Validating Youtube Playlist URL using Regex - regex

How to validate a youtube playlist url using regex?
I've found the answer for validating video from other question..
/^http:\/\/(?:www\.)?youtube.com\/watch\?(?=.*v=\w+)(?:\S+)?$/
But I've just unable to validate a url like this :
http://www.youtube.com/watch?list=PL1F9CA2A03CF286C2&v=pFS4zYWxzNA&
or
http://www.youtube.com/watch?v=pFS4zYWxzNA&list=PL1F9CA2A03CF286C2&

I ended up with a something similar but different:
/^.*(youtu.be\/|list=)([^#\&\?]*).*/
worked against the following urls:
https://www.youtube.com/watch?v=4jduuQh-Uho&list=PLBOh8f9FoHHjOz0vGrD20WcTtJar-LOrw&index=3
https://www.youtube.com/playlist?list=PLBOh8f9FoHHjOz0vGrD20WcTtJar-LOrw
Given the follow code:
var regExp = /^.*(youtu.be\/|list=)([^#\&\?]*).*/;
var match = url.match(regExp);
if (match && match[2]){
return match[2];
}
return null;
My return is PLBOh8f9FoHHjOz0vGrD20WcTtJar-LOrw

try this
^(https|http):\/\/(?:www\.)?youtube\.com\/watch\? #you forgot to mask the dot before com
( #may lead to wrong parsing
(v=.*&list=.*)| #v and then list
(list=.*&v=.*) #or versa verse - feel free to use better matching
) #for ids like "pFS4zYWxzNA"
(&.*)*$ #all other kinds of parameter
Edit:
i improved the matching
^(https|http):\/\/(?:www\.)?youtube\.com\/watch\?
(?:&.*)* #extra params at the beginning
(
(?:
v=([a-zA-Z0-9_\-]{11}) #propper mathing for id
(?:&.*)* #extras
&list=([a-zA-Z0-9_\-]{18}) #list
)
|
(?:
list=([a-zA-Z0-9_\-]{18})
(?:&.*)* #versa
&v=([a-zA-Z0-9_\-]{11})
)
)
(?:&.*)* #extras at the end
(?:\#.*)*$ #anchors

First validate YouTube url:
^https?\:\/\/(?:www\.youtube(?:\-nocookie)?\.com\/|m\.youtube\.com\/|youtube\.com\/)?(?:ytscreeningroom\?vi?=|youtu\.be\/|vi?\/|user\/.+\/u\/\w{1,2}\/|embed\/|watch\?(?:.*\&)?vi?=|\&vi?=|\?(?:.*\&)?vi?=)([^#\&\?\n\/<>\"']*)
then find the list:
[\?|&](list=)(.*)\&

I ended up with a something similar but different:
/[?&]list=([^#?&]*)/
worked against the following urls:
https://www.youtube.com/watch?v=HpqrKeD6ZTc&list=PLC18xlbCdwtTc61wuDviNoKmNUAVFT50D
https://www.youtube.com/playlist?list=PLC18xlbCdwtTc61wuDviNoKmNUAVFT50D
Given the follow code:
const listID = url.match(/[?&]list=([^#?&]*)/)?.[1];
console.log(listID);
My return is PLC18xlbCdwtTc61wuDviNoKmNUAVFT50D

Related

Angular2 pipe regex url detection

I would like to have a pipe which is detecting any url in a string and creating a link with it. At the moment I created this, which seems not working:
#Pipe({name:'matchUrl'})
export class MatchUrlPipe implements PipeTransform {
transform(value: string, arg?: any): any {
var exp = /https?:\/\/(www\.)?[-a-zA-Z0-9#:%._\+~#=]{2,256}\.[a-z]{2,4}\b([-a-zA-Z0-9#:%_\+.~#?&//=]*)/g;
return value.replace(exp, "<a href='$1'>$1</a>");
}
}
How can I fix it?
Seems like there are two problems with your implementation:
Your regex has the first capturing group ( $1 ) matching the 'www' part of the url. You want to change the regex like this for it to work (note the extra pair of parethesis at the start and end of the regex):
var exp = /(https?:\/\/(www\.)?[-a-zA-Z0-9#:%._\+~#=]{2,256}\.[a-z]{2,4}\b([-a-zA-Z0-9#:%_\+.~#?&//=]*))/g;
Pipes can't render html normally. You need a trick to do that as mentioned in other questione like this. You need to assign your 'piped value' to the attribute outerHTML of a span for example (the span will not be rendered).
Plunker example

