SunshinePHP 2020

Değişken değişkenler

Bazen değişken değişken isimlerine sahip olabilmek kullanışlı olur. Bu, devingen olarak belirtilebilen ve kullanılabilen bir değişken ismidir. Normal bir değişken şöyle bir deyimle atanır:

<?php
$a 
'merhaba';
?>

Bir değişken değişken bir değişkenin değerini alır ve bir değişkenin ismi gibi davranır. Yukarıdaki örnekteki, merhaba, iki tane dolar imi ile bir değişken ismi olarak kullanlabilir

<?php
$$a 'dünya';
?>

Bu noktada PHP sembol ağacında iki değişken tanımlanmış ve saklanmış olur: $a "merhaba" içerirken $merhaba ise "dünya" içerir. Bu nedenle, aşağıdaki iki örnek aynı sonucu üretir:

<?php
echo "$a ${$a}";
?>
<?php
echo "$a $merhaba";
?>

Her ikisi de 'merhaba dünya' üretir.

Değişken değişkenleri dizilerle kullanmak için, bir anlam karmaşası sorununu çözümlemeniz gerekir. $$a[1] yazdığınızda değişken olarak $a[1]'i mi kastettiğiniz, yoksa $$a'nın değişken olmasını isteyip [1] ile o değişkenin indisini mi kastettiğinizi çözümleyicinin bilmesi gerekir. Bu anlam karmaşasını çözümleyen söz dizimi, birinci durum için ${$a[1]} ve ikinci için ${$a}[1]'dir.

Sınıf özelliklerine değişken özellik isimlerinden de erişilebilir. Değişken özellik ismi çağrının yapıldığı etki alanı içinde çözümlenir. Örneğin, $foo->$zam gibi bir değişkeniniz olsun; burada $zam'ın etki alanı incelenir ve $foo özelliğinin ismi olarak kullanılır. $zam, bir diziye erişim için kullanılıyorsa yine aynı durum geçerlidir.

Dikkat

Bir değişken özelliğinin atıfla bırakılması bir dizi içindeyse PHP 5 ve PHP 7'de farklı sözdizimlerine sahiptir. PHP 7.0 göç rehberinde ifade türlerindeki değişiklikler ve karışıklıktan kaçınmak için kaşlı ayraçların nasıl kullanılacağı ile ilgili ayrıntıları bulabilirsiniz.

Kaşlı ayraçlar, özellik ismini açıkça sınırlamak için de kullanılabilir. Bir dizi içinde bulunan bir özelliğin değerlerine erişmek için ya da özellik ismi başka türlü geçerli olmayan karakterler (örn, json_decode() veya SimpleXML ile üretilmiş) içeriyorsa çok kullanışlıdır.

Örnek 1 - Değişken özellik örneği

<?php
class foo {
    var 
$bar 'I am bar.';
    var 
$arr = array('I am A.''I am B.''I am C.');
    var 
$r   'I am r.';
}

$foo = new foo();
$bar 'bar';
$baz = array('foo''bar''baz''quux');
echo 
$foo->$bar "\n";
echo 
$foo->{$baz[1]} . "\n";

$start 'b';
$end   'ar';
echo 
$foo->{$start $end} . "\n";

$arr 'arr';
echo 
$foo->{$arr[1]} . "\n";

?>

Yukarıdaki örneğin çıktısı:


I am bar.
I am bar.
I am bar.
I am r.

Uyarı

Değişken değişkenlerin işlev ve sınıf yöntemleri içinde PHP'nin Süper küresel dizileri ile kullanılamayacağını unutmayınız. $this değişkeni de devingen olarak gönderimli olamayan özel bir değişkendir.

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User Contributed Notes 28 notes

up
428
userb at exampleb dot org
9 years ago
<?php

 
//You can even add more Dollar Signs

 
$Bar = "a";
 
$Foo = "Bar";
 
$World = "Foo";
 
$Hello = "World";
 
$a = "Hello";

 
$a; //Returns Hello
 
$$a; //Returns World
 
$$$a; //Returns Foo
 
$$$$a; //Returns Bar
 
$$$$$a; //Returns a

 
$$$$$$a; //Returns Hello
 
$$$$$$$a; //Returns World

  //... and so on ...//

?>
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69
Anonymous
14 years ago
It may be worth specifically noting, if variable names follow some kind of "template," they can be referenced like this:

<?php
// Given these variables ...
$nameTypes    = array("first", "last", "company");
$name_first   = "John";
$name_last    = "Doe";
$name_company = "PHP.net";

