$_GET shows empty value - php

Question

I have a test.php and i have the below code
<?php
if(isset($_GET['p']) or $_GET['p'] != null) {
echo $_GET['p'];
} else {
echo "Not found";
}
?>
I have listed out below urls then required output are show
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
I accept that Test 1 and Test 2 are true
But when Test 2 and Test 3 fails out the solution.

Answer 1

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

Answer 2

if(isset($_GET['p']) and $_GET['p'] != null) {
echo $_GET['p'];
} else {
echo "Not found";
}
you need and as you want to check that it is set and is not null.
Like others have pointed out - you could use empty() :
if(!empty($_GET['p'])) {
echo $_GET['p'];
} else {
echo "Not found";
}
This will return true when the value is empty, the following is considered empty :
"" (an empty string)
0 (0 as an integer)
0.0 (0 as a float)
"0" (0 as a string)
NULL
FALSE
array() (an empty array)
var $var; (a variable declared, but without a value in a class)

Answer 3

<?php
if(isset($_GET['p']) and $_GET['p'] != null) {
echo $_GET['p'];
} else {
echo "Not found";
}
?>
The above code you mentioned as or so it will take any one of the following as true
try with and
<?php
if(isset($_GET['p']) and $_GET['p'] != null) {
echo $_GET['p'];
} else {
echo "Not found";
}
?>
And it will work perfect.

Answer 4

You are using an OR, so the second test will ALWAYS run. Meaning, if it's not set it will still try access it. You can use an AND instead (&&) so it will only check the value if it exists.

Answer 5

You should use empty() instead of isset().
<?php
if(empty($_GET['p'])) {
echo "Not found";
} else {
echo $_GET['p'];
}