Basics

Variables in PHP are represented by a dollar sign followed by the name of the variable. The variable name is case-sensitive.

Variable names follow the same rules as other labels in PHP. A valid variable name starts with a letter or underscore, followed by any number of letters, numbers, or underscores. As a regular expression, it would be expressed thus: '[a-zA-Z_\x7f-\xff][a-zA-Z0-9_\x7f-\xff]*'

Note: For our purposes here, a letter is a-z, A-Z, and the bytes from 127 through 255 (0x7f-0xff).

Note: $this is a special variable that can't be assigned.

Tip

See also the Userland Naming Guide.

For information on variable related functions, see the Variable Functions Reference.

<?php
$var 
'Bob';
$Var 'Joe';
echo 
"$var$Var";      // outputs "Bob, Joe"

$4site 'not yet';     // invalid; starts with a number
$_4site 'not yet';    // valid; starts with an underscore
$täyte 'mansikka';    // valid; 'ä' is (Extended) ASCII 228.
?>

By default, variables are always assigned by value. That is to say, when you assign an expression to a variable, the entire value of the original expression is copied into the destination variable. This means, for instance, that after assigning one variable's value to another, changing one of those variables will have no effect on the other. For more information on this kind of assignment, see the chapter on Expressions.

PHP also offers another way to assign values to variables: assign by reference. This means that the new variable simply references (in other words, "becomes an alias for" or "points to") the original variable. Changes to the new variable affect the original, and vice versa.

To assign by reference, simply prepend an ampersand (&) to the beginning of the variable which is being assigned (the source variable). For instance, the following code snippet outputs 'My name is Bob' twice:

<?php
$foo 
'Bob';              // Assign the value 'Bob' to $foo
$bar = &$foo;              // Reference $foo via $bar.
$bar "My name is $bar";  // Alter $bar...
echo $bar;
echo 
$foo;                 // $foo is altered too.
?>

One important thing to note is that only named variables may be assigned by reference.

<?php
$foo 
25;
$bar = &$foo;      // This is a valid assignment.
$bar = &(24 7);  // Invalid; references an unnamed expression.

function test()
{
   return 
25;
}

$bar = &test();    // Invalid.
?>

It is not necessary to initialize variables in PHP however it is a very good practice. Uninitialized variables have a default value of their type depending on the context in which they are used - booleans default to FALSE, integers and floats default to zero, strings (e.g. used in echo) are set as an empty string and arrays become to an empty array.

Example #1 Default values of uninitialized variables

<?php
// Unset AND unreferenced (no use context) variable; outputs NULL
var_dump($unset_var);

// Boolean usage; outputs 'false' (See ternary operators for more on this syntax)
echo($unset_bool "true\n" "false\n");

// String usage; outputs 'string(3) "abc"'
$unset_str .= 'abc';
var_dump($unset_str);

// Integer usage; outputs 'int(25)'
$unset_int += 25// 0 + 25 => 25
var_dump($unset_int);

// Float/double usage; outputs 'float(1.25)'
$unset_float += 1.25;
var_dump($unset_float);

// Array usage; outputs array(1) {  [3]=>  string(3) "def" }
$unset_arr[3] = "def"// array() + array(3 => "def") => array(3 => "def")
var_dump($unset_arr);

// Object usage; creates new stdClass object (see http://www.php.net/manual/en/reserved.classes.php)
// Outputs: object(stdClass)#1 (1) {  ["foo"]=>  string(3) "bar" }
$unset_obj->foo 'bar';
var_dump($unset_obj);
?>

Relying on the default value of an uninitialized variable is problematic in the case of including one file into another which uses the same variable name. It is also a major security risk with register_globals turned on. E_NOTICE level error is issued in case of working with uninitialized variables, however not in the case of appending elements to the uninitialized array. isset() language construct can be used to detect if a variable has been already initialized.

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User Contributed Notes 8 notes

up
16
jeff dot phpnet at tanasity dot com
4 years ago
This page should include a note on variable lifecycle:

Before a variable is used, it has no existence. It is unset. It is possible to check if a variable doesn't exist by using isset(). This returns true provided the variable exists and isn't set to null. With the exception of null, the value a variable holds plays no part in determining whether a variable is set.

Setting an existing variable to null is a way of unsetting a variable. Another way is variables may be destroyed by using the unset() construct.

<?php
print isset($a); // $a is not set. Prints false. (Or more accurately prints ''.)
$b = 0; // isset($b) returns true (or more accurately '1')
$c = array(); // isset($c) returns true
$b = null; // Now isset($b) returns false;
unset($c); // Now isset($c) returns false;
?>

is_null() is an equivalent test to checking that isset() is false.

The first time that a variable is used in a scope, it's automatically created. After this isset is true. At the point at which it is created it also receives a type according to the context.

