CascadiaPHP 2024

可变变量

有时候使用可变变量名是很方便的。就是说,一个变量的变量名可以动态的设置和使用。一个普通的变量通过声明来设置,例如:

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

一个可变变量获取了一个普通变量的值作为这个可变变量的变量名。在上面的例子中 hello 使用了两个美元符号($)以后,就可以作为一个可变变量的变量了。例如:

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

这时,两个变量都被定义了:$a 的内容是“hello”并且 $hello 的内容是“world”。因此,以下语句:

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

与以下语句输出完全相同的结果:

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

它们都会输出:hello world

要将可变变量用于数组,必须解决一个模棱两可的问题。这就是当写下 $$a[1] 时,解析器需要知道是想要 $a[1] 作为一个变量呢,还是想要 $$a 作为一个变量并取出该变量中索引为 [1] 的值。解决此问题的语法是,对第一种情况用 ${$a[1]},对第二种情况用 ${$a}[1]

类的属性也可以通过可变属性名来访问。可变属性名将在该调用所处的范围内被解析。例如,对于 $foo->$bar 表达式,则会在本地范围来解析 $bar 并且其值将被用于 $foo 的属性名。对于 $bar 是数组单元时也是一样。

也可使用花括号来给属性名清晰定界。最有用是在属性位于数组中,或者属性名包含有多个部分或者属性名包含有非法字符时(例如来自 json_decode()SimpleXML)。

示例 #1 可变属性示例

<?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";

?>

以上示例会输出:


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

警告

注意,在 PHP 的函数和类的方法中,超全局变量不能用作可变变量。$this 变量也是一个特殊变量,不能被动态引用。

add a note

User Contributed Notes 10 notes

up
546
userb at exampleb dot org
14 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
11
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
67
Anonymous
19 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
8
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
9
Sinured
17 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
16
mason
14 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
herebepost (ta at ta) [iwonderr] gmail dot com
8 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
5
nils dot rocine at gmail dot com
12 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;

?>
up
5
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();

?>
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