

$result = mysql_query("UPDATE $table ...");
}
?>
marcin
31-Dec-2005 01:07
Sometimes in PHP 4 you need static variabiles in class. You can do it by referencing static variable in constructor to the class variable:

<?php
class test  {

   var
$var;
   var
$static_var;
    function
test()
    {
        static
$s;
       
$this->static_var =& $s;
    }
 
}

 
$a=new test();

 
$a->static_var=4;
 
$a->var=4;
 
 
$b=new test();
 
 echo
$b->static_var; //this will output 4
 
echo $b->var; //this will output nul
?>
warhog at warhog dot net
13-Dec-2005 04:22
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.
tc underline at gmx TLD ch
14-Sep-2005 06:06
Pay attention while unsetting variables inside functions:

<?php
$a
= "1234";
echo
"<pre>";
echo
"outer: $a\n";
function
testa()
{
    global
$a;
    echo
"   inner testa: $a\n";
    unset (
$a);
    echo
"   inner testa: $a\n";
}
function
testb()
{
    global
$a;
    echo
"   inner testb: $a\n";
   
$a = null;
    echo
"   inner testb: $a\n";
}
testa();
echo
"outer: $a\n";
testb();
echo
"outer: $a\n";
echo
"</pre>";
?>

/***** Result:
outer: 1234
   inner testa: 1234
   inner testa:
outer: 1234
   inner testb: 1234
   inner testb:
outer:
******/

Took me 1 hour to find out why my variable was still there after unsetting it ...

Thomas Candrian
thomas at pixtur dot de
08-Aug-2005 11:02
Be careful with "require", "require_once" and "include" inside functions. Even if the included file seems to define global variables, they might not be defined as such.

consider those two files:

---index.php------------------------------
<?php
function foo() {
 require_once(
"class_person.inc");

 
$person= new Person();
 echo
$person->my_flag; // should be true, but is undefined
}

foo();
?>

---class_person.inc----------------------------
<?php
$seems_global
=true;

class
Person {
  public
$my_flag;

 public function 
__construct() {
   global
$seems_global;
  
$my_flag= $seems_global
 
}
}
?>

---------------------------------

The reason for this behavior is quiet obvious, once you figured it out. Sadly this might not be always as easy as in this example. A solution  would be to add the line...

<?php global $seems_global; ?>

at the beginning of "class_person.inc". That makes sure you set the global-var.

   best regards
    tom

ps: bug search time approx. 1 hour.
jameslee at cs dot nmt dot edu
17-Jun-2005 05:33
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
kouber at php dot net
28-Apr-2005 08:36
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

}
?>
27-Apr-2005 07:46
Be careful if your static variable is an array and you return
one of it's elements: Other than a scalar variable, elements
of an array are returned as reference (regardless if you
didn't define them to be returned by reference).

<?php
function incr(&$int) {
  return
$int++;
}

function
return_copyof_scalar() {
  static
$v;
  if (!
$v)  
   
$v = 1;
  return(
$v);
}

function
return_copyof_arrayelement() {
  static
$v;
  if (!
$v) {
   
$v = array();
   
$v[0] = 1;
  }
  return(
$v[0]);
}

echo
"scalar: ".
    
incr(return_copyof_scalar()).
    
incr(return_copyof_scalar()).
    
"\n";
echo
"arrayelement: ".
    
incr(return_copyof_arrayelement()).
    
incr(return_copyof_arrayelement()).
    
"\n";
?>

Should print

scalar: 11
arrayelement: 11

but it prints:

scalar: 11
arrayelement: 12

as in the second case the arrays element was returned by
reference. According to a guy from the bug reports the
explanation for this behaviour should be somewhere here in
the documentation (in 'the part with title: "References with
global and static variables"'). Unfortunately I can't find
anything about that here. As the guys from the bug reports
are surely right in every case, maybe there is something
missing in the documentation. Sadly I don't have a good
explanation why this happens, so I decided to document at
least the behaviour.
vdephily at bluemetrix dot com
22-Apr-2005 05:51
Be carefull about nested functions :
<?php
// won't work :
function foo1()
{
 
$who = "world";
  function
bar1()
  {
    global
$who;
    echo
"Hello $who";
  }
}

// will work :
function foo2()
{
 
$GLOBALS['who'] = "world";
  function
bar2()
  {
    global
$who;
    echo
"Hello $who";
  }
}

// also note, of course :
function foo3()
{
 
$GLOBALS['who'] = "world";

 
// won't work
 
echo "Hello $who";

 
// will work
 
global $who;
  echo
"Hello $who";
}
?>
pulstar at ig dot com dot br
09-Sep-2004 09:02
If you need all your global variables available in a function, you can use this:

<?php

function foo(parameters) {
  if(
version_compare(phpversion(),"4.3.0")>=0) {
    foreach(
$GLOBALS as $arraykey=>$arrayvalue) {
      global $
$arraykey;
    }
  }
 
// now all global variables are locally available...
}

?>
info AT SyPlex DOT net
01-Sep-2004 08:35
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
?>
Michael Bailey (jinxidoru at byu dot net)
05-Jun-2004 02:43
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.
Randolpho
03-Apr-2004 04:53
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. :)
jmarbas at hotmail dot com
17-Jan-2004 07:34
Whats good for the goose is not always good for the iterative gander. If you declare and initialize the static variable more than once inside a function ie.

