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.
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 $foo; ?> |
If a variable that is PASSED BY REFERENCE is unset() inside of a function, only the local variable is destroyed. The variable in the calling environment will retain the same value as before unset() was called.
<?php function foo(&$bar) { unset($bar); $bar = "blah"; } $bar = 'something'; echo "$bar\n"; foo($bar); echo "$bar\n"; ?> |
If a static variable is unset() inside of a function, unset() destroyes 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:
Note: Because this is a language construct and not a function, it cannot be called using variable functions
See also isset(), empty(), and array_splice().