I've got a Node module file containing about 100 exported methods, which looks something like this:
exports.methodOne = async user_id => {
// other method contents
};
exports.methodTwo = async user_id => {
// other method contents
fooMethod();
};
exports.methodThree = async user_id => {
// other method contents
fooMethod();
};
Goal: What I'd like to do is figure out how to grab the name of any method which contains a call to fooMethod, and return the correct method names: methodTwo and methodThree. I wrote a regex which gets kinda close:
exports\.(\w+).*(\n.*?){1,}fooMethod
Problem: using my example code from above, though, it would effectively match methodOne and methodThree because it finds the first instance of export and then the first instance of fooMethod and goes on from there. Here's a regex101 example.
I suspect I could make use of lookaheads or lookbehinds, but I have little experience with those parts of regex, so any guidance would be much appreciated!
Edit: Turns out regex is poorly-suited for this type of task. #ctcherry advised using a parser, and using that as a springboard, I was able to learn about Abstract Syntax Trees (ASTs) and the recast tool which lets you traverse the tree after using various tools (acorn and others) to parse your code into tree form.
With these tools in hand, I successfully built a script to parse and traverse my node app's files, and was able to find all methods containing fooMethod as intended.
Regex isn't the best tool to tackle all the parts of this problem, ideally we could rely on something higher level, a parser.
One way to do this is to let the javascript parse itself during load and execution. If your node module doesn't include anything that would execute on its own (or at least anything that would conflict with the below), you can put this at the bottom of your module, and then run the module with node mod.js.
console.log(Object.keys(exports).filter(fn => exports[fn].toString().includes("fooMethod(")));
(In the comments below it is revealed that the above isn't possible.)
Another option would be to use a library like https://github.com/acornjs/acorn (there are other options) to write some other javascript that parses your original target javascript, then you would have a tree structure you could use to perform your matching and eventually return the function names you are after. I'm not an expert in that library so unfortunately I don't have sample code for you.
This regex matches (only) the method names that contain a call to fooMethod();
(?<=exports\.)\w+(?=[^{]+\{[^}]+fooMethod\(\)[^}]+};)
See live demo.
Assuming that all methods have their body enclosed within { and }, I would make an approach to get to the final regex like this:
First, find a regex to get the individual methods. This can be done using this regex:
exports\.(\w+)(\s|.)*?\{(\s|.)*?\}
Next, we are interested in those methods that have fooMethod in them before they close. So, look for } or fooMethod.*}, in that order. So, let us name the group searching for fooMethod as FOO and the name of the method calling it as METH. When we iterate the matches, if group FOO is present in a match, we will use the corresponding METH group, else we will reject it.
exports\.(?<METH>\w+)(\s|.)*?\{(\s|.)*?(\}|(?<FOO>fooMethod)(\s|.)*?\})
Explanation:
exports\.(?<METH>\w+): Till the method name (you have already covered this)
(\s|.)*?\{(\s|.)*?: Some code before { and after, non-greedy so that the subsequent group is given preference
(\}|(?<FOO>fooMethod)(\s|.)*?\}): This has 2 parts:
\}: Match the method close delimiter, OR
(?<FOO>fooMethod)(\s|.)*?\}): The call to fooMethod followed by optional code and method close delimiter.
Here's a JavaScript code that demostrates this:
let p = /exports\.(?<METH>\w+)(\s|.)*?\{(\s|.)*?(\}|(?<FOO>fooMethod)(\s|.)*?\})/g
let input = `exports.methodOne = async user_id => {
// other method contents
};
exports.methodTwo = async user_id => {
// other method contents
fooMethod();
};
exports.methodThree = async user_id => {
// other method contents
fooMethod();
};';`
let match = p.exec( input );
while( match !== null) {
if( match.groups.FOO !== undefined ) console.log( match.groups.METH );
match = p.exec( input )
}
I'm pretty new to scala and basically I want to have a couple of functions coupled to a string in a hashmap.
