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(PHP 3, PHP 4, PHP 5 )

unset -- Unset a given variable

Description

void unset ( mixed var [, mixed var [, mixed ...]] )

unset() destroys the specified variables. Note that in PHP 3, unset() will always return TRUE (actually, the integer value 1). In PHP 4, however, unset() is no longer a true function: it is now a statement. As such no value is returned, and attempting to take the value of unset() results in a parse error.

Example 1. unset() example

<?php
// destroy a single variable
unset($foo);

// destroy a single element of an array
unset($bar['quux']);

// destroy more than one variable
unset($foo1, $foo2, $foo3);
?>

The behavior of unset() inside of a function can vary depending on what type of variable you are attempting to destroy.

If a globalized variable is unset() inside of a function, only the local variable is destroyed. The variable in the calling environment will retain the same value as before unset() was called.

<?php
function destroy_foo() 
{
    global $foo;
    unset($foo);
}

$foo = 'bar';
destroy_foo();
echo $foo;
?>

The above example will output:

bar

If a variable that is PASSED BY REFERENCE is unset() inside of a function, only the local variable is destroyed. The variable in the calling environment will retain the same value as before unset() was called.

<?php
function foo(&$bar) 
{
    unset($bar);
    $bar = "blah";
}

$bar = 'something';
echo "$bar\n";

foo($bar);
echo "$bar\n";
?>

The above example will output:

something
something

If a static variable is unset() inside of a function, unset() destroys the variable and all its references.

<?php
function foo() 
{
    static $a;
    $a++;
    echo "$a\n";
    unset($a);
}

foo();
foo();
foo();
?>

The above example will output:

1
2
3

If you would like to unset() a global variable inside of a function, you can use the $GLOBALS array to do so:

<?php
function foo() 
{
    unset($GLOBALS['bar']);
}

$bar = "something";
foo();
?>

Related Function(s)

isset()

empty()

array_splice()