Defining lists in VCL (Varnish Configuration Language) - list

Is there a way to define lists of domains in Varnish VCL language? I suppose something similar for ACLs. I would like to do something like this (using ACLs as an example).
acl website_list {
'(www\.)?domain.tld';
'(www\.)?domain2.tld';
}
...
if(req.http.Host ~ website_list) return(lookup);
I could just use separate RegEx tests but it isn't very re-usable if I want to use those domains somewhere else in the VCL.
Thanks!


You could have a test condition which sets a marker header, then test for that later on:
sub vcl_recv {
if (req.http.Host ~ "^(www\.)?domain.tld" ||
req.http.Host ~ "^(www\.)?domain2.tld") {
/* Set the magic marker */
set beresp.http.magicmarker = "1";
}
if (resp.http.magicmarker) {
return(lookup);
}
}

Related

ServiceStack: Routes.AddFromAssembly still uses /json/reply path and no URL-niceness for properties

I have a ServiceStack self-hosted webservice, using the AppSelfHostBase.
WHen the Configure method is executed, I have this:
public override void Configure(Container container)
{
Config.RouteNamingConventions = new List<RouteNamingConventionDelegate> {
RouteNamingConvention.WithRequestDtoName,
RouteNamingConvention.WithMatchingAttributes,
RouteNamingConvention.WithMatchingPropertyNames,
};
Routes.AddFromAssembly(typeof(ServiceStackHost).Assembly);
and I expected the following service to be executed under /StartBankIdAuthentication path, but it resides under /json/reply/StartBankIdAuthentication instead.
public class StartBankIdAuthentication : IReturn<StartBankIdAuthenticationResponse>
{
public string IdNbr { get; set; }
}
Also, is there an automatic way to make the properties in the DTO to be under "sub-paths", like /StartBankIdAuthentication/1234 instead of the /StartBankIdAuthentication?IdNbr=1234?
I know I can manually add the Route attribute, but it seems cumbersome and also messy in many ways (not Typed, error-prone etc).

I expected the following service to be executed under /StartBankIdAuthentication path, but it resides under /json/reply/StartBankIdAuthentication instead.
The /json/reply/StartBankIdAuthentication is a pre-defined route that's always available by default, they have no relation to Auto Generated Routes.
The default Route generation strategies you've listed are already registered by default and are what's applied when you use Routes.AddFromAssembly(). You should only override with route strategies you want in addition to the defaults, and you should use SetConfig() for any configuration in ServiceStack, e.g:
SetConfig(new HostConfig {
RouteNamingConventions = { MyCustomRouteStrategy }
});
The implementation for the different Route Strategies available in ServiceStack are in RouteNamingConvention.cs, you'll need to register your own strategy for anything additional Route strategies you want.
By default additional routes are generated for any Id or IDs property, the routing docs shows examples of how they can be customized:
The existing rules can be further customized by modifying the related static properties, e.g:
RouteNamingConvention.PropertyNamesToMatch.Add("UniqueId");
RouteNamingConvention.AttributeNamesToMatch.Add("DefaultIdAttribute");
Which will make these request DTOs:
class MyRequest1
{
public UniqueId { get; set;}
}
class MyRequest2
{
[DefaultId]
public CustomId { get; set;}
}
Generate the following routes:
/myrequest1
/myrequest1/{UniqueId}
/myrequest2
/myrequest2/{CustomId}
I know I can manually add the Route attribute, but it seems cumbersome and also messy in many ways (not Typed, error-prone etc).
If you really want you can use nameof() for Typed Routes:
[Route("/" + nameof(StartBankAuthentication) +"/"+ nameof(StartBankAuthentication.IdNbr))]

I'm not sure if Mythz will maybe come up with a different of better solution, but I managed to achieve what I wanted by overriding the GetRouteAttributes, and by using reflection, I could create what I wanted. It looks like this:
public override RouteAttribute[] GetRouteAttributes(Type requestType)
{
string fullname = requestType.FullName.Replace("AlfaOnlineServiceModel.Api.", "");
string path = "/" + fullname.ToLower().Replace(".", "/");
RouteAttribute[] routes = base.GetRouteAttributes(requestType);
if (routes.Length == 0)
{
routes = new RouteAttribute[1];
PropertyInfo[] pInfos = requestType.GetProperties(System.Reflection.BindingFlags.Public | System.Reflection.BindingFlags.Instance | System.Reflection.BindingFlags.DeclaredOnly);
foreach(PropertyInfo pi in pInfos)
{
path += "/{" + pi.Name + "}";
}
routes[0] = new RouteAttribute(path);
}
return routes;
}
Which will give for example:
MyMethodResult
The following routes are available for this service:
All Verbs /myCoolPath/mySubPath/myMethod/{MyProperty}