Regular expression to validate image

I am writing a web application in golang. I am using regular expression to validate the URL. But I am not able to validate image (abc.png) in the URL validation.
var validPath = regexp.MustCompile("^/$|/(home|about|badge)/(|[a-zA-Z0-9]+)$")
The above URL takes /home/, /about/ but could not make for /abc.png. I mean . itself not working
I tried the following regex, but it didn't help
var validPath = regexp.MustCompile("^/$|/(home|about|badge|.)/(|[a-zA-Z0-9]+)$")
var validPath = regexp.MustCompile("^/$|/(home|about|badge)(/|.)(|[a-zA-Z0-9]+)$")
And I am trying to match http://localhost:8080/badge.png
Could anyone please help me on this?
It appears
^/$|^(?:/(home|about|badge))?/((?:badge|abc)\.png|[a-zA-Z0-9]*)$
should work for you. See the regex demo.
The pattern breakdown:
^/$ - a / as a whole string
| - or...
^ - start of string
(?:/(home|about|badge))? - optional sequence of / + either home, or about or badge
/ - a / symbol followed with
((?:badge|abc)\.png|[a-zA-Z0-9]*) - Group 1 capturing:
(?:badge|abc)\.png - badge or abc followed with .png
| - or...
[a-zA-Z0-9]* - zero or more alphanumerics
$ - end of string
And here is the Go playground demo.
package main
import "fmt"
import "regexp"
func main() {
//var validPath = regexp.MustCompile("^/((home|about)(/[a-zA-Z0-9]*)?|[a-zA-Z0-9]+\\.[a-z]+)?$")
var validPath = regexp.MustCompile(`^/$|^(?:/(home|about|badge))?/((?:badge|abc)\.png|[a-zA-Z0-9]*)$`)
fmt.Println(validPath.MatchString("/"), validPath.MatchString("/home/"), validPath.MatchString("/about/"), validPath.MatchString("/home/13jia0"), validPath.MatchString("/about/1jnmjan"), validPath.MatchString("/badge.png"), validPath.MatchString("/abc.png"))
fmt.Println(validPath.MatchString("/nope/"), validPath.MatchString("/invalid.png"), validPath.MatchString("/test/test"))
m := validPath.FindStringSubmatch("/about/global")
fmt.Println("validate() :: URL validation path m[1] : ", m[1])
fmt.Println("validate() :: URL validation path m[2] : ", m[2])
if m == nil || m[2] != "global" {
fmt.Println("Not valid")
}
}
What you are looking for is the following (based off the example paths you posted):
var validPath = regexp.MustCompile("^/((home|about)(/[a-zA-Z0-9]*)?|[a-zA-Z0-9]+\\.[a-z]+)?$")
Playground with examples
You can validate with the following Regex:
var validPath = regexp.MustCompile("^\/(home|about|badge)\/[a-zA-Z0-9]+[.][a-z]+$")
Ps: I made a flexible Regex, so it accepts a lot of formats of images: png, jpg, jpeg and so on..
You can test it here: Regex

Using Regular Expression in Mvc 4 Razor View's cshttml page for jquery email validation

My Jquery Regular expression for email validation throwing syntax error.
Error : "Unexpected character \". Below is my code. please anyone give me right solution.
function validateEmail(sEmail) {
var filter = /^([\w-\.]+)#((\[[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.)|(([\w-]+\.)+))([a-zA-Z]{2,4}|[0-9]{1,3})(\]?)$/;
if (filter.test(sEmail)) {
return true;
}
else {
return false;
}
}​
You have to escape the # sign with two ## alike so :
var filter = /^([\w-\.]+)##((\[[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.)|(([\w-]+\.)+))([a-zA-Z]{2,4}|[0-9]{1,3})(\]?)$/;
try
var filter = /^([-\w\.]+)#...
note the - upfront instead of \w-\.
- in between means range as in [a-z] here with `\w' it does not make sense.
btw whats with \[ and \] after #
Try this:
(?:.*)#(?:.*).(?:.*)

