(PHP 4, PHP 5)
isset — Determine whether a variable is set
Determine whether a variable is set.
If a variable has been unset with unset(), it will no longer be set. isset() will return FALSE if testing a variable that has been set to NULL. Also note that a NULL byte ("\0") is not equivalent to the PHP NULL constant.
If multiple parameters are supplied then isset() will return TRUE only if all of the parameters are set. Evaluation goes from left to right and stops as soon as an unset variable is encountered.
The variable to be checked.
Another variable ..
Returns TRUE if var exists; FALSE otherwise.
Example#1 isset() Examples
<?php
$var = '';
// This will evaluate to TRUE so the text will be printed.
if (isset($var)) {
echo "This var is set so I will print.";
}
// In the next examples we'll use var_dump to output
// the return value of isset().
$a = "test";
$b = "anothertest";
var_dump(isset($a)); // TRUE
var_dump(isset($a, $b)); // TRUE
unset ($a);
var_dump(isset($a)); // FALSE
var_dump(isset($a, $b)); // FALSE
$foo = NULL;
var_dump(isset($foo)); // FALSE
?>
This also work for elements in arrays:
<?php
$a = array ('test' => 1, 'hello' => NULL);
var_dump(isset($a['test'])); // TRUE
var_dump(isset($a['foo'])); // FALSE
var_dump(isset($a['hello'])); // FALSE
// The key 'hello' equals NULL so is considered unset
// If you want to check for NULL key values then try:
var_dump(array_key_exists('hello', $a)); // TRUE
?>
isset() only works with variables as passing anything else will result in a parse error. For checking if constants are set use the defined() function.
Note: Поскольку это языковая конструкция, а не функция, она не может вызываться при помощи переменных функций