Function preg_match() returns nothing - php

Question

I want to use preg_match() in my code, but the result is nothing ... (or null or empty ?)
$domain = "stackoverflow.com";
$uriToTest = "http://stackoverflow.com/";
$pattern = "/^http(s)?://(([a-z]+)\.)*".$domain."/";
echo preg_match($pattern, $uriToTest);
What is the problem?

Answer 1

If you take a look at your pattern, it's this
/^http(s)?://(([a-z]+)\.)*stackoverflow.com/
The delimiter is used as a matching character, and if you had errors turned on, you'd get a "Unknown modifier" error. So first tip: TURN ERROR REPORTING ON!
To fix it, try using a different delimiter, e.g. {}, as it's easier to read than loads of leaning toothpicks...
{^http(s)?://(([a-z]+)\.)*stackoverflow.com}
The other problem is the dot in the $domain becomes a wildcard match - anytime you insert unknown data into a regex, get in the habit of using preg_quote to escape it, e.g.
$pattern = "{^http(s)?://(([a-z]+)\.)*" . preg_quote($domain, '{') . "}";
(Note - nice catch from stema in the comments: if you use a different delimiter, you must pass that preg_quote. It's clever enough to spot paired delimiters, so if you pass { it will also escape }.)

Answer 2

You're most likely getting an error and preg_match is returning false, as you are not escaping your forward slashes in your expression. Either use something else like a # as the expression delimeter or escape any forward slashes to stop the parser from trying to end the expression (/ should be \/ - or change the / at either end to be #)
//Quick fix to yours
$pattern = "/^http(s)?:\/\/(([a-z]+)\.)*".preg_quote($domain,'/')."/";
//More legible fix
$pattern = '#^https?://(([a-z]+)\.)*'.preg_quote($domain,'#').'#';
Note that you don't need parenthesis around the s in https (unless you're hoping to capture it)

Answer 3

You need to escape your forward slashes and the . in the domain name
$domain = "stackoverflow.com";
$uriToTest = "http://stackoverflow.com/";
$escapedDomain = str_replace('.', '\.', $domain);
$pattern = "/^http(s)?:\/\/(([a-z]+)\.)*".$escapedDomain."/";
echo preg_match($pattern, $uriToTest);

Answer 4

If you were using T-Regx, then this exception would be thrown immediately:
$domain = "stackoverflow.com";
$uriToTest = "http://stackoverflow.com/";
try
{
pattern("/^http(s)?://(([a-z]+)\.)*" . $domain . '/')->match($uriToTest);
}
catch (SafeRegexException $e) {
echo $e->getMessage(); // `Unknown modifier '/'`
}
But also!! T-Regx can automatically add delimiters, so you can go
pattern("^http(s)?://(([a-z]+)\.)*" . $domain)->match($uriToTest);
and it would automatically add a suitable delimiter for you.

Answer 5

$domain = "stackoverflow.com";
$uriToTest = "http://stackoverflow.com/";
$pattern = "^http(s)?://(([a-z]+)\.)*" . $domain . "^";
preg_match($pattern, $uriToTest, $matches);
print_r($matches);