Your PHP tutorial is pure bollocks, simple as that. When running with error reporting set to all errors, warnings and notice, accessing an undefined variable or undefined member of an array will throw the notice you are seeing right now. This is to force developers to declare their variables.
Whoever wrote this code had no clue about this, and probably came from JavaScript, where if (var === undefined) does not throw an error or a warning (as variables default to undefined if not defined, which is logical).
In PHP, if you want to do this, you need to merge and change the two lines of code to this:
[php]
<?php
$my_name = (!empty($_GET['name'])) ? $_GET['name'] : "Steve";
[/php]
Which is exactly equivalent to this:
[php]<?php
if (!empty($_GET['name'])) {
$my_name = $_GET['name'];
}
else {
$my_name = "Steve";
}[/php]
Here, the [b]empty()[/b] function is not actually a function (though it looks like one): it is something called a [u]language construct[/u] and is not bound by the same rules as functions. For a start, passing it an undefined variable does not throw a notice - this is the very purpose of it. There are two constructs with this property:
- [b]isset()[/b], which returns true if a variable is defined
- [b]empty()[/b], which returns true if a variable is defined [u]and non-falsy[/u] (no false, no empty string)
The code I've put above will clear the notice out.
