Javascript debugger
Website design
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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.
<?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.
Because this is a language construct and not a function, it cannot be called using variable functions
empty() |
unset() |
defined() |
the type comparison tables |
array_key_exists() |
is_null() |
the error control @ operator |