However I get an error at subscribers.get(e.key)(e.EventArgs); stating Option[EventArgs => Unit] does not take parameters...
Example code:
object Monitor {
val subscribers = HashMap.empty[String, (EventArgs) => Unit ]
def trigger(e : Event){
subscribers.get(e.key)(e.EventArgs);
}
def subscribe(key: String, e: (EventArgs) => Unit) {
subscribers += key -> e;
}
}
The get method of a Map gives you an Option of the value, not the value. Thus, if the key if found in the map, you get Some(value), if not, you get None. So you need to first "unroll" that option to make sure there is actually a value of a function which you can invoke (call apply on):
def trigger(e: Event): Unit =
subscribers.get(e.key).foreach(_.apply(e.EventArgs))
or
def trigger(e: Event): Unit =
subscribers.get(e.key) match {
case Some(value) => value(e.EventArgs)
case None =>
}
There are many posts around explaining Scala's Option type. For example this one or this one.
Also note Luigi's remark about using an immutable map (the default Map) with a var instead.
Since the get method returns Option, you can use 'map' on that:
subscribers.get(e.key).map(f => f(e.EventArgs))
or even shorter:
subscribers.get(e.key) map (_(e.EventArgs))
get only takes one argument. So subscribers.get(e.key) returns an Option, and you're trying to feed (e.EventArgs) to that Option's apply method (which doesn't exist).
Also, try making the subscribers a var (or choosing a mutable collection type). At the moment you have an immutable collection and an immutable variable, so your map cannot change. A more idiomatic way to declare it would be
var subscribers = Map[String, EventArgs => Unit]()
HashMap.get() in Scala works in a bit different way, than in Java. Instead of returning value itself, get() returns Option. Option is a special type, that can have 2 values - Some(x) and None. In first case it tells "there's some value with such a key in a map". In second case it tells "nope, there's nothing (none) for this key in a map". This is done to force programmers check whether map actually has an object or not and avoid NullPointerException, which appears so frequently in Java code.
So you need something like this:
def trigger(e: Event) {
val value = subscribers.get(e.key)
value match {
case None => throw new Exception("Oops, no such subscriber...")
case Some(f) => f(e.EventArgs)
}
}
You can find more info about Option type and pattern matching in Scala here.
How can I fetch a domain name from a URL String?
Examples:
+----------------------+------------+
| input | output |
+----------------------+------------+
| www.google.com | google |
| www.mail.yahoo.com | mail.yahoo |
| www.mail.yahoo.co.in | mail.yahoo |
| www.abc.au.uk | abc |
+----------------------+------------+
Related:
Matching a web address through regex
I once had to write such a regex for a company I worked for. The solution was this:
Get a list of every ccTLD and gTLD available. Your first stop should be IANA. The list from Mozilla looks great at first sight, but lacks ac.uk for example so for this it is not really usable.
Join the list like the example below. A warning: Ordering is important! If org.uk would appear after uk then example.org.uk would match org instead of example.
Example regex:
.*([^\.]+)(com|net|org|info|coop|int|co\.uk|org\.uk|ac\.uk|uk|__and so on__)$
This worked really well and also matched weird, unofficial top-levels like de.com and friends.
The upside:
Very fast if regex is optimally ordered
The downside of this solution is of course:
Handwritten regex which has to be updated manually if ccTLDs change or get added. Tedious job!
Very large regex so not very readable.
A little late to the party, but:
const urls = [
'www.abc.au.uk',
'https://github.com',
'http://github.ca',
'https://www.google.ru',
'http://www.google.co.uk',
'www.yandex.com',
'yandex.ru',
'yandex'
]
urls.forEach(url => console.log(url.replace(/.+\/\/|www.|\..+/g, '')))
Extracting the Domain name accurately can be quite tricky mainly because the domain extension can contain 2 parts (like .com.au or .co.uk) and the subdomain (the prefix) may or may not be there. Listing all domain extensions is not an option because there are hundreds of these. EuroDNS.com for example lists over 800 domain name extensions.
