检查number = ONE，TWO或THREE +如果number = Is Not Set，则添加

Question

Heres my code so far:

$onetwothree = (isset($_GET['number']) && (
    $_GET['number'] == 'one' ||
    $_GET['number'] == 'two' ||
    $_GET['number'] == 'three'));

This displays content if ?number=one, ?number=two, ?number=three but how do i also display content if there is no number= in the url?

Answer 1

The easiest method would be to go with an extra condition:

if (!isset($_GET['country']) || $onetwothree) {
    // display content
}

If you want to display different content based on the value:

if (isset($_GET['number'])) {
    if (!isset($_GET['country'])) {
        // display content
    } else if (in_array($_GET['country'], array('one', 'two', 'three')) {
        // display other content
    }
}

I took a shortcut and used in_array() in the second example; it will make things a bit easier in the future if you ever want to expand the list of accepted values (instead of adding more $_GET['country'] == 'xxxx' lines)