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This below does not seem to work how I would expect it, event though $_GET['friendid'] = 55 it is returning NULL
<?PHP
$_GET['friendid'] = 55;
$friendid = (!isset($_GET['friendid'])) ? $_GET['friendid'] : 'empty';
echo $friendid;
exit;
?>
438
PHP isset() Function The isset() function checks whether a variable is set, which means that it has to be declared and is not NULL. This function returns true if the variable exists and is not NULL, otherwise it returns false.
isset( $_POST['submit'] ) : This line checks if the form is submitted using the isset() function, but works only if the form input type submit has a name attribute (name=”submit”).
The isset() function is an inbuilt function in PHP which checks whether a variable is set and is not NULL. This function also checks if a declared variable, array or array key has null value, if it does, isset() returns false, it returns true in all other possible cases.
The Scope Resolution Operator (also called Paamayim Nekudotayim) or in simpler terms, the double colon, is a token that allows access to static, constant, and overridden properties or methods of a class.
As of PHP 7's release, you can use the null-coalescing operator (double "?") for this:
$var = $array["key"] ?? "default-value";
// which is synonymous to:
$var = isset($array["key"]) ? $array["key"] : "default-value";
In PHP 5.3+, if all you are checking on is a "truthy" value, you can use the "Elvis operator" (note that this does not check isset).
$var = $value ?: "default-value";
// which is synonymous to:
$var = $value ? $value : "default-value";
Remove the !. You don't want to negate the expression.
$friendid = isset($_GET['friendid']) ? $_GET['friendid'] : 'empty';
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