Regular expression for youtube links

Does someone have a regular expression that gets a link to a Youtube video (not embedded object) from (almost) all the possible ways of linking to Youtube?
I think this is a pretty common problem and I'm sure there are a lot of ways to link that.
A starting point would be:
http://www.youtube.com/watch?v=iwGFalTRHDA
http://www.youtube.com/watch?v=iwGFalTRHDA&feature=related
http://youtu.be/iwGFalTRHDA
http://youtu.be/n17B_uFF4cA
http://www.youtube.com/embed/watch?feature=player_embedded&v=r5nB9u4jjy4
http://www.youtube.com/watch?v=t-ZRX8984sc
http://youtu.be/t-ZRX8984sc
... please add more possible links and/or regular expressions to detect them.
So far I got this Regular expression working for the examples I posted, and it gets the ID on the first group:
http(?:s?):\/\/(?:www\.)?youtu(?:be\.com\/watch\?v=|\.be\/)([\w\-\_]*)(&(amp;)?‌​[\w\?‌​=]*)?
You can use this expression below.
(?:https?:\/\/)?(?:www\.)?youtu\.?be(?:\.com)?\/?.*(?:watch|embed)?(?:.*v=|v\/|\/)([\w\-_]+)\&?
I'm using it, and it cover the most used URLs.
I'll keep updating it on This Gist.
You can test it on this tool.
I like #brunodles's solution the most but you can still match non video links like https://www.youtube.com/feed/subscriptions
I went with this solution
(?:https?:\/\/)?(?:www\.)?youtu(?:\.be\/|be.com\/\S*(?:watch|embed)(?:(?:(?=\/[-a-zA-Z0-9_]{11,}(?!\S))\/)|(?:\S*v=|v\/)))([-a-zA-Z0-9_]{11,})
It can also be used to match multiple whitespace separated links.
The video id will be captured in the first group.
Tested with the following urls:
youtu.be/iwGFalTRHDA
youtube.com/watch?v=iwGFalTRHDA
www.youtube.com/watch?v=iwGFalTRHDA
http://www.youtube.com/watch?v=iwGFalTRHDA
https://www.youtube.com/watch?v=iwGFalTRHDA
https://www.youtube.com/watch?v=MoBL33GT9S8&feature=share
https://www.youtube.com/embed/watch?feature=player_embedded&v=iwGFalTRHDA
https://www.youtube.com/embed/watch?v=iwGFalTRHDA
https://www.youtube.com/embed/v=iwGFalTRHDA
https://www.youtube.com/watch/iwGFalTRHDA
http://www.youtube.com/attribution_link?u=/watch?v=aGmiw_rrNxk&feature=share
https://m.youtube.com/watch?v=iwGFalTRHDA
// will not match
https://www.youtube.com/feed/subscriptions
https://www.youtube.com/channel/UCgc00bfF_PvO_2AvqJZHXFg
https://www.youtube.com/c/NatGeoEdOrg/videos
https://regex101.com/r/rq2KLv/1
I improved the links posted above with a friend for a script I wrote for IRC to recognize even links without http at all. It worked on all stress tests I got so far, including garbled text with barely recognizable youtube urls, so here it is:
~(?:https?://)?(?:www\.)?youtu(?:be\.com/watch\?(?:.*?&(?:amp;)?)?v=|\.be/)([\w\-]+)(?:&(?:amp;)?[\w\?=]*)?~
I testet all the regular expressions that are shown here and none could cover all url types that my client was using.
I built this pretty much through trial and error, but it seems to work with all the patterns that Poppy Deejay posted.
