Detect if localhost/Xampp/Laragon is running - php

Question

I am building a project where an online php script needs to loads files from the local server (this is not a public website).
Is it possible to detect if the local server is running or not and display a message. Something like this (C# Check If Xampp Server/Localhost is Running) but with php.
gethostbyname will not work.
$domain = '127.0.0.1/info.php'; // or $domain = 'localhost/info.php';
if (gethostbyname($domain) != $domain ) {
echo 'Up and running';}
else {
echo 'Run xampp first';
}
This will not work too
$file = '127.0.0.1/info.php';
$file_headers = #get_headers($file);
if(!$file_headers || $file_headers[0] == 'HTTP/1.1 404 Not Found') {
echo 'Run xampp first';
}
else {
echo 'Up and running';
}

Answer 1

This is not a URL, it's a local filename:
$file = '127.0.0.1/info.php';
It lacks the leading protocol specifier, so it's looking for a file named info.php in a directory named 127.0.0.1.
You will need:
$file = 'http://127.0.0.1/info.php';
Then, assuming that http://127.0.0.1/info.php is a valid URL that will be served if the web service is running, you can use file_get_contents() to try and load the page. This will issue a warning and return false on failure.
if (#file_get_contents('http://127.0.0.1/info.php')) {
echo "server is up";
} else {
echo "server is down";
}
You could also use get_headers() as in your exmaple, but note that if you get a 404 Not Found, that still means the server is up and running.