I need to use a regex to pull a value out a url domain that will exclude everything but the host (ex: wordpress) and domain type (ex .com). The urls are dynamic and contain 2-3 values for each result (www.example.com or example.org). I am trying to use this expression, but I am only getting back the first letter of every item I am attempting to exclude:
Expresssion
(?!wordpress|com|www)(\w+|\d+)
String
example.wordpress.com
Results
example
ordpress
om
Desired Result
example
Any assistance would be greatly appreciated
Anchor your regular expression:
\b(?!wordpress|com|www)(\w+|\d+)\b
You might also want to consider whether (\w+|\d+) is really what you mean. \w already includes digits. Also, there are other characters allowed in URLs such as -. Do you need to handle this?
If I was to do thing like that, I would take advantage of the format of the url: anything (dot) 2nd-level-domain (dot) 1st-level-domain:
^(?<level3>.*)[.]?(?<level2>.+)[.](?<level1>.+)$
Is it so that you are only after what is after the domain part??
(/\/(?!\/).*?\/(.*)/).exec("http://www.google.com/sdfsdf/fdsff")[1]
// returns sdfsdf/fdsff
Related
I've racked my brain over this JS regex and have so far only managed to get parts of it to work or the whole thing to work in certain circumstances.
I have a string like this:
Some string<br>http://anysubdomain.particulardomain.com<br>Rest of string
The goal is to move the domain part to the end of the string, if it's there. The http part is also optional and can also be https. The TLD is always particulardomain.com, the subdomain can be anything.
I've managed to get everything into capture groups when the domain with protocol is present with this regex:
(.*)(https?\:\/\/[a-z\d\-]*\.particulardomain\.com)(.*)
But any attempt at making the domain part and the protocol part within it optional has resulted in no or the wrong matches.
The end result I'm looking for is to have the three parts of the string – beginning, domain, end – in separate capture groups so I can move capture group 2 (the domain part) to the end, or, if there's no domain present, the whole string in the first capture group.
To clarify, here are some examples with the expected output/capture groups:
INPUT:
Some string<br>http://anysubdomain.particulardomain.com<br>Rest of string
OR (no protocol):
Some string<br>anysubdomain.particulardomain.com<br>Rest of string
OUTPUT:
$1: Some string<br>
$2: http://anysubdomain.particulardomain.com
$3: <br>Rest of string
INPUT:
Some string<br>Rest of string
OUTPUT:
$1: Some string<br>Rest of string
$2: empty
$3: empty
One mistake in your regex is that it contains only particular whereas
the source text contains particulardomain, but this is a detail.
Now let's move to the protocol part. You put only one ? (after s),
which means that only s is optional, but both http and :
are still required.
To make the whole protocol optional, you must:
enclose it with a group (either capturing or not),
make this group optional (put ? after it).
And now maybe the most important thing: Your regex starts with (.*).
Note that it is greedy version, which:
initially tries to capture the whole rest of source string,
then moves back one char by one, to allow matching by the
following part of regex.
Change it to reluctant version (.*?) and then optional
group (https?:)? will match as expected.
Another detail: \ before : is not needed. It does not do
any harm either, but due to the principle "Keep It Simple...",
I recommend to delete it (as I did above).
One more detail: After [a-z\d\-] (subdomain part) you should put
+, not *, as this part may not be empty.
So the whole regex can be:
(.*?)((https?:)?\/\/[a-z\d\-]+\.particulardomain\.com)(.*)
And the last remark: I am in doubt, whether you really need three
capturing groups. Maybe it would be enough to leave only the content
of the middle capturing group, i.e.:
(https?:)?\/\/[a-z\d\-]+\.particulardomain\.com
Found a solution. Since, as stated, the goal is to move the domain to the end of the string, if it's present, I'm just matching the domain and anything after it. If there's no domain, nothing matches and hence nothing gets replaced. The problem was the two .* both at the beginning and the end of the regex. Only the one at the end is needed.
REGEX:
([a-z\d\-:\/]+\.particulardomain\.com)(.*)
Works for the following strings:
Domain present:
Start of string 1234<br>https://subdomain.particulardomain.com<br>End of string 999
Domain without protocol:
Start of string 1234<br>subdomain.particulardomain.com<br>End of string 999
No domain:
Start of string 1234<br>End of string 999
Thanks everyone for helping me rethink the problem!