"(?:.+?)?(?:\/v\/|watch\/|\?v=|\&v=|youtu\.be\/|\/v=|^youtu\.be\/)([a-zA-Z0-9_-]{11})+"
Maybe it helps someone who is in a similar situation that I had today ;)
Piggy backing on Fanmade, this covers the below links including the url encoded version of attribution_links:
(?:.+?)?(?:\/v\/|watch\/|\?v=|\&v=|youtu\.be\/|\/v=|^youtu\.be\/|watch\%3Fv\%3D)([a-zA-Z0-9_-]{11})+
https://www.youtube.com/attribution_link?a=tolCzpA7CrY&u=%2Fwatch%3Fv%3DMoBL33GT9S8%26feature%3Dshare
https://www.youtube.com/watch?v=MoBL33GT9S8&feature=share
http://www.youtube.com/watch?v=iwGFalTRHDA
https://www.youtube.com/watch?v=iwGFalTRHDA
http://www.youtube.com/watch?v=iwGFalTRHDA&feature=related
http://youtu.be/iwGFalTRHDA
http://www.youtube.com/embed/watch?feature=player_embedded&v=iwGFalTRHDA
http://www.youtube.com/embed/watch?v=iwGFalTRHDA
http://www.youtube.com/embed/v=iwGFalTRHDA
http://www.youtube.com/watch?feature=player_embedded&v=iwGFalTRHDA
http://www.youtube.com/watch?v=iwGFalTRHDA
www.youtube.com/watch?v=iwGFalTRHDA
www.youtu.be/iwGFalTRHDA
youtu.be/iwGFalTRHDA
youtube.com/watch?v=iwGFalTRHDA
http://www.youtube.com/watch/iwGFalTRHDA
http://www.youtube.com/v/iwGFalTRHDA
http://www.youtube.com/v/i_GFalTRHDA
http://www.youtube.com/watch?v=i-GFalTRHDA&feature=related
http://www.youtube.com/attribution_link?u=/watch?v=aGmiw_rrNxk&feature=share&a=9QlmP1yvjcllp0h3l0NwuA
http://www.youtube.com/attribution_link?a=fF1CWYwxCQ4&u=/watch?v=qYr8opTPSaQ&feature=em-uploademail
http://www.youtube.com/attribution_link?a=fF1CWYwxCQ4&feature=em-uploademail&u=/watch?v=qYr8opTPSaQ
I've been having problems lately with the atttribution_link urls so i tried making my own regex that works for those too.
Here is my regex string:
(https?://)?(www\\.)?(yotu\\.be/|youtube\\.com/)?((.+/)?(watch(\\?v=|.+&v=))?(v=)?)([\\w_-]{11})(&.+)?
and here are some test cases i've tried:
http://www.youtube.com/watch?v=iwGFalTRHDA
https://www.youtube.com/watch?v=iwGFalTRHDA
http://www.youtube.com/watch?v=iwGFalTRHDA&feature=related
http://youtu.be/iwGFalTRHDA
http://www.youtube.com/embed/watch?feature=player_embedded&v=iwGFalTRHDA
http://www.youtube.com/embed/watch?v=iwGFalTRHDA
http://www.youtube.com/embed/v=iwGFalTRHDA
http://www.youtube.com/watch?feature=player_embedded&v=iwGFalTRHDA
http://www.youtube.com/watch?v=iwGFalTRHDA
www.youtube.com/watch?v=iwGFalTRHDA
www.youtu.be/iwGFalTRHDA
youtu.be/iwGFalTRHDA
youtube.com/watch?v=iwGFalTRHDA
http://www.youtube.com/watch/iwGFalTRHDA
http://www.youtube.com/v/iwGFalTRHDA
http://www.youtube.com/v/i_GFalTRHDA
http://www.youtube.com/watch?v=i-GFalTRHDA&feature=related
http://www.youtube.com/attribution_link?u=/watch?v=aGmiw_rrNxk&feature=share&a=9QlmP1yvjcllp0h3l0NwuA
http://www.youtube.com/attribution_link?a=fF1CWYwxCQ4&u=/watch?v=qYr8opTPSaQ&feature=em-uploademail
http://www.youtube.com/attribution_link?a=fF1CWYwxCQ4&feature=em-uploademail&u=/watch?v=qYr8opTPSaQ