// Then this loop is ...
foreach($nameTypes as $type)
  print ${
"name_$type"} . "\n";

// ... equivalent to this print statement.
print "$name_first\n$name_last\n$name_company\n";
?>

This is apparent from the notes others have left, but is not explicitly stated.
up
14
php at ianco dot co dot uk
10 years ago
<?php
// $variable-name = 'parse error';
// You can't do that but you can do this:
$a = 'variable-name';
$
$a = 'hello';
echo
$variable-name . ' ' . $$a; // Gives     0 hello
?>

For a particular reason I had been using some variable names with hyphens for ages. There was no problem because they were only referenced via a variable variable. I only saw a parse error much later, when I tried to reference one directly. It took a while to realise that illegal hyphens were the cause because the parse error only occurs on assignment.
up
10
jefrey.sobreira [at] gmail [dot] com
4 years ago
If you want to use a variable value in part of the name of a variable variable (not the whole name itself), you can do like the following:

<?php
$price_for_monday
= 10;
$price_for_tuesday = 20;
$price_for_wednesday = 30;

$today = 'tuesday';

$price_for_today = ${ 'price_for_' . $today};
echo
$price_for_today; // will return 20
?>
up
22
mason
9 years ago
PHP actually supports invoking a new instance of a class using a variable class name since at least version 5.2

<?php
class Foo {
   public function
hello() {
      echo
'Hello world!';
   }
}
$my_foo = 'Foo';
$a = new $my_foo();
$a->hello(); //prints 'Hello world!'
?>

Additionally, you can access static methods and properties using variable class names, but only since PHP 5.3

<?php
class Foo {
   public static function
hello() {
      echo
'Hello world!';
   }
}
$my_foo = 'Foo';
$my_foo::hello(); //prints 'Hello world!'
?>
up
20
J. Dyer
17 years ago
Another use for this feature in PHP is dynamic parsing.. 

Due to the rather odd structure of an input string I am currently parsing, I must have a reference for each particular object instantiation in the order which they were created.  In addition, because of the syntax of the input string, elements of the previous object creation are required for the current one. 

Normally, you won't need something this convolute.  In this example, I needed to load an array with dynamically named objects - (yes, this has some basic Object Oriented programming, please bare with me..)

<?php
  
include("obj.class");

  
// this is only a skeletal example, of course.
  
$object_array = array();

  
// assume the $input array has tokens for parsing.
  
foreach ($input_array as $key=>$value){
     
// test to ensure the $value is what we need.
        
$obj = "obj".$key;
         $
$obj = new Obj($value, $other_var);
        
Array_Push($object_array, $$obj);
     
// etc..
  
}

?>

Now, we can use basic array manipulation to get these objects out in the particular order we need, and the objects no longer are dependant on the previous ones.

I haven't fully tested the implimentation of the objects.  The  scope of a variable-variable's object attributes (get all that?) is a little tough to crack.  Regardless, this is another example of the manner in which the var-vars can be used with precision where tedious, extra hard-coding is the only alternative.

Then, we can easily pull everything back out again using a basic array function: foreach.

<?php
//...
  
foreach($array as $key=>$object){

      echo
$key." -- ".$object->print_fcn()." <br/>\n";

   }
// end foreach  

?>

Through this, we can pull a dynamically named object out of the array it was stored in without actually knowing its name.
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12
antony dot booth at nodomain dot here
17 years ago
You may think of using variable variables to dynamically generate variables from an array, by doing something similar to: -

<?php
foreach ($array as $key => $value)
{
  $
$key= $value;
}

?>

This however would be reinventing the wheel when you can simply use:

<?php
extract
( $array, EXTR_OVERWRITE);
?>

Note that this will overwrite the contents of variables that already exist.

Extract has useful functionality to prevent this, or you may group the variables by using prefixes too, so you could use: -

EXTR_PREFIX_ALL

<?php
$array
=array("one" => "First Value",
"two" => "2nd Value",
"three" => "8"
               
);
          
extract( $array, EXTR_PREFIX_ALL, "my_prefix_");
  
?>

This would create variables: -
$my_prefix_one
$my_prefix_two
$my_prefix_three

containing: -
"First Value", "2nd Value" and "8" respectively
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2
houssemzitoun91 at gmail dot com
1 year ago
Even outside a function or a class method  variable variables cannot be used with PHP's Superglobal arrays. See the example bellow:
<?php

$a
= '$GLOBALS["a"]';

function
foo()
{
        global
$a;
        echo $
$a;
}

echo
$GLOBALS["a"], PHP_EOL; // $GLOBALS["a"]

echo ${$a}; // PHP Notice:  Undefined variable: $GLOBALS["a"]
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4
Aycan Yat
6 years ago
Sometimes you might wish to modify value of an existing variable by its name. This is easily accomplishable with a combination of using "passing by reference" and "variable variables".