I see good answer here, as you explained you need three group and set the domain to the back of the string(to be clear the entire url or only the domain e.g particulardomain.com)
You can do this:
//Don't know if the <br> tag matter for you problem, suppose it not
//this is you input
let str = "Start of string 1234<br>https://subdomain.particulardomain.com<br>End of string 99";
let group = str.split(<br>);
let indexOfDomain;
/*moere code like a for loop or work with a in-build funcion of the array with the regExp you made /[a-z\d\-:\/]+\.particulardomain\.com/ you can validated the domain separately.
}
TO HAVE IN MIND:
With your solution will not work at 100%, why?
your regExp:
([a-z\d\-:\/]+\.particulardomain\.com)(.*)
will mach a http, https, *(any other thing that is not a protocol) and will not work for this input you can test if you like and do a comment
Start of string 1234<br>End of string 999
The regExp that #Valdi_Bo answer:
(.*?)((https?:)?\/\/[a-z\d\-]+\.particulardomain\.com)(.*)
will fit to the what you described in the question
This regExp don't fit all yours input maybe he did not test it for all your input as you did not explained in your question like you did in your own answer
In conclusion at the end you need to extract the domain (wich don't know if is the entire url as you mix up the idea). If you are not going to use the do a split and then validated the regExp it will be more easy
I have to set some rules on not accepting wrong url for my project. I am using regex for this.
My Url is "http ://some/resource/location".
This url should not allow space in beginning or middle or in end.
For example these spaces are invalid:
"https ://some/(space here in middle) resource/location"
"https ://some/resource/location (space in end)"
"(space in starting) https ://some/resource/location"
"https ://(space here) some/resource/location"
Also these scenario's are invalid.
"httpshttp ://some/resource/location"
"https ://some/resource/location,https ://some/resource/location"
Currently I am using a regex
var regexp = /(ftp|http|https):\/\/(\w+:{0,1}\w*#)?(\S+)(:[0-9]+)?(\/|\/([\w#!:.?+=&%#!\-\/]))?/;
This regex accepts all those invalid scenarios. I am unable to find the correct matching regex which will accept only if the url is valid. Can anyone help me out on this?
We need to validate n number of scenarios for URL validation. If your particular about your given pattern then above regex expression from other answer looks good.
Or
If you want to take care of all the URL validation scenarios please refer In search of the perfect URL validation regex
/(ftp|http|https){1}:\/\/(?:.(?! ))+$/
is this regex OK ?
use this
^\?([\w-]+(=[\w-]*)?(&[\w-]+(=[\w-]*)?)*)?$
See live demo
This considers each "pair" as a key followed by an optional value (which maybe blank), and has a first pair, followed by an optional & then another pair,and the whole expression (except for the leading?) is optional. Doing it this way prevents matching ?&abc=def
Also note that hyphen doesn't need escaping when last in the character class, allowing a slight simplification.
You seem to want to allow hyphens anywhere in keys or values. If keys need to be hyphen free:
^\?(\w+(=[\w-]*)?(&\w+(=[\w-]*)?)*)?$
I'm trying to find a simple regexp for url validation, but not very good in regexing..
Currently I have such regexp: (/^https?:\/\/\w/).test(url)
So it's allowing to validate urls as http://localhost:8080 etc.
What I want to do is NOT to validate urls if they have some long special characters at the end like: http://dodo....... or http://dododo&&&&&
Could you help me?
How about this?
/^http:\/\/\w+(\.\w+)*(:[0-9]+)?\/?(\/[.\w]*)*$/
Will match: http://domain.com:port/path or just http://domain or http://domain:port
/^http:\/\/\w+(\.\w+)*(:[0-9]+)?\/?$/
match URLs without path
Some explanations of regex blocks:
Domain: \w+(\.\w+)* to match text with dots: localhost or www.yahoo.com (could be as long as Path or Port section begins)
Port: (:[0-9]+)? to match or to not match a number starting with semicolon: :8000 (and it could be only one)
Path: \/?(\/[.\w]*)* to match any alphanums with slashes and dots: /user/images/0001.jpg (until the end of the line)
(path is very interesting part, now I did it to allow lone or adjacent dots, i.e. such expressions could be possible: /. or /./ or /.../ and etc. If you'd like to have dots in path like in domain section - without border or adjacent dots, then use \/?(\/\w+(.\w+)*)* regexp, similar to domain part.)
