Validate the path of an url - regex

I'm trying to validate an url without domain, just the path and params.
The regular expression that I'm using do most of the work, but It has some errors that I dont know how to prevent (I'm pretty noob with regexp):
/^(\/([\w#!:.?+=&%#!\-\/])+)$/i
The next example are correctly validated
/asd.jsp -> true
/asd/asd.jsp -> true
/asd/asd.jsp?bar=baz&inga=42&quux -> true
/asd/asd.jsp?bar=ba z&inga=42&quux -> false
But this ones arent correct ulrs and them gives me true too:
/asd/asd./jsp -> true :(
/asd/asd.jsp/ -> true :(
/asd./asd.jsp -> true :(
/asd///asd.jsp -> true :(
/asd/asd.jsp&bar=baz?inga=42?quux -> true :(
Do you recommend to use a function instead of a regex?
Very much thanks!


Try this:
^(\\/\\w+)+\\.\\w+(\\?(\\w+=[\\w\\d]+(&\\w+=[\\w\\d]+)+)+)*$
I already escaped special characters, so you can directly use it in java.
By the way, /asd/asd.jsp?bar=baz&inga=42&quux is not a valid URL.
Unescaped Regex:
^(\/\w+)+\.\w+(\?(\w+=[\w\d]+(&\w+=[\w\d]+)+)+)*$


^(\/\w+)+(\.)?\w+(\?(\w+=[\w\d]+(&\w+=[\w\d]+)*)+){0,1}$
VALID:
/test
/test/test
/test.jsp
/test.jsp?v=1
/test.jsp?v=1&v=2
/test/test.jsp?v=1&v=2
INVALID:
/test/test./jsp
/test/test.jsp/
/test./test.jsp
/test///test.jsp

Related

Check value extracted with JSONPath and compare with regex using Gatling

Is there a way of doing something like this in Gatling:
scenario("Check UUID with regex")
.exec(http("Get UUID")
.get("http://myapp/api/v1/goal/a24e210c-0fc1-44a0-a5ca-9bd5d8d71916")
.check(jsonPath("$.id").is(regex("[0-9a-fA-F]{8}\-[0-9a-fA-F]{4}\-[0-9a-fA-F]{4}\-[0-9a-fA-F]{4}\-[0-9a-fA-F]{12}")))
Basically I want to apply a regex comparison on something returned in a check JSONPath construct.
Peace out!
P.S. I know I can do:
.check(regex("\"id\": \"[0-9a-fA-F]{8}\\-[0-9a-fA-F]{4}\\-[0-9a-fA-F]{4}\\-[0-9a-fA-F]{4}\\-[0-9a-fA-F]{12}\"").exists)

There's nothing quite like what you describe, but you can kind of fake it using a transform
.check(jsonPath("$.id")
.find
.transform(id => "[0-9a-fA-F]{8}\-[0-9a-fA-F]{4}\-[0-9a-fA-F]{4}\-[0-9a-fA-F]{4}\-[0-9a-fA-F]{12}".r
.findFirstMatchIn(id) match {
case Some(value) => true
case None => false
}
).is(true)
but if the straight regex check works... I'd go with that

Regex - string excluded both before and after main expression

Given three URLS:
http://stackoverflow.com/questionsss/3003310 -> true
http://stackoverflow.com/questionsss/3003310?s=1 -> false (s=1 found)
http://stackoverflow.com/questionsss?s=1&N=3003310 -> false (s=1 found)
How can I write a Regex in JS that returns true if 3003310 is found in the URL and false if s=1 is found anywhere in the URL.
Thanks!

maybe something like this:
^(?=.*3003310)(?!.*\bs=1\b).*
a positive lookahead, and a negative one, for each condition. the \b word boundaries around s=1 prevent false positives such as sass=1 or s=100.

REGEXP:
(.*\?s=1.*)
u should do a method, if result is equal to regex match. it will return FALSE else TRUE.
Original Text:
http://stackoverflow.com/questionsss/3003310 -> true
http://stackoverflow.com/questionsss/3003310?s=1 -> false (s=1 found)
http://stackoverflow.com/questionsss?s=1&N=3003310 -> false (s=1 found)
Result:
http://stackoverflow.com/questionsss/3003310?s=1 -> false (s=1 found)
http://stackoverflow.com/questionsss?s=1&N=3003310 -> false (s=1 found)
Algorithm:
Print(!OriginalText.equals(MethodCheck(Result)) // if empty. it will be TRUE
See: https://regex101.com/r/335HNW/1

Codeigniter - Regex Expression to include a period - preg_match

I have a regex expression as follows:
return ( ! preg_match("/^([-a-z_ -])+$/i", $str)) ? FALSE : TRUE;
I need to add to it to allow a period, so that the input could be something like this:
"st. austell" or "st. thomas"
I have tried various ways to add the period to the above rule, but the page either crashes out and just displays a blank page, or my validation errors are triggered.
I have tried things like this, but to no avail.
([-a-z_ -.])
([-a-z_ -\.])
([-a-z_ -])\.
(\.[-a-z_ -]) etc etc...
I have tried everything and don't seem to bee having any luck - any ideas
Many Thanks
James

Three points:
The initial regex is redundant, since it specifies - twice.
[a-z_ .-] should work, you just have to let a-z together and - just after [ or just before ] (otherwise, - will be considered as a metachar).
return !condition ? false : true; should be simplified to return condition;, since condition already returns true or false (same logic for if(!condition) {var = false;} else {var = true;} which should be simplified to var = condition;).

Rails 3: validate REGEX fails if no value input

I am using the following REGEX
VALID_WEBSITE_REGEX = /^(http|https):\/\/[a-z0-9]+([\-\.]{1}[a-z0-9]+)*\.[a-z]{2,6}(:[0-9]{1,5})?(\/.*)?$/ix
to validate a website entry with this rule:
validates :website, length: { maximum: 150 }, format: { with: VALID_WEBSITE_REGEX }
(The 150 is arbitrary).
However, when I save / update the form I get a validation error "website is invalid". How do I ensure that the 'format' section of the validation rule is processed only if there is content to process?

You can use allow_blank option for validation
:allow_blank => true
This option will let validation pass if the attribute’s value is blank?, like nil or an empty string for example.
read more:
http://guides.rubyonrails.org/active_record_validations_callbacks.html#allow_blank

Enclose the entire thing with the ? operator, e.g.
VALID_WEBSITE_REGEX = /^((http|https):\/\/[a-z0-9]+([\-\.]{1}[a-z0-9]+)*\.[a-z]{2,6}(:[0-9]{1,5})?(\/.*)?)?$/ix
If you want to allow whitespace too, then add \s* on each end, e.g.
VALID_WEBSITE_REGEX = /^\s*((http|https):\/\/[a-z0-9]+([\-\.]{1}[a-z0-9]+)*\.[a-z]{2,6}(:[0-9]{1,5})?(\/.*)?)?\s*$/ix

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?