$first_var = 1;
$second_var = 2;
$third_var = 3;

$which_one = array_rand('first', 'second', 'third');
//Let's consider the result is "second".

$modifier = $$which_one;  //Now $modifier has value 2.
$modifier++; //Now $modifier's value is 3.
echo $second_var; //Prints out 2

//Consider we wish to modify the value of $second_var
$modifier = &$$which_one;  //Simply passing by reference
$modifier++; //Now value of $second_var is 3 too.
echo $second_var; //Prints out 3

It's that simple!
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9
Nathan Hammond
11 years ago
These are the scenarios that you may run into trying to reference superglobals dynamically. Whether or not it works appears to be dependent upon the current scope.

<?php

$_POST
['asdf'] = 'something';

function
test() {
   
// NULL -- not what initially expected
   
$string = '_POST';
   
var_dump(${$string});

   
// Works as expected
   
var_dump(${'_POST'});

   
// Works as expected
   
global ${$string};
   
var_dump(${$string});

}

// Works as expected
$string = '_POST';
var_dump(${$string});

test();

?>
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11
dlorre at yahoo dot com
9 years ago
Adding an element directly to an array using variables:

<?php
$tab
= array("one", "two", "three") ;
$a = "tab" ;
$
$a[] ="four" ; // <==== fatal error
print_r($tab) ;
?>
will issue this error:

Fatal error: Cannot use [] for reading

This is not a bug, you need to use the {} syntax to remove the ambiguity.

<?php
$tab
= array("one", "two", "three") ;
$a = "tab" ;
${
$a}[] =  "four" ; // <==== this is the correct way to do it
print_r($tab) ;
?>
up
7
nullhility at gmail dot com
11 years ago
It's also valuable to note the following:

<?php
${date("M")} = "Worked";
echo ${
date("M")};
?>

This is perfectly legal, anything inside the braces is executed first, the return value then becomes the variable name. Echoing the same variable variable using the function that created it results in the same return and therefore the same variable name is used in the echo statement. Have fun ;).
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4
sir_hmba AT yahoo DOT com
16 years ago
This is somewhat redundant, but I didn't see an example that combined dynamic reference of *both* object and attribute names.

Here's the code:

<?php
class foo
{
    var
$bar;
    var
$baz;

    function
foo()
    {
       
$this->bar = 3;
       
$this->baz = 6;
    }
}

$f = new foo();
echo
"f->bar=$f->bar  f->baz=$f->baz\n";

$obj  = 'f';
$attr = 'bar';
$val  = $$obj->{$attr};

echo
"obj=$obj  attr=$attr  val=$val\n";
?>

And here's the output:

f->bar=3  f->baz=6
$obj=f  $attr=bar  $val=3
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9
herebepost (ta at ta) [iwonderr] gmail dot com
3 years ago
While not relevant in everyday PHP programming, it seems to be possible to insert whitespace and comments between the dollar signs of a variable variable.  All three comment styles work. This information becomes relevant when writing a parser, tokenizer or something else that operates on PHP syntax.

<?php

    $foo
= 'bar';
    $

   
/*
        I am complete legal and will compile without notices or error as a variable variable.
    */
       
$foo = 'magic';

    echo
$bar; // Outputs magic.

?>

Behaviour tested with PHP Version 5.6.19
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4
chrisNOSPAM at kampmeier dot net
18 years ago
Note that normal variable variables will not be parsed in double-quoted strings. You'll have to use the braces to make it work, to resolve the ambiguity. For example:

<?php
$varname
= "foo";
$foo = "bar";

print $
$varname// Prints "bar"
print "$$varname"// Prints "$foo"
print "${$varname}"; // Prints "bar"
?>
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5
nils dot rocine at gmail dot com
7 years ago
Variable Class Instantiation with Namespace Gotcha:

Say you have a class you'd like to instantiate via a variable (with a string value of the Class name)

<?php

class Foo
{
    public function
__construct()
    {
        echo
"I'm a real class!" . PHP_EOL;
    }
}

$class = 'Foo';

$instance = new $class;

?>

The above works fine UNLESS you are in a (defined) namespace. Then you must provide the full namespaced identifier of the class as shown below. This is the case EVEN THOUGH the instancing happens in the same namespace. Instancing a class normally (not through a variable) does not require the namespace. This seems to establish the pattern that if you are using an namespace and you have a class name in a string, you must provide the namespace with the class for the PHP engine to correctly resolve (other cases: class_exists(), interface_exists(), etc.)

