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Manuale PHP


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 "truen" "falsen");

// 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.