I therefore wrote a short php function that uses 'parse_url()' and some observations about domain extensions to accurately extract the url components AND the domain name. The function is as follows:
function parse_url_all($url){
$url = substr($url,0,4)=='http'? $url: 'http://'.$url;
$d = parse_url($url);
$tmp = explode('.',$d['host']);
$n = count($tmp);
if ($n>=2){
if ($n==4 || ($n==3 && strlen($tmp[($n-2)])<=3)){
$d['domain'] = $tmp[($n-3)].".".$tmp[($n-2)].".".$tmp[($n-1)];
$d['domainX'] = $tmp[($n-3)];
} else {
$d['domain'] = $tmp[($n-2)].".".$tmp[($n-1)];
$d['domainX'] = $tmp[($n-2)];
}
}
return $d;
}
This simple function will work in almost every case. There are a few exceptions, but these are very rare.
To demonstrate / test this function you can use the following:
$urls = array('www.test.com', 'test.com', 'cp.test.com' .....);
echo "<div style='overflow-x:auto;'>";
echo "<table>";
echo "<tr><th>URL</th><th>Host</th><th>Domain</th><th>Domain X</th></tr>";
foreach ($urls as $url) {
$info = parse_url_all($url);
echo "<tr><td>".$url."</td><td>".$info['host'].
"</td><td>".$info['domain']."</td><td>".$info['domainX']."</td></tr>";
}
echo "</table></div>";
The output will be as follows for the URL's listed:
As you can see, the domain name and the domain name without the extension are consistently extracted whatever the URL that is presented to the function.
I hope that this helps.
/^(?:www\.)?(.*?)\.(?:com|au\.uk|co\.in)$/
There are two ways
Using split
Then just parse that string
var domain;
//find & remove protocol (http, ftp, etc.) and get domain
if (url.indexOf('://') > -1) {
domain = url.split('/')[2];
} if (url.indexOf('//') === 0) {
domain = url.split('/')[2];
} else {
domain = url.split('/')[0];
}
//find & remove port number
domain = domain.split(':')[0];
Using Regex
var r = /:\/\/(.[^/]+)/;
"http://stackoverflow.com/questions/5343288/get-url".match(r)[1]
=> stackoverflow.com
Hope this helps
I don't know of any libraries, but the string manipulation of domain names is easy enough.
The hard part is knowing if the name is at the second or third level. For this you will need a data file you maintain (e.g. for .uk is is not always the third level, some organisations (e.g. bl.uk, jet.uk) exist at the second level).
The source of Firefox from Mozilla has such a data file, check the Mozilla licensing to see if you could reuse that.
import urlparse
GENERIC_TLDS = [
'aero', 'asia', 'biz', 'com', 'coop', 'edu', 'gov', 'info', 'int', 'jobs',
'mil', 'mobi', 'museum', 'name', 'net', 'org', 'pro', 'tel', 'travel', 'cat'
]
def get_domain(url):
hostname = urlparse.urlparse(url.lower()).netloc
if hostname == '':
# Force the recognition as a full URL
hostname = urlparse.urlparse('http://' + uri).netloc
# Remove the 'user:passw', 'www.' and ':port' parts
hostname = hostname.split('#')[-1].split(':')[0].lstrip('www.').split('.')
num_parts = len(hostname)
if (num_parts < 3) or (len(hostname[-1]) > 2):
return '.'.join(hostname[:-1])
if len(hostname[-2]) > 2 and hostname[-2] not in GENERIC_TLDS:
return '.'.join(hostname[:-1])
if num_parts >= 3:
return '.'.join(hostname[:-2])
This code isn't guaranteed to work with all URLs and doesn't filter those that are grammatically correct but invalid like 'example.uk'.
However it'll do the job in most cases.