<?php
function Test(){
   static
$count = 0;
   static
$count = 1;
   static
$count = 2;
   echo
$count;
}
?>

the variable will take the value of the last declaration. In this case $count=2.

But! however when you make that function recursive ie.

<?php
 
function Test(){
   static
$count = 0;
   static
$count = 1;
   static
$count = 2;

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

Every call to the function Test() is a differenct SCOPE and therefore the static declarations and initializations are NOT executed again. So what Im trying to say is that its OK to declare and initialize a static variable multiple times if you are in one function... but its NOT OK to declare and initialize a static variable multiple times if you call that same function multiple times. In other words the static variable is set once you LEAVE a function (even if you go back into that very same function).
Jack at soinsincere dot com
15-Nov-2003 02:11
Alright, so you can't set a static variable with a reference.
However, you can set a static variable to an array with an element that is a reference:
<?php

class myReference {
    function
getOrSet($array = null) {
        static
$myValue;
        if (!
$array) {
            return
$myValue[0];     //Return reference in array
       
}
       
$myValue = $array;          //Set static variable with array
       
static $myValue;
    }
}

$static = "Dummy";

$dummy = new myReference;
$dummy->getOrSet(array(&$static));

$static = "Test";
print
$dummy->getOrSet();

?>
flobee at gmx dot net
06-Nov-2003 04:26
i found out that on any (still not found) reason the <?php static $val =NULL; ?> is not working when trying to extract the data form the $var with a while statment
e.g.:
<?php
funktion get_data
() {
static
$myarray = null;
   if(
$myarray == NULL) {
    
//get some info in an array();
    
$myarray = array('one','two');
   }
   while(list(
$key,$val) = each( $myarray ) ) {
  
// do something
  
echo "x: $key , y: $val";
   }
}
?>
when using foreach($myarray AS $key => $val) { .... instad of while then i see the result!
ppo at beeznest dot net
09-Jul-2003 09:59
Even if an included file return a value using return(), it's still sharing the same scope as the caller script!

<?php
$foo
= 'aaa';
$bar = include('include.php');
echo(
$foo.' / '.$bar);
?>

where include.php is
<?php
$foo
= 'bbb';
return
$foo;
?>

The output is: bbb / bbb
Not: aaa / bbb
jg at nerd-boy dot net
08-Feb-2003 08:10
It's possible to use a variable variable when specifying a variable as global in a function. That way your function can decide what global variable to access in run-time.

<?php
function func($varname)
{
   global $
$varname;

   echo $
$varname;
}

$hello = "hello world!";
func("hello");
?>

This will print "hello world!", and is roughly the same as passing by reference, in the case when the variable you want to pass is global. The advantage over references is that they can't have default parameters. With the method above, you can do the following.

<?php
function func($varname = FALSE)
{
   if (
$varname === FALSE)
     echo
"No variable.";
   else
   {
     global $
$varname;

     echo $
$varname;
   }
}

$hello = "hello world!";
func("hello");                   // prints "hello world!"
func();                          // prints "No variable."
?>
wjs@sympaticoDOTca
11-Dec-2002 01:03
Becareful where you define your global variables:

This will work:
<?php
  $MyArray
= array("Dog");

  function
SeeArray(){
    global
$MyArray;
    if (
in_array("Dog",$MyArray)){
      foreach (
$MyArray as $Element){
        echo
"$Element <hr/>";
      }
    }
  }

 
SeeArray();
?>

while this will not:
<?php
  SeeArray
();
 
$MyArray = array("Dog");

  function
SeeArray(){
    global
$MyArray;
    if (
in_array("Dog",$MyArray)){ // an error will generate here
     
foreach ($MyArray as $Element){
        echo
"$Element <hr/>";
      }
    }
  }

?>
heatwave at fw dot hu
15-Oct-2002 08:12
Some people (including me) had a problem with defining a long GLOBAL variable list in functions (very error prone). Here is a possible solution. My program parses php file for functions, and compiles GLOBAL variable lists. Then you can just remove from the list those variables which need not be global.