<?php

namespace MyNamespace;

class
Foo
{
    public function
__construct()
    {
        echo
"I'm a real class!" . PHP_EOL;
    }
}

$class = 'MyNamespace\Foo';

$instance = new $class;

?>
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3
jupp-mueller at t-online dot de
17 years ago
I found another undocumented/cool feature: variable member variables in classes. It's pretty easy:

<?php
class foo {
  function
bar() {
   
$bar1 = "var1";
   
$bar2 = "var2";
   
$this->{$bar1}= "this ";
   
$this->{$bar2} = "works";
  }
}

$test = new foo;
$test->bar();
echo
$test->var1 . $test->var2;
?>
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4
the_tevildo at yahoo dot com
12 years ago
This is a handy function I put together to allow variable variables to be used with arrays.

To use the function, when you want to reference an array, send it in the form 'array:key' rather than 'array[key]'.

For example:

<?php

function indirect ($var, $value)     // Replaces $$var = $value
{
  
$var_data = $explode($var, ':');
   if (isset(
$var_data[1]))
   {
      ${
$var_data[0]}[$var_data[1]] = $value;
   }
   else
   {
      ${
$var_data[0]} = $value;
   }
}

$temp_array = array_fill(0, 4, 1);
$temp_var = 1;
$int_var_list = array('temp_array[2]', 'temp_var');

while (list(
$key, $var_name) = each($int_var_list))
{
  
//  Doesn't work - creates scalar variable called "$temp_array[2]"
  
$$var_name = 0;
}

var_dump($temp_array);
echo
'<br>';
var_dump($temp_var);
echo
'<br>';

//  Does work!

$int_var_list = array('temp_array:2', 'temp_var');

while (list(
$key, $var_name) = each($int_var_list))
{
  
indirect($var_name, 2);
}

var_dump($temp_array);
echo
'<br>';
var_dump($temp_var);
echo
'<br>';
?>
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4
Sinured
12 years ago
One interesting thing I found out: You can concatenate variables and use spaces. Concatenating constants and function calls are also possible.

<?php
define
('ONE', 1);
function
one() {
    return
1;
}
$one = 1;

${
"foo$one"} = 'foo';
echo
$foo1; // foo
${'foo' . ONE} = 'bar';
echo
$foo1; // bar
${'foo' . one()} = 'baz';
echo
$foo1; // baz
?>

This syntax doesn't work for functions:

<?php
$foo
= 'info';
{
"php$foo"}(); // Parse error

// You'll have to do:
$func = "php$foo";
$func();
?>

Note: Don't leave out the quotes on strings inside the curly braces, PHP won't handle that graciously.
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2
correojulian33-php at yahoo dot es
12 years ago
This example may help to overcome the limitation on $this.

Populate automatically fields of an object form a $_GET variable.

<?php
class pp{
   var
$prop1=1,$prop2=2,$prop3=array(3,4,5);

   function
fun1(){
     
$vars=get_class_vars('pp');
      while(list(
$var,$value)=each($vars)){
              
$ref=& $this->$var;
              
$ref=$_GET[$var];

      }
// while
     
var_dump($this);
   }
}

$_GET['prop1']="uno";
$_GET['prop2']="dos";
$_GET['prop3']=array('tres','cuatro','cinco','seis');

$p=new pp();
$p->fun1();
?>

output is ...

object(pp)#1 (3) {
  ["prop1"]=>
  &string(3) "uno"
  ["prop2"]=>
  &string(3) "dos"
  ["prop3"]=>
  &array(4) {
    [0]=>
    string(4) "tres"
    [1]=>
    string(6) "cuatro"
    [2]=>
    string(5) "cinco"
    [3]=>
    string(4) "seis"
  }
}
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2
Anonymous
17 years ago
The 'dollar dereferencing' (to coin a phrase) doesn't seem to be limited to two layers, even without curly braces.  Observe:

