Variable variables

Sometimes it is convenient to be able to have variable variable names. That is, a variable name which can be set and used dynamically. A normal variable is set with a statement such as:

<?php
$a
= 'hello';
?>

A variable variable takes the value of a variable and treats that as the name of a variable. In the above example, hello, can be used as the name of a variable by using two dollar signs. i.e.

<?php
$$a = 'world';
?>

At this point two variables have been defined and stored in the PHP symbol tree: $a with contents "hello" and $hello with contents "world". Therefore, this statement:

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

produces the exact same output as:

<?php
echo "$a $hello";
?>

i.e. they both produce: hello world.

In order to use variable variables with arrays, you have to resolve an ambiguity problem. That is, if you write $$a[1] then the parser needs to know if you meant to use $a[1] as a variable, or if you wanted $$a as the variable and then the [1] index from that variable. The syntax for resolving this ambiguity is: ${$a[1]} for the first case and ${$a}[1] for the second.

Class properties may also be accessed using variable property names. The variable property name will be resolved within the scope from which the call is made. For instance, if you have an expression such as $foo->$bar, then the local scope will be examined for $bar and its value will be used as the name of the property of $foo. This is also true if $bar is an array access.

Curly braces may also be used, to clearly delimit the property name. They are most useful when accessing values within a property that contains an array, when the property name is made of multiple parts, or when the property name contains characters that are not otherwise valid (e.g. from json_decode() or SimpleXML).

Example #1 Variable property example

<?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";

?>

The above example will output:


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

Warning

Please note that variable variables cannot be used with PHP's Superglobal arrays within functions or class methods. The variable $this is also a special variable that cannot be referenced dynamically.

add a note

User Contributed Notes 10 notes

up
546
userb at exampleb dot org
13 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 ...//

?>
up
7
sebastopolys at gmail dot com
1 year ago
In addition, it is possible to use associative array to secure name of variables available to be used within a function (or class / not tested).

This way the variable variable feature is useful to validate variables; define, output and manage only within the function that receives as parameter
an associative array :
array('index'=>'value','index'=>'value');
index = reference to variable to be used within function
value = name of the variable to be used within function
<?php

$vars
= ['id'=>'user_id','email'=>'user_email'];

validateVarsFunction($vars);

function
validateVarsFunction($vars){

//$vars['id']=34; <- does not work
// define allowed variables
$user_id=21;
$user_email='email@mail.com';

echo
$vars['id']; // prints name of variable: user_id
echo ${$vars['id']}; // prints 21
echo 'Email: '.${$vars['email']}; // print email@mail.com

// we don't have the name of the variables before declaring them inside the function
}
?>
up
68
Anonymous
18 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
9
marcin dot dzdza at gmail dot com
5 years 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;
up
12
jefrey.sobreira [at] gmail [dot] com
9 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
8
Sinured
16 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.
up
9
herebepost (ta at ta) [iwonderr] gmail dot com
7 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
up
16
mason
13 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
8
Nathan Hammond
16 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();

?>
up
4
nils dot rocine at gmail dot com
11 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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