<?php
   
//parser for GLOBAL variable list
   
$pfile=file("myfile.php4");
   
    for(
$i=0;$i<sizeof($pfile);$i++) {
     if(
eregi("function",$pfile[$i])) {
      list(
$part1,$part2)=sscanf($pfile[$i],"%s %s");
      echo
"\n\n $part1 $part2:\nGLOBAL ";
     
     
$varlist=array();
     
$level=0; $end=$i;
      do {
      
$lpar=explode("{",$pfile[$end]);
      
$level+=sizeof($lpar)-1;
      
$lpar=explode("}",$pfile[$end]);
      
$level-=sizeof($lpar)-1;
      
$end++;
      } while((
$end<sizeof($pfile))&&($level>0));
     
$pstr="";
      for(
$j=$i;$j<=$end;$j++) $pstr.=$pfile[$j];
     
$lpar=explode("$",$pstr);
      for(
$j=1;$j<sizeof($lpar);$j++) {
         
eregi('[a-zA-Z_][a-zA-Z0-9_]*',$lpar[$j],$cvar);
       
$varlist[$cvar[0]]=1;
      }
     
array_walk($varlist,'var_print');
     }
    }
function
var_print ($item, $key) {
     echo
"$key,";
 }
?>
01-May-2002 04:14
Seems as though when a cookie is saved and referenced as a variable of the same name as the cookie, that variable is NOT global.  If you make a function ro read the value of the cookie, the cooke variable name must be declared as a global.

example:

<?php
function ReturnCookie()
{
       
$cookieName = "Test_Cookie";
        global $
$cookieName;
        if (isset($
$cookieName))
        {
                echo (
"$cookieName is set");
               
$returnvalue = $$cookieName;
        }
        else
        {
               
$newCookieValue = "Test Value";
               
setcookie("$cookieName","$newCookieValue", (time() + 3153600));
                echo (
"made a cookie:" . $newCookieValue ."<BR>");
               
$returnvalue = $newCookieValue;
        }
        echo (
"the cookie that was set is now $returnvalue <BR>");
        return
$returnvalue;
}
?>
huntsbox at pacbell dot net
03-Apr-2002 12:11
Not sure of the implications of this but...
You can create nested functions within functions but you must make sure they aren't defined twice, e.g.:

<?php
function norm($a, $b) {
    static
$first_time = true;
    if (
$first_time) {
        function
square($x) {
            return
$x * $x;
        }
       
$first_time = false;
    }
    return
sqrt(square($a) + square($b));
}

print
square(5); // error, not defined yet
print norm(5,4);
print
"<br>";
print
norm(3,2);
print
square(5); // OK
?>

If you don't include the if ($first_time) you get an error saying you can't define square() twice.  Note that square is not local to the function it just appears there.  The last line successfully accesses square in the page scope.  This is not terribly useful, but interesting.
jochen_burkhard at web dot de
30-Mar-2002 03:47
Please don't forget:
values of included (or required) file variables are NOT available in the local script if the included file resides on a remote server:

remotefile.php:

<?PHP
$paramVal
=10;
?>

localfile.php:

<?PHP
include "http://example.com/remotefile.php";
echo
"remote-value= $paramVal";
?>

Will not work (!!)
steph_rondinaud at club-internet dot fr
09-Feb-2002 08:41
I'm using PHP 4.1.1

While designing a database access class, I needed a static variable that will be incremented for all instances of the class each time the class connected to the database. The obvious solution was to declare a "connection" class variable with static scope. Unfortunatly, php doesn't allow such a declaration.
So I went back to defining a static variable in the connect method of my class. But it seems that the static scope is not inherited: if class "a" inherit the "db access" class, then the "connection" variable is shared among "a" instances, not among both "a" AND "db access" instances.
Solution is to declare the static variable out of the db access class, and declare "global" said variable in the connect method.
admin at essentialhost dot com
04-Feb-2002 10:30
Quick tip for beginners just to speed things up:
If you have a bunch of global variables to import into a function, it's best to put them into a named array like $variables[stuff].
When it's time to import them you just so the following;

<?php
function here() {
 
$vars = $GLOBALS['variables'];
  print
$vars[stuff];

}
?>

This really helps with big ugly form submissions.
tomek at pluton dot pl
11-Dec-2001 02:53
When defining static variables you may use such declarations:

<?php
static $var = 1; //numbers
static $var = 'strings';
static
$var = array(1,'a',3); //array construct
?>

but these ones would produce errors:

<?php
static $var = some_function('arg');
static
$var = (some_function('arg'));
static
$var = 2+3; //any expression
static $var = new object;
?>
danno at wpi dot edu
24-Jul-2001 03:28
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
carpathia_uk at mail dot com
08-May-2001 05:21
On confusing aspect about global scope...

If you want to access a variable such as a cookie inside a function, but theres a chance it may not even be defined, you need to access it using he GLOBALS array, not by defining it as global.

This wont work correctly....

<?php
function isLoggedin()
{
global
$cookie_username;
if (isset(
$cookie_username)
echo
"blah..";
}
?>

This will..

<?php
function isLoggedin()
{
if (isset(
$GLOBALS["cookie_username"]))
echo
"blah..";
}
?>
shevek at anarres dot org
05-Feb-2000 08:51
If you include a file from within a function using include(), the included file inherits the function scope as its own global scope, it will not be able to see top level globals unless they are explicit in the function.

<?php
$foo
= "bar";
function
baz() {
    global
$foo; # NOTE THIS
   
include("qux");
}
?>

可变变量> <预定义变量
Last updated: Sun, 25 Nov 2007
 
 
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