It is not possible without using a TLD list to compare with as their exist many cases like http://www.db.de/ or http://bbc.co.uk/ that will be interpreted by a regex as the domains db.de (correct) and co.uk (wrong).
But even with that you won't have success if your list does not contain SLDs, too. URLs like http://big.uk.com/ and http://www.uk.com/ would be both interpreted as uk.com (the first domain is big.uk.com).
Because of that all browsers use Mozilla's Public Suffix List:
https://en.wikipedia.org/wiki/Public_Suffix_List
You can use it in your code by importing it through this URL:
http://mxr.mozilla.org/mozilla-central/source/netwerk/dns/effective_tld_names.dat?raw=1
Feel free to extend my function to extract the domain name, only. It won't use regex and it is fast:
http://www.programmierer-forum.de/domainnamen-ermitteln-t244185.htm#3471878
Basically, what you want is:
google.com -> google.com -> google
www.google.com -> google.com -> google
google.co.uk -> google.co.uk -> google
www.google.co.uk -> google.co.uk -> google
www.google.org -> google.org -> google
www.google.org.uk -> google.org.uk -> google
Optional:
www.google.com -> google.com -> www.google
images.google.com -> google.com -> images.google
mail.yahoo.co.uk -> yahoo.co.uk -> mail.yahoo
mail.yahoo.com -> yahoo.com -> mail.yahoo
www.mail.yahoo.com -> yahoo.com -> mail.yahoo
You don't need to construct an ever-changing regex as 99% of domains will be matched properly if you simply look at the 2nd last part of the name:
(co|com|gov|net|org)
If it is one of these, then you need to match 3 dots, else 2. Simple. Now, my regex wizardry is no match for that of some other SO'ers, so the best way I've found to achieve this is with some code, assuming you've already stripped off the path:
my #d=split /\./,$domain; # split the domain part into an array
$c=#d; # count how many parts
$dest=$d[$c-2].'.'.$d[$c-1]; # use the last 2 parts
if ($d[$c-2]=~m/(co|com|gov|net|org)/) { # is the second-last part one of these?
$dest=$d[$c-3].'.'.$dest; # if so, add a third part
};
print $dest; # show it
To just get the name, as per your question:
my #d=split /\./,$domain; # split the domain part into an array
$c=#d; # count how many parts
if ($d[$c-2]=~m/(co|com|gov|net|org)/) { # is the second-last part one of these?
$dest=$d[$c-3]; # if so, give the third last
$dest=$d[$c-4].'.'.$dest if ($c>3); # optional bit
} else {
$dest=$d[$c-2]; # else the second last
$dest=$d[$c-3].'.'.$dest if ($c>2); # optional bit
};
print $dest; # show it
I like this approach because it's maintenance-free. Unless you want to validate that it's actually a legitimate domain, but that's kind of pointless because you're most likely only using this to process log files and an invalid domain wouldn't find its way in there in the first place.
If you'd like to match "unofficial" subdomains such as bozo.za.net, or bozo.au.uk, bozo.msf.ru just add (za|au|msf) to the regex.
I'd love to see someone do all of this using just a regex, I'm sure it's possible.
/[^w{3}\.]([a-zA-Z0-9]([a-zA-Z0-9\-]{0,65}[a-zA-Z0-9])?\.)+[a-zA-Z]{2,6}/gim
usage of this javascript regex ignores www and following dot, while retaining the domain intact. also properly matches no www and cc tld
Could you just look for the word before .com (or other) (the order of the other list would be the opposite of the frequency see here
and take the first matching group
i.e.
window.location.host.match(/(\w|-)+(?=(\.(com|net|org|info|coop|int|co|ac|ie|co|ai|eu|ca|icu|top|xyz|tk|cn|ga|cf|nl|us|eu|de|hk|am|tv|bingo|blackfriday|gov|edu|mil|arpa|au|ru)(\.|\/|$)))/g)[0]
You can test it could by copying this line into the developers' console on any tab
This example works in the following cases:
So if you just have a string and not a window.location you could use...