* UPDATED *
Also, if you would like to have (it is valid) - characters in your URL (or any other), you should simply expand character class for "URL text matching", i.e. \w+ should become [\-\w]+ and so on.
If you want to match ABCD then you may leave the start part..
For Example to match http://localhost:8080
' just write
/(localhost).
if you want to match specific thing then please focus the term that you want to search, not the starting and ending of sentence.
Regular expression is for searching the terms, until we have a rigid rule for the same. :)
i hope this will do..
It depends on how complex you need the Regex to be. A simple way would be to just accept words (and the port/domain):
^https?:\/\/\w+(:[0-9]*)?(\.\w+)?$
Remember you need to use the + character to match one or more characters.
Of course, there are far better & more complicated solutions out there.
^https?:\/\/localhost:[0-9]{1,5}\/([-a-zA-Z0-9()#:%_\+.~#?&\/=]*)
match:
https://localhost:65535/file-upload-svc/files/app?query=abc#next
not match:
https://localhost:775535/file-upload-svc/files/app?query=abc#next
explanation
it can only be used for localhost
it also check the value for port number since it should be less than 65535 but you probably need to add additional logic
You can use this. This will allow localhost and live domain as well.
^https?:\/\/\w+(\.\w+)*(:[0-9]+)?(\/.*)?$
I'm pretty late to the party but now you should consider validating your URL with the URL class. Avoid the headache of regex and rely on standard
let isValid;
try {
new URL(endpoint); // Will throw if URL is invalid
isValid = true;
} catch (err) {
isValid = false;
}
^https?:\/\/(localhost:([0-9]+\.)+[a-zA-Z0-9]{1,6})?$
Will match the following cases :
http://localhost:3100/api
http://localhost:3100/1
http://localhost:3100/AP
http://localhost:310
Will NOT match the following cases :
http://localhost:3100/
http://localhost:
http://localhost
http://localhost:31
I 'borrowed' a regex from this website : http://daringfireball.net/2010/07/improved_regex_for_matching_urls that is almost complete but i want to match exemple.com
I know that stackoverflow is not doyourhomework.com but I passed a long time thinking without results. Here is a fiddle to test : http://jsfiddle.net/BGnMm/25/ and you can see at the end that exemple.com is not a link.
var reg=/\b((?:[a-z][\w-]+:(?:\/*)|www\d{0,3}[.]|[a-z0-9.\-]+[.][a-z]{2,4}\/)(?:[^\s()<>]+|\(([^\s()<>]+|(\([^\s()<>]+\)))*\))+(?:\(([^\s()<>]+|(\([^\s()<>]+\)))*\)|[^\s`!()\[\]{};:'".,<>?«»“”‘’]))/gi;
var allurl="http:foo.com/blah_blah http://foo.com/blah_blah/ (Something like http://foo.com/blah_blah) http://foo.com/blah_blah_(wikipedia) http://foo.com/more_(than)_one_(parens) (Something like http://foo.com/blah_blah_(wikipedia)) http://foo.com/blah_(wikipedia)#cite-1 http://foo.com/blah_(wikipedia)_blah#cite-1 http://foo.com/unicode_(✪)_in_parens http://foo.com/(something)?after=parens http://foo.com/blah_blah. http://foo.com/blah_blah/. <http://foo.com/blah_blah> <http://foo.com/blah_blah/> http://foo.com/blah_blah, http://www.extinguishedscholar.com/wpglob/?p=364. http://✪df.ws/1234 rdar://1234 rdar:/1234 x-yojimbo-item://6303E4C1-6A6E-45A6-AB9D-3A908F59AE0E message://%3c330e7f840905021726r6a4ba78dkf1fd71420c1bf6ff#mail.gmail.com%3e http://➡.ws/䨹 www.c.ws/䨹 <tag>http://example.com</tag> Just a www.example.com link. http://example.com/something?with,commas,in,url, but not at end What about <mailto:gruber#daringfireball.net?subject=TEST> (including brokets). mailto:name#example.com bit.ly/foo “is.gd/foo/” WWW.EXAMPLE.COM http://www.asianewsphoto.com/(S(neugxif4twuizg551ywh3f55))/Web_ENG/View_DetailPhoto.aspx?PicId=752 http://www.asianewsphoto.com/(S(neugxif4twuizg551ywh3f55)) http://lcweb2.loc.gov/cgi-bin/query/h?pp/horyd:#field(NUMBER+#band(thc+5a46634)) 6:00p filename.txt http://example.com/quotes-are-“part” ✪df.ws/1234 example.com example.com/";
document.write(allurl.replace(reg,"<a href='$1' >$1</a><br />"));
Add an alternation operator (|) after the {2,4}\/, i.e.