<?php
$a_bool
= true;   // a boolean
$a_str = 'foo';    // a string
?>

If it is used without having been given a value then it is uninitalized and it receives the default value for the type. The default values are the _empty_ values. E.g  Booleans default to FALSE, integers and floats default to zero, strings to the empty string '', arrays to the empty array.

A variable can be tested for emptiness using empty();

<?php
$a
= 0; //This isset, but is empty
?>

Unset variables are also empty.

<?php
empty($vessel); // returns true. Also $vessel is unset.
?>

Everything above applies to array elements too.

<?php
$item
= array();
//Now isset($item) returns true. But isset($item['unicorn']) is false.
//empty($item) is true, and so is empty($item['unicorn']

$item['unicorn'] = '';
//Now isset($item['unicorn']) is true. And empty($item) is false.
//But empty($item['unicorn']) is still true;

$item['unicorn'] = 'Pink unicorn';
//isset($item['unicorn']) is still true. And empty($item) is still false.
//But now empty($item['unicorn']) is false;
?>

For arrays, this is important because accessing a non-existent array item can trigger errors; you may want to test arrays and array items for existence with isset before using them.
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7
megan at voices dot com
2 years ago
"Note: $this is a special variable that can't be assigned."

While the PHP runtime generates an error if you directly assign $this in code, it doesn't for $$name when name is 'this'.

<?php

$this
= 'text'; // error

$name = 'this';
$
$name = 'text'; // sets $this to 'text'

?>
up
6
maurizio dot domba at pu dot t-com dot hr
3 years ago
If you need to check user entered value for a proper PHP variable naming convention you need to add ^ to the above regular expression so that the regular expression should be '^[a-zA-Z_\x7f-\xff][a-zA-Z0-9_\x7f-\xff]*'.

Example

<?php
$name
="20011aa";
if(!
preg_match('/[a-zA-Z_\x7f-\xff][a-zA-Z0-9_\x7f-\xff]*/',$name))
   echo
$name.' is not a valid PHP variable name';
else
   echo
$name.' is valid PHP variable name';
?>

Outputs: 2011aa is valid PHP variable name

but

<?php
$name
="20011aa";
if(!
preg_match('/^[a-zA-Z_\x7f-\xff][a-zA-Z0-9_\x7f-\xff]*/',$name))
   echo
$name.' is not a valid PHP variable name';
else
   echo
$name.' is valid PHP variable name';
?>

Outputs: 2011aa is not a valid PHP variable name
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0
karst at onlinq dot nl
8 days ago
From the "Properties" page, about initialising properties:
"This declaration may include an initialization, but this initialization must be a constant value--that is, it must be able to be evaluated at compile time and must not depend on run-time information in order to be evaluated. "

So if you define a variable in a class (as a member variable/property), things like public <?php $test = 2+3; ?> are invalid, because logic has to be performed on the right hand side.
It HAS to be either a constant, or a scalar value (int/string/bool/float). Not even another variable which is a constant (so <?php public $test = "test"; public $invalid = $test; ?> would not work.

Just thought this should be mentioned here for all those like me who get here before getting to the "Properties" page.
up
-7
Edoxile
4 years ago
When wanting to switch two variables from content, you can use the XOR operator:

<?PHP
$a
=5;
$b=3;

//Please mind the order of these, as it's important for the outcome.

$a^=$b;
$b^=$a;
$a^=$b;

echo
$a.PHP_EOL.$b;
/* prints:
3
5
*/
?>

This will also work on strings, but it won't work on arrays and objects, so for them you'll have to use the serialize() function before the operation, and the unserialize() function after.
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-10
php at richardneill dot org
1 year ago
Note that "$1" is not a variable name. PHP treats it literally, even when it is in double quotes. Eg:

$fruit="apple";
echo "This $fruit costs $1 ";

This is especially notable when using $1, $2 etc inside parameterised queries in SQL.
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-5
jeff12 at fastpitchcentral dot com
8 months ago
While the recommendation here is to initialize variables, there is a very good reason to definitely initialize variables.

I just had the unpleasant task of making numerous folks happy that the Apache "error_log" is now much smaller.  We had over 1500 variables that showed up as "error" in the error_log file.  In many cases just one uninitialized variable might cause 10s or 100s of thousands of records to be entered in the "error_log".

While it may not be necessary to initialize variables, there can be a significant cost if you do not do so.

And no, that website was not willing to lower the bar and not write to the "error_log" file if it was simply a case of uninitialized variables.  It was also the case that in a small percentage of cases  the variables newly initialized caused me to scratch my head and conclude that a blank or zero might cause different logic to occur.
up
-17
tymac at hotmail dot com
1 year ago
Hi,

something like $foo = &myfunc(); seems to work fine.

Regards.
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