Also remember to check the string you get for your video url, sometimes it may get the percent characters. If so just do this
url = [url stringByReplacingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
and it should fix it.
Remember also that the index of the youtube key is now index 9.
NSRange youtubeKey = [result rangeAtIndex:9]; //the youtube key
NSString * strKey = [url substringWithRange:youtubeKey] ;
It'd be the longest RegEx in the world if you managed to cover all link formats, but here's one to get you started which will cover the first couple of link formats:
http://(www\.)?youtube\.com/watch\?.*v=([a-zA-Z0-9]+).*
The second group will match the video ID if you need to get that out.
(?:http?s?:\/\/)?(?:www.)?(?:m.)?(?:music.)?youtu(?:\.?be)(?:\.com)?(?:(?:\w*.?:\/\/)?\w*.?\w*-?.?\w*\/(?:embed|e|v|watch|.*\/)?\??(?:feature=\w*\.?\w*)?&?(?:v=)?\/?)([\w\d_-]{11})(?:\S+)?
https://regex101.com/r/nJzgG0/3
Detects YouTube and YouTube Music link in any string
I took all variants from here:
https://gist.github.com/rodrigoborgesdeoliveira/987683cfbfcc8d800192da1e73adc486#file-youtubeurlformats-txt
And built this regexp (YouTube ID is in group 2):
(\/|%3D|v=|vi=)([0-9A-z-_]{11})[%#?&\s]
Check it here: https://regexr.com/4u4ud
Edit: Works for any single string w/o breaks.
I'm working with that kind of links:
http://www.youtube.com/v/M-faNJWc9T0?fs=1&rel=0
And here's the regEx I'm using to get ID from it:
"(.+?)(\/v/)([a-zA-Z0-9_-]{11})+"
This is iterating on the existing answers and handles edge cases better. (for example http://thisisnotyoutu.be/thing)
/(?:https?:\/\/|www\.|m\.|^)youtu(?:be\.com\/watch\?(?:.*?&(?:amp;)?)?v=|\.be\/)([\w‌​\-]+)(?:&(?:amp;)?[\w\?=]*)?/
here is the complete solution for getting youtube video id for java or android, i didn't found any link which doesn't work with this function
public static String getValidYoutubeVideoId(String youtubeUrl)
{
if(youtubeUrl == null || youtubeUrl.trim().contentEquals(""))
{
return "";
}
youtubeUrl = youtubeUrl.trim();
String validYoutubeVideoId = "";
String regexPattern = "^(?:https?:\\/\\/)?(?:[0-9A-Z-]+\\.)?(?:youtu\\.be\\/|youtube\\.com\\S*[^\\w\\-\\s])([\\w\\-]{11})(?=[^\\w\\-]|$)(?![?=&+%\\w]*(?:['\"][^<>]*>|<\\/a>))[?=&+%\\w]*";
Pattern regexCompiled = Pattern.compile(regexPattern, Pattern.CASE_INSENSITIVE);
Matcher regexMatcher = regexCompiled.matcher(youtubeUrl);
if(regexMatcher.find())
{
try
{
validYoutubeVideoId = regexMatcher.group(1);
}
catch(Exception ex)
{
}
}
return validYoutubeVideoId;
}
This is my answer to use in Scala. This is useful to extract 11 digits from Youtube's URL.
"https?://(?:[0-9a-zA-Z-]+.)?(?:www.youtube.com/|youtu.be\S*[^\w-\s])([\w -]{11})(?=[^\w-]|$)(?![?=&+%\w](?:[\'"][^<>]>|))[?=&+%\w-]*"
def getVideoLinkWR: UserDefinedFunction = udf(f = (videoLink: String) => {
val youtubeRgx = """https?://(?:[0-9a-zA-Z-]+\.)?(?:youtu\.be/|youtube\.com\S*[^\w\-\s])([\w \-]{11})(?=[^\w\-]|$)(?![?=&+%\w]*(?:[\'"][^<>]*>|</a>))[?=&+%\w-./]*""".r
