How to get word between particular symbols in string - php

Question

My source string could be:
example.com or http://example.com or www.example.com or https://example.com or http://www.example.com or https://www.example.com
or
example.abc.com or http://example.abc.com or www.example.abc.com or https://example.abc.com or http://www.example.abc.com or https://www.example.abc.com
I want the result: example
How can we do this using php string functions? or in other way?

Answer 1

Try this
$str = 'http://example.abc.com';
$last = explode("/", $str, 3);
$ans = explode('.',$last[2]);
echo $ans[0];

Answer 2

You can use parse_url
<?php
// Real full current URL, this can be useful for a lot of things
$url = 'http'.((isset($_SERVER['HTTPS']) && $_SERVER['HTTPS'] == 'on') ? 's' : '').'://'.$_SERVER['HTTP_HOST'].$_SERVER['REQUEST_URI'];
// Or you can put another url
$url = 'https://www.example.foo.biz/';
// Get the host name
$hostName = parse_url($url, PHP_URL_HOST);
// Get the first part of the host name
$host = substr($hostName, 0, strpos($hostName, '.'));
print_r($url);
print_r($hostName);
// Here is what you want
print_r($host);
?>

Answer 3

you can use strpos:
<?php
$url = "http://www.example.com";
/* Use any of you want.
$url = "https://example.com";
$url = "https://www.example.abc.com";
$url = "https://www.www.example.com"; */
if ($found = strpos($url,'example') !== false) {
echo "it exists";
}
?>
EDIT:
So this is what I cam up with now, using explode and substr:
$url = "http://www.example.com";
/* Use any of you want.
$url = "https://example.com";
$url = "https://www.example.abc.com";
$url = "https://www.www.example.com"; */
$exp ='example';
if ($found = strpos($url, $exp) !== false) {
echo $str = substr($url, strpos($url, $exp));
echo "<br>". "it exists" . "<br>";
$finalword = explode(".", $str);
var_dump($finalword);
}
?>