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.
Пример 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.
The above example would output:
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.
The above example would output:
If a static variable is unset() inside of a
function, unset() destroys the variable and all
its references.
The above example would output:
If you would like to unset() a global variable
inside of a function, you can use
the $GLOBALS array to do so:
Замечание: Поскольку это языковая
конструкция, а не функция, она не может вызываться при помощи
переменных функций
Смотрите также isset(),
empty(), and
array_splice().