String.prototype.toUrl = function(){
if(!this && 0 < this.length)
{
return undefined;
}
var original = this.toString();
var s = original;
if(!original.toLowerCase().startsWith('http'))
{
s = 'http://' + original;
}
s = this.split('/');
var protocol = s[0];
var host = s[2];
var relativePath = '';
if(s.length > 3){
for(var i=3;i< s.length;i++)
{
relativePath += '/' + s[i];
}
}
s = host.split('.');
var domain = s[s.length-2] + '.' + s[s.length-1];
return {
original: original,
protocol: protocol,
domain: domain,
host: host,
relativePath: relativePath,
getParameter: function(param)
{
return this.getParameters()[param];
},
getParameters: function(){
var vars = [], hash;
var hashes = this.original.slice(this.original.indexOf('?') + 1).split('&');
for (var i = 0; i < hashes.length; i++) {
hash = hashes[i].split('=');
vars.push(hash[0]);
vars[hash[0]] = hash[1];
}
return vars;
}
};};
How to use.
var str = "http://en.wikipedia.org/wiki/Knopf?q=1&t=2";
var url = str.toUrl;
var host = url.host;
var domain = url.domain;
var original = url.original;
var relativePath = url.relativePath;
var paramQ = url.getParameter('q');
var paramT = url.getParamter('t');
For a certain purpose I did this quick Python function yesterday. It returns domain from URL. It's quick and doesn't need any input file listing stuff. However, I don't pretend it works in all cases, but it really does the job I needed for a simple text mining script.
Output looks like this :
http://www.google.co.uk => google.co.uk
http://24.media.tumblr.com/tumblr_m04s34rqh567ij78k_250.gif => tumblr.com
def getDomain(url):
parts = re.split("\/", url)
match = re.match("([\w\-]+\.)*([\w\-]+\.\w{2,6}$)", parts[2])
if match != None:
if re.search("\.uk", parts[2]):
match = re.match("([\w\-]+\.)*([\w\-]+\.[\w\-]+\.\w{2,6}$)", parts[2])
return match.group(2)
else: return ''
Seems to work pretty well.
However, it has to be modified to remove domain extensions on output as you wished.
how is this
=((?:(?:(?:http)s?:)?\/\/)?(?:(?:[a-zA-Z0-9]+)\.?)*(?:(?:[a-zA-Z0-9]+))\.[a-zA-Z0-9]{2,3})
(you may want to add "\/" to end of pattern
if your goal is to rid url's passed in as a param you may add the equal sign as the first char, like:
=((?:(?:(?:http)s?:)?//)?(?:(?:[a-zA-Z0-9]+).?)*(?:(?:[a-zA-Z0-9]+)).[a-zA-Z0-9]{2,3}/)
and replace with "/"
The goal of this example to get rid of any domain name regardless of the form it appears in.
(i.e. to ensure url parameters don't incldue domain names to avoid xss attack)
All answers here are very nice, but all will fails sometime.
So i know it is not common to link something else, already answered elsewhere, but you'll find that you have to not waste your time into impossible thing.
This because domains like mydomain.co.uk there is no way to know if an extracted domain is correct.
If you speak about to extract by URLs, something that ever have http or https or nothing in front (but if it is possible nothing in front, you have to remove
filter_var($url, filter_var($url, FILTER_VALIDATE_URL))
here below, because FILTER_VALIDATE_URL do not recognize as url a string that do not begin with http, so may remove it, and you can also achieve with something stupid like this, that never will fail:
$url = strtolower('hTTps://www.example.com/w3/forum/index.php');
if( filter_var($url, FILTER_VALIDATE_URL) && substr($url, 0, 4) == 'http' )
{
// array order is !important
$domain = str_replace(array("http://www.","https://www.","http://","https://"), array("","","",""), $url);
$spos = strpos($domain,'/');
if($spos !== false)
{
$domain = substr($domain, 0, $spos);
} } else { $domain = "can't extract a domain"; }
echo $domain;
Check FILTER_VALIDATE_URL default behavior here
But, if you want to check a domain for his validity, and ALWAYS be sure that the extracted value is correct, then you have to check against an array of valid top domains, as explained here:
https://stackoverflow.com/a/70566657/6399448
or you'll NEVER be sure that the extracted string is the correct domain. Unfortunately, all the answers here sometime will fails.