var reg=/\b((?:[a-z][\w-]+:(?:\/*)|www\d{0,3}[.]|[a-z0-9.\-]+[.][a-z]{2,4}\/|)(?:[^\s()<>]+|\(([^\s()<>]+|(\([^\s()<>]+\)))*\))+(?:\(([^\s()<>]+|(\([^\s()<>]+\)))*\)|[^\s`!()\[\]{};:'".,<>?«»“”‘’]))/gi;
There's something you should understand about this. The first non-captured group, (?: … ), looks for "indicators" of URLs. One indicator, for example, is the www (followed by up to 3 digits of numbers). You however are asking for a way to identify URLs without any indicator at all. So, what we've done above is we've added a clause, "or an empty match," as a "valid" indicator. The consequence of this is that your regular expression is less selective now: all sorts of strings, not only example.com but also filename.txt, 3.141593, and omg...really are identified as URLs! Your only other (readily available) option is to be more selective about suffixes, e.g. require specific suffixes (com|org|net), but then this takes away from the generality of the original regex, which doesn't specify any suffixes at all.
In other words, you are probably faced with a limitation of logic, not a limitation of regex-writing skills or the regex language itself.
Please check if
var reg=/\b((?:[a-z][\w-]+:(?:\/*)|(?:www\d{0,3}[.])|[a-z0-9.\-]+[.][a-z]{2,4}\/{0,1})(?:[^\s()<>]+|\(([^\s()<>]+|(\([^\s()<>]+\)))*\))*(?:\(([^\s()<>]+|(\([^\s()<>]+\)))*\)|[^\s`!()\[\]{};:'".,<>?«»“”‘’]))/gi;
suits your needs. www(anyNumber) has just been put to appear one or zero times. Sorry for the first answer, did not notice the texts.
If I am correct, the following code will only match a URL that is exactly as presented.
However, what would it look like if you wanted to identify subdomains as well as urls that contain various different query strings - in other words, any address that contains this domain:
var url = /test.com/
if (window.location.href.match(url)){
alert("match!");
}
If you want this regex to match "test.com" you need to escape the "." and both of the "/" that means any character in regex syntax.
Escaped : \/test\.com\/
Take a look for here for more info
No, your pattern will actually match on all strings containing test.com.
The regular expresssion /test.com/ says to match for test[ANY CHARACTER]com anywhere in the string
Better to use example.com for example links. So I replaces test with example.
Some example matches could be
http://example.com
http://examplexcom.xyz
http://example!com.xyz
http://example.com?q=123
http://sub.example.com
http://fooexample.com
http://example.com/asdf/123
http://stackoverflow.com/?site=example.com
I think you need to use /g. /g enables "global" matching. When using the replace() method, specify this modifier to replace all matches, rather than only the first one:
var /test.com/g;
If you want to test if an URL is valid this is the one I use. Fairly complex, because it takes care also of numeric domain & a few other peculiarities :
var urlMatcher = /(([\w]+:)?\/\/)?(([\d\w]|%[a-fA-f\d]{2,2})+(:([\d\w]|%[a-fA-f\d]{2,2})+)?#)?([\d\w][-\d\w]{0,253}[\d\w]\.)+[\w]{2,4}(:[\d]+)?(\/([-+_~.\d\w]|%[a-fA-f\d]{2,2})*)*(\?(&?([-+_~.\d\w]|%[a-fA-f\d]{2,2})=?)*)?(#([-+_~.\d\w]|%[a-fA-f\d]{2,2})*)?/;
Takes care of parameters and anchors etc... dont ask me to explain the details pls.