<?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
$$$$a; //Returns Bar
$$$$$a; //Returns a
$$$$$$a; //Returns Hello
$$$$$$$a; //Returns World
//... and so on ...//
?>
PHP - Manual: 可变变量
2024-11-15
可变变量
有时候使用可变变量名是很方便的。就是说，一个变量的变量名可以动态的设置和使用。一个普通的变量通过声明来设置，例如：
<?php
$a = 'hello';
?>
一个可变变量获取了一个普通变量的值作为这个可变变量的变量名。在上面的例子中 hello 使用了两个美元符号（$）以后，就可以作为一个可变变量的变量了。例如：
<?php
$$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() 或 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";
?>
以上例程会输出：
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 30 notes
userb at exampleb dot org ¶
12 years ago
Anonymous ¶
17 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.
wuhanqinb at gmail dot com ¶
4 years ago
For example #1，line：
echo $foo->$baz[1] . "\n";
It should be:
echo $foo->{$baz[1]} . "\n";
to run correctly.
J. Dyer ¶
19 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.
mason ¶
11 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!'
?>
herebepost (ta at ta) [iwonderr] gmail dot com ¶
5 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
Nathan Hammond ¶
14 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();
?>
Sinured ¶
15 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.
marcin dot dzdza at gmail dot com ¶
3 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 $$a;
antony dot booth at nodomain dot here ¶
19 years ago
You may think of using variable variables to dynamically generate variables from an array, by doing something similar to: -
<?php
foreach ($array as $key => $value)
{
$$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
jefrey.sobreira [at] gmail [dot] com ¶
7 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
?>
php at ianco dot co dot uk ¶
13 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.
Anonymous ¶
20 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";
$$$one = "four";
$$$$one = "five";
echo $$$$one; //still prints 'five'.
?>
NOTE: Tested on PHP 4.2.1, Apache 2.0.36, Red Hat 7.2
nils dot rocine at gmail dot com ¶
9 years ago
Variable Class Instantiation with Namespace Gotcha:
Say you have a class you'd like to instantiate via a variable (with a string value of the Class name)
<?php
class Foo
{
public function __construct()
{
echo "I'm a real class!" . PHP_EOL;
}
}
$class = 'Foo';
$instance = new $class;
?>
The above works fine UNLESS you are in a (defined) namespace. Then you must provide the full namespaced identifier of the class as shown below. This is the case EVEN THOUGH the instancing happens in the same namespace. Instancing a class normally (not through a variable) does not require the namespace. This seems to establish the pattern that if you are using an namespace and you have a class name in a string, you must provide the namespace with the class for the PHP engine to correctly resolve (other cases: class_exists(), interface_exists(), etc.)
<?php
namespace MyNamespace;
class Foo
{
public function __construct()
{
echo "I'm a real class!" . PHP_EOL;
}
}
$class = 'MyNamespace\Foo';
$instance = new $class;
?>
nullhility at gmail dot com ¶
14 years ago
It's also valuable to note the following:
<?php
${date("M")} = "Worked";
echo ${date("M")};
?>
This is perfectly legal, anything inside the braces is executed first, the return value then becomes the variable name. Echoing the same variable variable using the function that created it results in the same return and therefore the same variable name is used in the echo statement. Have fun ;).
dlorre at yahoo dot com ¶
12 years ago
Adding an element directly to an array using variables:
<?php
$tab = array("one", "two", "three") ;
$a = "tab" ;
$$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) ;
?>
sir_hmba AT yahoo DOT com ¶
19 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 = $$obj->{$attr};
echo "obj=$obj attr=$attr val=$val\n";
?>
And here's the output:
f->bar=3 f->baz=6
$obj=f $attr=bar $val=3
chrisNOSPAM at kampmeier dot net ¶
20 years ago
Note that normal variable variables will not be parsed in double-quoted strings. You'll have to use the braces to make it work, to resolve the ambiguity. For example:
<?php
$varname = "foo";
$foo = "bar";
print $$varname; // Prints "bar"
print "$$varname"; // Prints "$foo"
print "${$varname}"; // Prints "bar"
?>
coviex at gmail dot com ¶
8 years ago
In 5.4 "Dynamic class references require the fully qualified class name (with the namespace in it) because at runtime there is no information about the current namespace." is still true.
Neither simple class name nor containing subnamespace works.
Initial source: https://bugs.php.net/bug.php?id=45197
Omar Juvera ¶
10 years ago
The example given in the php manual is confusing!
I think this example it's easier to understand:
<?php
//Let's create a new variable: $new_variable_1
$var_name = "new_variable_1"; //$var_name will store the NAME of the new variable
//Let's assign a value to that [$new_variable_1] variable:
$$var_name = "value 1"; //Value of $new_variable_1 = "value 1"
echo "VARIABLE: " . $var_name;
echo "<br />";
echo "VALUE: " . $$var_name;
?>
The OUTPUT is:
VARIABLE: new_variable_1
VALUE: value 1
You can also create new variables in a loop:
<?php
for( $i = 1; $i < 6; $i++ )
{
$var_name[] = "new_variable_" . $i; //$var_name[] will hold the new variable NAME
}
${$var_name[0]} = "value 1"; //Value of $new_variable_1 = "value 1"
${$var_name[1]} = "value 2"; //Value of $new_variable_2 = "value 2"
${$var_name[2]} = "value 3"; //Value of $new_variable_3 = "value 3"
${$var_name[3]} = "value 4"; //Value of $new_variable_4 = "value 4"
${$var_name[4]} = "value 5"; //Value of $new_variable_5 = "value 5"
echo "VARIABLE: " . $var_name[0] . "\n";
echo "<br />";
echo "VALUE: " . ${$var_name[0]};
?>
The OUTPUT is:
VARIABLE: new_variable_1
VALUE: value 1
mstearne at entermix dot com ¶
21 years ago
Variable variables techniques do not work when one of the "variables" is a constant. The example below illustrates this. This is probably the desired behavior for constants, but was confusing for me when I was trying to figure it out. The alternative I used was to add the variables I needed to the $GLOBALS array instead of defining them as constants.
