Reputation: 51
I need to get part from url, i know lots tutorial in this website but i'm still unable to get it...
this sample url :
https://www.domain.com/want-to-keep-35/?idku=rbxbbgh3dKqn
I only need this part only from that url, without domain, slash and end url
want-to-keep-35
already try this code :
preg_replace('|https?://www\.[a-z\.0-9]+|i', '', $serp);
but not working. let me know if anyone in here can help me and thanks
Upvotes: 1
Views: 397
Reputation: 17626
Only if you know the domain name, why not simply use:
$url = 'https://www.domain.com/want-to-keep-35/?idku=rbxbbgh3dKqn';
$keep = explode("https://www.domain.com/", $url);
$keep2 = explode("/", $keep[0]);
$keep[0] will contain 'want-to-keep-35/?idku=rbxbbgh3dKqn'
$keep2[0] will contain 'want-to-keep-35'
Upvotes: 0
Reputation: 3158
Try this
<?php
$url = 'https://www.domain.com/want-to-keep-35/?idku=rbxbbgh3dKqn';
$parsedURL = parse_url($url);
echo "<pre>";
var_dump(parse_url($url));
echo "</pre>";
$result = trim($parsedURL['path'], '/');
echo $result;
Upvotes: 0
Reputation: 1675
If it's that simple, you may use the following simple approach to get what you want:
$url = 'https://www.domain.com/want-to-keep-35/?idku=rbxbbgh3dKqn';
$re = '#https://www\..*?\..*?/(.*?)/#i';
preg_match($re, $url, $matches);
echo $matches[1]; // want-to-keep-35
Upvotes: 1
Reputation: 4564
If you use two patterns and run them in succession, you will be able to eliminate the first part and then the second part.
pattern 1 will find http or https and then the ://. Then it will read everything that is not a slash until the first slash effectively removing the domain from the url. Finally, it will lump in the the slash you don't want.
pattern 2 will take from a string starting with want-to...
and find the first slash and everything after it.
$url = "https://www.domain.com/want-to-keep-35/?idku=rbxbbgh3dKqn";
$pattern1 = "/https?\:\/\/[^\/]+\//";
$pattern2 = "/\/.*/";
$url = preg_replace($pattern1, '', $url);
$url = preg_replace($pattern2, '', $url);
echo $url;
yields
want-to-keep-35
You can also pass in an array of patterns and it will evaluate them in order
$url = "https://www.domain.com/want-to-keep-35/?idku=rbxbbgh3dKqn";
$patterns = array("/https?\:\/\/[^\/]+\//", "/\/.*/");
$url = preg_replace($patterns, '', $url);
echo $url;
Upvotes: 0