Removing "www." from search query

Question

Currently I'm removing https:// and http:// with following code:

$r=\Input::get('searchQuery');
        
        $Query = preg_replace( "#^[^:/.]*[:/]+#i", "", $r);
        $Query=rtrim($Query, "/");

I need to remove www. as well. What should I add to the code inorder to remove www.
Eg: When I search for https://www.example.com/ It should only search for example.com

check this https://stackoverflow.com/questions/16027102/get-domain-name-from-full-url @stanev01 answer — John Lobo, Jul 18 '21 at 13:11
you can simply do separate if the string is either `https://www.example.com/` or `http://www.example.com/`, use `Str::of($str)->after('https://www.');}` see here https://laravel.com/docs/8.x/helpers#method-fluent-str-after — bhucho, Jul 18 '21 at 13:18
Does this answer your question? [Get domain name from full URL](https://stackoverflow.com/questions/16027102/get-domain-name-from-full-url) — nice_dev, Jul 18 '21 at 13:27
Are you just wanting the www. removing or does it also need to remove other subdomains? i.e. `www.example.com` becomes `example.com`, should `sub1.example.com` remain the same or become `example.com` — mic, Jul 18 '21 at 13:49
Could you please check it - https://stackoverflow.com/questions/6336281/php-remove-www-from-url-inside-a-string — Dmitry Leiko, Jul 26 '21 at 08:08

score 3 · Answer 1 · answered Jul 18 '21 at 19:22

3

It is really easy, you can use preg_replace with a simple regex:

$query = preg_replace('/^((http:\/\/|https:\/\/)?(www\.)?)/i', '', $text);

See this example, so you can understand my regex.

answered Jul 18 '21 at 19:22

matiaslauriti

7,065
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score 0 · Answer 2 · answered Jul 20 '21 at 17:22

The following regex works on any string with an optional "http(s)", "www." and trailing slash:

$r=\Input::get('searchQuery');
$Query = preg_replace('/^(https?\:\/\/)?(www.)?|\/$/', "", $r);

How it works:

^ start of the string.

(https?\:\/\/)? optionally match "http(s)://" (the "s" is also optional).

(www.)? optionally match: "www.".

|\/$ match "/" if it is the last character of the string.

score 0 · Answer 3 · answered Jul 24 '21 at 05:07

You can use Laravel Helpers Method preg_replace_array. preg_replace_array function replaces a given pattern in the string sequentially useing an array:

$string = 'https://www.example.com/'
$query = preg_replace_array('#^[^:/.]*[:/www.?]+#i', [''], $string); // example.com/
$query = rtrim($query, '/'); // example.com

Input: https://www.example.com/
Output: example.com

Rakesh kumar Oad · Answer 4 · 2021-07-26T10:29:49.050

0

$r=\Input::get('searchQuery');
$query = preg_replace_array('#^[^:/.]*[:/www.?]+#i', [''], $r);  
$query = rtrim($query, '/'); // google.com

Your OUT PUT will be google.com

I hope You Got Your Solution .

edited Jul 26 '21 at 10:29

answered Jul 26 '21 at 10:18

Rakesh kumar Oad

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score 0 · Answer 5 · answered Jul 27 '21 at 06:56

code might get bit long but this works everytime, you can add more fuctionality by getting position of / inside of string , working code as below

 <?php
       // $r=\Input::get('searchQuery');
       // $searchQuery = $r; 

       $searchQuery = "www.google.com";

        $searchQuery = str_replace("https://","",$searchQuery);
        $searchQuery = str_replace("http://","",$searchQuery);
      $searchQuery = str_replace("www.","",$searchQuery);
            ?>
<?php #to remove only last / from string
         $lastchar = substr($searchQuery , -1);
    $length = strlen($searchQuery);

    if ($lastchar == "/")
    {
    $searchQuery  = substr_replace($searchQuery, "" , -1);
    }
    else {
    $searchQuery  = $searchQuery ;
    }
    echo $searchQuery; ?>

Removing "www." from search query

5 Answers5