<?php
$one
= "two";
$two = "three";
$three = "four";
$four = "five";
echo $$$
$one; //prints 'five'.
?>

This works for L-values as well.  So the below works the same way:

<?php
$one
= "two";
$
$one = "three";
$$
$one = "four";
$$$
$one = "five";
echo $$$
$one; //still prints 'five'.
?>

NOTE: Tested on PHP 4.2.1, Apache 2.0.36, Red Hat 7.2
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2
dnl at au dot ru
19 years ago
By the way...
Variable variables can be used as pointers to objects' properties:

<?php
class someclass {
  var
$a = "variable a";
  var
$b = "another variable: b";
  }

$c = new someclass;
$d = "b";
echo
$c->{$d};
?>

outputs: another variable: b
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2
marcin dot dzdza at gmail dot com
11 months ago
The feature of variable variable names is welcome, but it should be avoided when possible. Modern IDE software fails to interpret such variables correctly, regular find/replace also fails. It's a kind of magic :) This may really make it hard to refactor code. Imagine you want to rename variable $username to $userName and try to find all occurrences of $username in code by checking "$userName". You may easily omit:
$a = 'username';
echo $$a;
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0
wuhanqinb at gmail dot com
1 year ago
For example #1,line:
echo $foo->$baz[1] . "\n";
It should be:
echo $foo->{$baz[1]} . "\n";
to run correctly.
up
0
coviex at gmail dot com
6 years ago
In 5.4 "Dynamic class references require the fully qualified class name (with the namespace in it) because at runtime there is no information about the current namespace." is still true.
Neither simple class name nor containing subnamespace works.
Initial source: https://bugs.php.net/bug.php?id=45197
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0
Omar Juvera
8 years ago
The example given in the php manual is confusing!
I think this example it's easier to understand:

<?php

//Let's create a new variable: $new_variable_1
$var_name = "new_variable_1"; //$var_name will store the NAME of the new variable

//Let's assign a value to that [$new_variable_1] variable:
$$var_name  = "value 1"; //Value of $new_variable_1 = "value 1"

echo "VARIABLE: " . $var_name;
echo
"<br />";
echo
"VALUE: " . $$var_name;
?>

The OUTPUT is:
VARIABLE: new_variable_1
VALUE: value 1

You can also create new variables in a loop:
<?php

for( $i = 1; $i < 6; $i++ )
{
$var_name[] = "new_variable_" . $i; //$var_name[] will hold the new variable NAME
}

${
$var_name[0]}  = "value 1"; //Value of $new_variable_1 = "value 1"
${$var_name[1]}  = "value 2"; //Value of $new_variable_2 = "value 2"
${$var_name[2]}  = "value 3"; //Value of $new_variable_3 = "value 3"
${$var_name[3]}  = "value 4"; //Value of $new_variable_4 = "value 4"
${$var_name[4]}  = "value 5"; //Value of $new_variable_5 = "value 5"

echo "VARIABLE: " . $var_name[0] . "\n";
echo
"<br />";
echo
"VALUE: " . ${$var_name[0]};
?>

The OUTPUT is:
VARIABLE: new_variable_1
VALUE: value 1
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0
mstearne at entermix dot com
18 years ago
Variable variables techniques do not work when one of the "variables" is a constant.  The example below illustrates this.  This is probably the desired behavior for constants, but was confusing for me when I was trying to figure it out.  The alternative I used was to add the variables I needed to the $GLOBALS array instead of defining them as constants.

<?php

define
("DB_X_NAME","database1");
define("DB_Y_NAME","database2");
$DB_Z_NAME="database3";


function
connectTo($databaseName){
global
$DB_Z_NAME;

$fullDatabaseName="DB_".$databaseName."_NAME";
return ${
$fullDatabaseName};

}

print
"DB_X_NAME is ".connectTo("X")."<br>";
print
"DB_Y_NAME is ".connectTo("Y")."<br>";
print
"DB_Z_NAME is ".connectTo("Z")."<br>";

?>
[Editor Note: For variable constants, use constant() --Philip]
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0
mccoyj at mail dot utexas dot edu
19 years ago
There is no need for the braces for variable object names...they are only needed by an ambiguity arises concerning which part of the reference is variable...usually with arrays.

<?php
class Schlemiel {
var
$aVar = "foo";
}

$schlemiel = new Schlemiel;
$a = "schlemiel";
echo $
$a->aVar;
?>

This code outputs "foo" using PHP 4.0.3.

Hope this helps...
- Jordan
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