Spray Routing not matching HTTP Method correctly

I am using Spray Routing to try to match routes using different HTTP method but when I do a GET request it actually goes through DELETE, PUT and GET. I thought delete and put rejects all requests that are not HTTP DELETE or HTTP PUT.
This is my basic code:
path(Segment ~ Slash.?) { id =>
delete {
println("Hello from DELETE")
//do stuff for delete
complete("done for DELETE")
} ~
put {
println("Hello from PUT")
//do stuff for put
complete("done for PUT")
} ~
get {
println("Hello from GET")
//do stuff for get
complete("done for GET")
}
}
If I trigger a GET request I can see the app printing:
Hello from DELETE
Hello from PUT
Hello from GET
Am I missing any return call or something?

No, your code is (almost) correct.
The issue is that, in spray, the code that lives in a method matcher but does not live under an extraction (one of the directives "extracting" something, such as "parameters" or "segment") is executed all times.
In your case, you correctly match the path extractor, but after that the route executes for all get put delete etc.
The solution for this is to add the "dynamic" keyword right below your get/put etc. The downside is that you lose some performance.
path(...) {
get {
dynamic {
...
}
}
}
Alternatively, you can reshuffle your code so that the method matcher is at the top level, and the path extractor under it
get {
path(...) {
...
}
}

Using Varnish with multiple named vhosts using regex

I'm implementing Varnish (4.0) for a server with lots of named virtual hosts. Using a name
convention would make configuration easy. I want rename foo.com to foo.myhost.com.
sub vcl_recv {
if (req.http.host == "foo.com") {
set req.http.host = "foo.myhost.com";
}
set req.http.host = regsub(req.http.host, "(*.).com$","\1.myhost.com");
}
I cant get the regular expression right.

Symfony 2 Controller as a Service: container not injected

I want to be able to inject services into my controllers, so I had a look at http://symfony.com/doc/current/cookbook/controller/service.html and after some fiddling with the notation (could be a little more consistent but whatever) I have my WebTestCase using the service definition entry.
But the controller needs the container itself injected (and does indeed extend ContainerAware via the default framework controller), and the ControllerResolver in the FrameworkBundle does not do that.
Looking at the code (Symfony\Bundle\FrameworkBundle\Controller\ControllerResolver::createController()) this is not suprising:
protected function createController($controller)
{
if (false === strpos($controller, '::')) {
$count = substr_count($controller, ':');
if (2 == $count) {
// controller in the a:b:c notation then
$controller = $this->parser->parse($controller);
} elseif (1 == $count) {
// controller in the service:method notation
list($service, $method) = explode(':', $controller, 2);
return array($this->container->get($service), $method);
} else {
throw new \LogicException(sprintf('Unable to parse the controller name "%s".', $controller));
}
}
list($class, $method) = explode('::', $controller, 2);
if (!class_exists($class)) {
throw new \InvalidArgumentException(sprintf('Class "%s" does not exist.', $class));
}
$controller = new $class();
if ($controller instanceof ContainerAwareInterface) {
$controller->setContainer($this->container);
}
return array($controller, $method);
}
Apparently when using service:method notation,, it directly returns the controller from the container, not injecting the container itself.
Is this is a bug or am I missing something?

This is not a bug. It works as expected. This workflow generally "protects" the Controller as a Service concept. This way You need to look at Controller as a regular Service. In regular Service You inject everything what You need - if You need the controller itself - inject it explicitly.
To explain it more clearly, this "protection" I mentioned helps to avoid using service:method notation in one place and controller::method or bundle:controller:method in another.
So if not this "protection" it would be difficult to state that particular Controller is described as a Service or not as this would depend on which notation will be called first in Container build-up.

How to get domain name from URL

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?