php regex preg_match on a variable containing a url

Question

I'm trying to run a regex on a url to extract all the segments after the host. I can't get it working when the host segment is in a variable and i'm not sure how to get it working

// this works
if(preg_match("/^http\:\/\/myhost(\/[a-z0-9A-Z-_\/.]*)$/", $url, $matches)) {
  return $matches[2];
}

// this doesn't work
$siteUrl = "http://myhost";
if(preg_match("/^$siteUrl(\/[a-z0-9A-Z-_\/.]*)$/", $url, $matches)) {
  return $matches[2];
}

// this doesn't work
$siteUrl = preg_quote("http://myhost");
if(preg_match("/^$siteUrl(\/[a-z0-9A-Z-_\/.]*)$/", $url, $matches)) {
  return $matches[2];
}

Shankar Narayana Damodaran · Accepted Answer · 2013-11-16 06:24:28Z

4

In PHP, there is a function called parse_url. (Something similar to what you are trying to achieve through your code).

<?php
$url = 'http://username:password@hostname/path?arg=value#anchor';

print_r(parse_url($url));

echo parse_url($url, PHP_URL_PATH);
?>

OUTPUT :

Array
(
    [scheme] => http
    [host] => hostname
    [user] => username
    [pass] => password
    [path] => /path
    [query] => arg=value
    [fragment] => anchor
)
/path

answered Nov 16, 2013 at 6:24

Shankar Narayana Damodaran

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dlock · Accepted Answer · 2013-11-16 06:27:30Z

2

You forgot to escape the / in your variable declaration. One quick fix is to change your regex delimiter from / to #. Try:

$siteUrl = "http://myhost";
if(preg_match("#^$siteUrl(\/[a-z0-9A-Z-_\/.]*)$#", $url, $matches)) { //note the hashtags!
  return $matches[2];
}

Or without changing the regex delimiter:

$siteUrl = "http:\/\/myhost"; //note how we escaped the slashes
if(preg_match("/^$siteUrl(\/[a-z0-9A-Z-_\/.]*)$/", $url, $matches)) { //note the hashtags!
  return $matches[2];
}

answered Nov 16, 2013 at 6:27

dlock

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