P.s the unique answer that make sense here seem to me this (i did not read it before sorry. It provide the same solution, even if do not provide an example as mine above mentioned or linked):
https://stackoverflow.com/a/569219/6399448
I know you actually asked for Regex and were not specific to a language. But In Javascript you can do this like this. Maybe other languages can parse URL in a similar way.
Easy Javascript solution
const domain = (new URL(str)).hostname.replace("www.", "");
Leave this solution in js for completeness.
In Javascript, the best way to do this is using the tld-extract npm package. Check out an example at the following link.
Below is the code for the same:
var tldExtract = require("tld-extract")
const urls = [
'http://www.mail.yahoo.co.in/',
'https://mail.yahoo.com/',
'https://www.abc.au.uk',
'https://github.com',
'http://github.ca',
'https://www.google.ru',
'https://google.co.uk',
'https://www.yandex.com',
'https://yandex.ru',
]
const tldList = [];
urls.forEach(url => tldList.push(tldExtract(url)))
console.log({tldList})
which results in the following output:
0: Object {tld: "co.in", domain: "yahoo.co.in", sub: "www.mail"}
1: Object {tld: "com", domain: "yahoo.com", sub: "mail"}
2: Object {tld: "uk", domain: "au.uk", sub: "www.abc"}
3: Object {tld: "com", domain: "github.com", sub: ""}
4: Object {tld: "ca", domain: "github.ca", sub: ""}
5: Object {tld: "ru", domain: "google.ru", sub: "www"}
6: Object {tld: "co.uk", domain: "google.co.uk", sub: ""}
7: Object {tld: "com", domain: "yandex.com", sub: "www"}
8: Object {tld: "ru", domain: "yandex.ru", sub: ""}
Found a custom function which works in most of the cases:
function getDomainWithoutSubdomain(url) {
const urlParts = new URL(url).hostname.split('.')
return urlParts
.slice(0)
.slice(-(urlParts.length === 4 ? 3 : 2))
.join('.')
}
You need a list of what domain prefixes and suffixes can be removed. For example:
Prefixes:
www.
Suffixes:
.com
.co.in
.au.uk
#!/usr/bin/perl -w
use strict;
my $url = $ARGV[0];
if($url =~ /([^:]*:\/\/)?([^\/]*\.)*([^\/\.]+)\.[^\/]+/g) {
print $3;
}
/^(?:https?:\/\/)?(?:www\.)?([^\/]+)/i
Just for knowledge:
'http://api.livreto.co/books'.replace(/^(https?:\/\/)([a-z]{3}[0-9]?\.)?(\w+)(\.[a-zA-Z]{2,3})(\.[a-zA-Z]{2,3})?.*$/, '$3$4$5');
# returns livreto.co
I know the question is seeking a regex solution but in every attempt it won't work to cover everything
I decided to write this method in Python which only works with urls that have a subdomain (i.e. www.mydomain.co.uk) and not multiple level subdomains like www.mail.yahoo.com
def urlextract(url):
url_split=url.split(".")
if len(url_split) <= 2:
raise Exception("Full url required with subdomain:",url)
return {'subdomain': url_split[0], 'domain': url_split[1], 'suffix': ".".join(url_split[2:])}
Let's say we have this: http://google.com
and you only want the domain name
let url = http://google.com;
let domainName = url.split("://")[1];
console.log(domainName);
Use this
(.)(.*?)(.)
then just extract the leading and end points.
Easy, right?