videoLink match {
case youtubeRgx(a) => s"$a".toString
case _ => videoLink.toString
}
}
Youtube video URL Change to iframe supported link:
REGEX: https://regex101.com/r/LeZ9WH/2/
http://www.youtube.com/watch?v=iwGFalTRHDA
http://www.youtube.com/watch?v=iwGFalTRHDA&feature=related
http://youtu.be/iwGFalTRHDA
http://youtu.be/n17B_uFF4cA
http://www.youtube.com/embed/watch?feature=player_embedded&v=r5nB9u4jjy4
http://www.youtube.com/watch?v=t-ZRX8984sc
http://youtu.be/t-ZRX8984sc
https://youtu.be/2sFlFPmUfNo?t=1
Php function example:
if (!function_exists('clean_youtube_link')) {
/**
* #param $link
* #return string|string[]|null
*/
function clean_youtube_link($link)
{
return preg_replace(
'#(.+?)(\/)(watch\x3Fv=)?(embed\/watch\x3Ffeature\=player_embedded\x26v=)?([a-zA-Z0-9_-]{11})+#',
"https://www.youtube.com/embed/$5",
$link
);
}
}
This should work for almost all youtube links when extracting from a string:
((?:https?:)?\/\/)?((?:www|m)\.)?((?:youtube\.com|youtu.be))(\/(?:[\w\-]+\?v=|embed\/|v\/)?)([\w\-]{10}).\b
var isValidYoutubeLink: Bool{
// working for all the youtube url's
NSPredicate(format: "SELF MATCHES %#", "(?:http?s?:\\/\\/)?(?:www.)?(?:m.)?(?:music.)?youtu(?:\\.?be)(?:\\.com)?(?:(?:\\w*.?:\\/\\/)?\\w*.?\\w*-?.?\\w*\\/(?:embed|e|v|watch|.*\\/)?\\??(?:feature=\\w*\\.?\\w*)?&?(?:v=)?\\/?)([\\w\\d_-]{11})(?:\\S+)?").evaluate(with: self)
}
With this Javascript Regex, the first capture is a video ID :
^(?:https?:)?(?:\/\/)?(?:www\.)?(?:youtu\.be\/|youtube(?:\-nocookie)?\.(?:[A-Za-z]{2,4}|[A-Za-z]{2,3}\.[A-Za-z]{2})\/)(?:watch|embed\/|vi?\/)*(?:\?[\w=&]*vi?=)?([^#&\?\/]{11}).*$
(?-s)^https?\W+(?:www\.|m\.|music\.)*youtu\.?be(?:\.com|\/watch|\/o?embed|\/shorts|\/attribution_link\?[&\w\-=]*[au]=|\/ytsc\w+|[\?&\/]+[ve]i?\b|\?feature=\w+|-nocookie)*[\/=]([a-z\d\-_]{11})[\?&#% \t ] *.*$
or
(?-s)^(?:(?!https?[:\/]|www\.|m\.yo|music\.yo|youtu\.?be[\/\.]|watch[\/\?]|embed\/)\V)*(?:https?[:\/]+|www\.|m\.|music\.)+youtu\.?be(?:\.com\/|watch|o?embed(?:\/|\?url=\S+?)?|shorts|attribution_link\?[&\w\-=]*[au]=\/?|ytsc\w+|[\?&]*[ve]i?\b|\?feature=\w+|[\?&]time_continue=\d+|-nocookie|%[23][56FD])*(?:[\/=]|%2F|%3D)([a-z\d\-_]{11})[\?&#% \t ]? *.*$
(the part >>#% \t⠀ ]<< should contain continuous space, which is Alt+255, but stackoverflow-com can't print it)
(this string may be replaced to \1, sorted and abbreviated with: )
V█(?-i)^([A-Za-z\d\-_]{11})(?:\v+\1)*$
>█https:\/\/youtu\.be\/\1
(./dot can take up any symbol; \V or [^\r\n] can any except special, emoji and others; this >> [^!-⠀:/‽|\s] << can grab some emoji)
https://youtu.be/x26ANNC3C-8 • ♾ 𝕳𝕰𝕽𝕰𝕿𝕳𝕰𝖄𝕮𝕺𝕸𝕰 - 𝔩𝔢𝔞𝔳𝔢 𝔪𝔢 𝔞𝔩𝔬𝔫𝔢 • 7:15
This regex solve my problem, I can get youtube link having watch, embed or shared link
(?:http(?:s)?:\/\/)?(?:www\.)?(?:youtu\.be\/|youtube\.com\/(?:(?:watch)?\?(?:.*&)?v(?:i)?=|(?:embed|v|vi|user)\/))([^\?&\"'<> #]+)
You can check here https://regex101.com/r/Kvk0nB/1

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?