$ _GET显示空值

Question

我有一个test.php,我有以下代码

<?php

    if(isset($_GET['p']) or $_GET['p'] != null) {
        echo $_GET['p'];
    } else {
        echo "Not found";
    }

?>

我列出了下面的网址然后要求输出显示

Test 1 : http://localhost/example/test.php
output : Notice: Undefined index: p in R:\xampp\htdocs\example\test.php on line 3
        Not found


Test 2 : http://localhost/example/test.php?p
output : blank page



Test 3 : http://localhost/example/test.php?p=
output : blank page


Test 4 : http://localhost/example/test.php?p=1
output : 1

我承认,Test 1和Test 2是true 但当Test 2并Test 3失败了解决办法.

Answer 1

<?php

if(!empty($_GET['p']))
    echo $_GET['p'];
else
    echo "Not found";
?>