$ and $$ Variables

We know that PHP uses the convention of prefixing the variable names by the “$” symbol. PHP also has the provision of declaring dynamic variables by prefixing two dollar symbols ($$) to the name. A variable variable (or a dynamic variable) can be set and used dynamically.

The declaration of a normal variable is like this −

$a='good';

A dynamic variable takes the value of a normal variable and treats that as the name of the variable. In the above example, “good” can be used as the name of a variable by using two dollar signs “$$” −

$$a='morning';

We now have two variables: “$a” with contents “good” and “$$a” with contents “morning”. As a result, the following echo statements will produce the same output −

echo"$a {$$a}";echo"$a $good";

Both produce the same output −

good morning

Example 1

Take a look at this following example −

Open Compiler

<?php  
   $a = 'good';
   $$a = 'morning';

   echo "$a {$$a}\n";
   echo "$a $good";
?>

It will produce the following output −

good morning
good morning

Example 2

Let’s take a look at another example −

Open Compiler

<?php  
   $x = "foo";  
   $$x = "bar";  
   echo "Value of x = " .$x . "\n";  
   echo 'Value of $$x = ' . $$x . "\n";  
   echo 'Value of foo = ' . $foo;  
?>

Here, you will get the following output −

Value of x = foo
Value of $$x = bar
Value of foo = bar

Explore our latest online courses and learn new skills at your own pace. Enroll and become a certified expert to boost your career.

Using Multiple “$” Symbols

Note that the use of “$” symbol is not restricted to two. Any number of dollar symbols can be prefixed.

Suppose there is a variable “$x” with “a” as its value. Next, we define $$x=’as’, then “$$x” as well as “$a” will have the same value. Similarly, the statement $$$x=’and’ effectively declares a “$as” variable whose value is ‘and’.

Example

Here is a complete example that shows the use of multiple “$” symbols.

Open Compiler

<?php  
   $php = "a";
   $lang = "php";
   $World = "lang";
   $Hello = "World";
   $a = "Hello";
   echo '$a= ' . $a;
   echo "\n";
   echo '$$a= ' . $$a;
   echo "\n";
   echo '$$$a= ' . $$$a;
   echo "\n";
   echo '$$$$a= ' . $$$$a;
   echo "\n";
   echo '$$$$$a= ' . $$$$$a;
?>

When you run this code, it will produce the following output −

$a= Hello
$$a= World
$$$a= lang
$$$$a= php
$$$$$a= a

Using Dynamic Variables with Arrays

Using dynamic variables with arrays may lead to certain ambiguous situations. With an array “a”, if you write $$a[1], then the parser needs to know if you are refering to “$a[1]” as a variable or if you want “$$a” as the variable and then the [1] index from that variable.

To resolve this ambiguity, use ${$a[1]} for the first case and ${$a}[1] for the second.

Example

Take a look at the following example −

Open Compiler

<?php  
   $vars = array("hw", "os", "lang");
   $var_hw="Intel";
   $var_lang="PHP";
   $var_os="Linux";

   foreach ($vars as $var)
      echo ${"var_$var"} . "\n";

   print "$var_hw\n$var_os\n$var_lang";
?>

It will produce the following output −

Intel
Linux
PHP
Intel
Linux
PHP

It may be noted that this technique cannot be used with PHP’s Superglobal arrays (Several predefined variables in PHP are “superglobals”, which means they are available in all scopes throughout a script) within functions or class methods. The variable “$this” is a special variable in PHP and it cannot be referenced dynamically.