Website/URL Validation Regex in JAVA

Question

I need a regex string to match URL starting with "http://", "https://", "www.", "google.com"

the code i tried using is:

//Pattern to check if this is a valid URL address
    Pattern p = Pattern.compile("(http://|https://)(www.)?([a-zA-Z0-9]+).[a-zA-Z0-9]*.[a-z]{3}.?([a-z]+)?");
    Matcher m;
    m=p.matcher(urlAddress);

but this code only can match url such as "http://www.google.com"

I know this ma be a dupicate question but i have tried all of the regex provided and it does not suit my requirement. Willl someone please help me? Thank you.

Is your requirement that the url must start with either one of these "http://" "https://" "www." "google.com" ? — Scary Wombat
– Scary Wombat, Commented Jul 24, 2014 at 2:19
No. not only google website. the google.com is jus an exampl for websites without "www." or any "http or https" in their url. — Hao Ting
– Hao Ting, Commented Jul 24, 2014 at 2:37

Avinash Raj · Accepted Answer · 2014-07-24 03:45:48Z

19

You need to make (http://|https://) part in your regex as optional one.

^(http:\/\/|https:\/\/)?(www.)?([a-zA-Z0-9]+).[a-zA-Z0-9]*.[a-z]{3}.?([a-z]+)?$

DEMO

answered Jul 24, 2014 at 3:45

Avinash Raj

175k32 gold badges247 silver badges289 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

Avinash Raj Over a year ago

even more simpler ^(https?:\/\/)?(www.)?([a-zA-Z0-9]+).[a-zA-Z0-9]*.[a-z]{3}.?([a-z]+)?$

Ananda Over a year ago

correct is ^(http:\/\/|https:\/\/)?(www.)?([a-zA-Z0-9]+).[a-zA-Z0-9]*.[a-z]{3}\.([a-z]+)?$

Akash Tomar Over a year ago

This regex does not accept slash eg. https://www.google.com/123. It also does not accept multiple key value pairs, Eg: https://www.google.com?key1=value1&&key2=value2.

Raj Hassani · Accepted Answer · 2016-06-16 16:20:05Z

11

You can use the Apache commons library(org.apache.commons.validator.UrlValidator) for validating a url:

String[] schemes = {"http","https"}.
UrlValidator urlValidator = new UrlValidator(schemes);

And use :-

 urlValidator.isValid(your url)

Then there is no need of regex.

Link:- https://commons.apache.org/proper/commons-validator/apidocs/org/apache/commons/validator/routines/UrlValidator.html

answered Jun 16, 2016 at 16:20

Raj Hassani

1,6771 gold badge21 silver badges26 bronze badges

2 Comments

Chargnn Over a year ago

You might need a regex to avoid an exception if someone tries to enter "http:\\" or "http:/"

Udit Kumawat Over a year ago

this validator doesn't allow underscore in host names

KnechtRootrecht · Accepted Answer · 2019-07-15 10:25:59Z

If you use Java, I recommend use this RegEx (I wrote it by myself):

^(https?:\/\/)?(www\.)?([\w]+\.)+[‌\w]{2,63}\/?$
"^(https?:\\/\\/)?(www\.)?([\\w]+\\.)+[‌\\w]{2,63}\\/?$" // as Java-String

to explain:

^ = line start
(https?://)? = "http://" or "https://" may occur.
(www.)? = "www." may orrur.
([\w]+.)+ = a word ([a-zA-Z0-9]) has to occur one or more times. (extend here if you need special characters like ü, ä, ö or others in your URL - remember to use IDN.toASCII(url) if you use special characters. If you need to know which characters are legal in general: https://kb.ucla.edu/articles/what-characters-can-go-into-a-valid-http-url
[‌\w]{2,63} = a word ([a-zA-Z0-9]) with 2 to 63 characters has to occur exactly one time. (a TLD (top level domain (for example .com) can not be shorter than 2 or longer than 63 characters)
/? = a "/"-character may occur. (some people or servers put a / at the end... whatever)
$ = line end

-

If you extend it by special characters it could look like this:

^(https?:\/\/)?(www\.)?([\w\Q$-_+!*'(),%\E]+\.)+[‌\w]{2,63}\/?$
"^(https?:\\/\\/)?(www\.)?([\\w\\Q$-_+!*'(),%\\E]+\\.)+[‌\\w]{2,63}\\/?$" // as Java-String

The answer of Avinash Raj is not fully correct.

^(http:\/\/|https:\/\/)?(www.)?([a-zA-Z0-9]+).[a-zA-Z0-9]*.[a-z]{3}.?([a-z]+)?$

The dots are not escaped what means it matches with any character. Also my version is simpler and I never heard of a domain like "test..com" (which actually matches...)

Demo: https://regex101.com/r/vM7wT6/279

Edit: As I saw some people needing a regex which also matches servers directories I wrote this:

^(https?:\/\/)?([\w\Q$-_+!*'(),%\E]+\.)+(\w{2,63})(:\d{1,4})?([\w\Q/$-_+!*'(),%\E]+\.?[\w])*\/?$

while this may not be the best one, since I didn't spend too much time with it, maybe it helps someone. You can see how it works here: https://regex101.com/r/vM7wT6/700 It also matches urls like "hello.to/test/whatever.cgi"

raghavsood33 · Accepted Answer · 2015-11-01 12:19:18Z

3

Java compatible version of @Avinash's answer would be

//Pattern to check if this is a valid URL address
Pattern p = Pattern.compile("^(http://|https://)?(www.)?([a-zA-Z0-9]+).[a-zA-Z0-9]*.[a-z]{3}.?([a-z]+)?$");
Matcher m;
m=p.matcher(urlAddress);
boolean matches = m.matches();

answered Nov 1, 2015 at 12:19

raghavsood33

7787 silver badges18 bronze badges

Comments

Sagar V · Accepted Answer · 2017-07-28 11:37:52Z

1

pattern="w{3}\.[a-z]+\.?[a-z]{2,3}(|\.[a-z]{2,3})"

this will only accept addresses like e.g www.google.com & www.google.co.in

edited Jul 28, 2017 at 11:37

Sagar V

12.5k8 gold badges46 silver badges71 bronze badges

answered Jul 28, 2017 at 11:37

darkwinter

191 bronze badge

Comments

Adil B · Accepted Answer · 2018-12-09 03:28:09Z

-1

//I use that

static boolean esURL(String cadena){

    boolean bandera = false;

    bandera = cadena.matches("\\b(https://?|ftp://|file://|www.)[-a-zA-Z0-9+&@#/%?=~_|!:,.;]*[-a-zA-Z0-9+&@#/%=~_|]");

    return bandera;
}

edited Dec 9, 2018 at 3:28

Adil B

17.2k12 gold badges72 silver badges92 bronze badges

answered Dec 8, 2018 at 21:15

A ch b

13 bronze badges

Collectives™ on Stack Overflow

Website/URL Validation Regex in JAVA

6 Answers 6

3 Comments

2 Comments

Comments

Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

6 Answers 6

3 Comments

2 Comments

Comments

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related