<?php
define("DB_X_NAME","database1");
define("DB_Y_NAME","database2");
$DB_Z_NAME="database3";
function connectTo($databaseName){
global $DB_Z_NAME;
$fullDatabaseName="DB_".$databaseName."_NAME";
return ${$fullDatabaseName};
}
print "DB_X_NAME is ".connectTo("X")."<br>";
print "DB_Y_NAME is ".connectTo("Y")."<br>";
print "DB_Z_NAME is ".connectTo("Z")."<br>";
?>
[Editor Note: For variable constants, use constant() --Philip]
the_tevildo at yahoo dot com ¶
14 years ago
This is a handy function I put together to allow variable variables to be used with arrays.
To use the function, when you want to reference an array, send it in the form 'array:key' rather than 'array[key]'.
For example:
<?php
function indirect ($var, $value) // Replaces $$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]"
$$var_name = 0;
}
var_dump($temp_array);
echo '<br>';
var_dump($temp_var);
echo '<br>';
// Does work!
$int_var_list = array('temp_array:2', 'temp_var');
while (list($key, $var_name) = each($int_var_list))
{
indirect($var_name, 2);
}
var_dump($temp_array);
echo '<br>';
var_dump($temp_var);
echo '<br>';
?>
Aycan Yat ¶
8 years ago
Sometimes you might wish to modify value of an existing variable by its name. This is easily accomplishable with a combination of using "passing by reference" and "variable variables".
$first_var = 1;
$second_var = 2;
$third_var = 3;
$which_one = array_rand('first', 'second', 'third');
//Let's consider the result is "second".
$modifier = $$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 = &$$which_one; //Simply passing by reference
$modifier++; //Now value of $second_var is 3 too.
echo $second_var; //Prints out 3
It's that simple!
dnl at au dot ru ¶
21 years ago
By the way...
Variable variables can be used as pointers to objects' properties:
<?php
class someclass {
var $a = "variable a";
var $b = "another variable: b";
}
$c = new someclass;
$d = "b";
echo $c->{$d};
?>
outputs: another variable: b
correojulian33-php at yahoo dot es ¶
14 years ago
This example may help to overcome the limitation on $this.
Populate automatically fields of an object form a $_GET variable.
<?php
class pp{
var $prop1=1,$prop2=2,$prop3=array(3,4,5);
function fun1(){
$vars=get_class_vars('pp');
while(list($var,$value)=each($vars)){
$ref=& $this->$var;
$ref=$_GET[$var];
} // while
var_dump($this);
}
}
$_GET['prop1']="uno";
$_GET['prop2']="dos";
$_GET['prop3']=array('tres','cuatro','cinco','seis');
$p=new pp();
$p->fun1();
?>
output is ...
object(pp)#1 (3) {
["prop1"]=>
&string(3) "uno"
["prop2"]=>
&string(3) "dos"
["prop3"]=>
&array(4) {
[0]=>
string(4) "tres"
[1]=>
string(6) "cuatro"
[2]=>
string(5) "cinco"
[3]=>
string(4) "seis"
}
}
jupp-mueller at t-online dot de ¶
19 years ago
I found another undocumented/cool feature: variable member variables in classes. It's pretty easy:
<?php
class foo {
function bar() {
$bar1 = "var1";
$bar2 = "var2";
$this->{$bar1}= "this ";
$this->{$bar2} = "works";
}
}
$test = new foo;
$test->bar();
echo $test->var1 . $test->var2;
?>
houssemzitoun91 at gmail dot com ¶
4 years ago
Even outside a function or a class method variable variables cannot be used with PHP's Superglobal arrays. See the example bellow:
<?php
$a = '$GLOBALS["a"]';
function foo()
{
global $a;
echo $$a;
}
echo $GLOBALS["a"], PHP_EOL; // $GLOBALS["a"]
echo ${$a}; // PHP Notice: Undefined variable: $GLOBALS["a"]
mccoyj at mail dot utexas dot edu ¶
21 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
derek at deperu dot com ¶
1 year ago
How to use variable variables with for and foreach():
<?php
$obj_1 = array("one" => "One", "two" => "Two", "three" => "Three");
$obj_2 = array("one" => "Four", "two" => "Five", "three" => "Six");
$obj_3 = array("one" => "Seven", "two" => "Eight", "three" => "Nine");
for ($i=1; $i < 4; $i++) {
foreach (${"obj_".$i} as $key => $value) {
echo $key . ': ' . $value."\n";
}
}
?>
Result:
one: One
two: Two
three: Three
one: Four
two: Five
three: Six
one: Seven
two: Eight
three: Nine
Alex ¶
2 years ago
If you want to use Variable Variables with $_POST,
you may follow this example:
<?php
// you created a form with variable names:
$_POST['content_1'] = "blabla";
$_POST['content_2'] = "blublu";
// ...
$_POST['content_99'] = "bleble";
// Then, you can use a loop:
for ($i=1; $i<99;$i++)
{
echo $_POST["content_$i"];
}
?>
官方地址：https://www.php.net/manual/en/language.variables.variable.php
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