Validate URL in java

Question

Does anyone know how to validate the format of a url in Java using regex?

Thanks

See also this question: stackoverflow.com/questions/27745/getting-parts-of-a-url-regex — Pops
– Pops, Commented Jun 7, 2010 at 20:45
Use regex. Don't use the classes shared below. They may open network connections. — MikeNereson
– MikeNereson, Commented Nov 24, 2011 at 4:56
@MikeNereson that's incorrect. java.net.URI will never accesses the network, it's the best way to validate URIs in Java. The equals() and hadhCode() methods of java.net.URL try to access the network, that's correct, but the java.net.URL constructor most likely won't. — jcsahnwaldt Reinstate Monica
– jcsahnwaldt Reinstate Monica, Commented May 11, 2012 at 21:54

kldavis4 · Accepted Answer · 2012-12-05 19:22:24Z

7

A very sneaky way is to do:

try {
    new java.net.URI(myUrl);
} catch(URISyntaxException e) {
    // url badly formed
}

edited Dec 5, 2012 at 19:22

kldavis4

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answered Jun 7, 2010 at 20:40

Finbarr

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5 Comments

dplass Over a year ago

I'd advise against using the URL class (see my answer below for URI) - it does a network lookup to resolve server names.

Finbarr Over a year ago

I actually meant to use the URI class, thanks for pointing this out.

dplass Over a year ago

Thanks for fixing the example.

Ben Clayton Over a year ago

The java.net.URI constructor throws a URISyntaxException not a MalformedURLException. At least on the version of Java on Android!

Pantelis Sopasakis Over a year ago

Firstly, it should be URISyntaxException and not MalformedURLException. And second, using exceptions for normal flow control is not a good practice. Additionally, if you don't intend to use the object returned by new java.net.URI(myUrl), then better avoid the invocation whatsoever. Checking the validity of a URI calls for a URIValidator class which should - to the best of my knowledge - make use of regex (which I'm still looking for around the net. I'll let you know as soon as I find something).

dplass · Accepted Answer · 2010-06-07 20:46:12Z

7

You might be better off using the URI class in Java: http://java.sun.com/j2se/1.5.0/docs/api/java/net/URI.html

It will throw an exception in the constructor if it doesn't parse correctly.

edited Jun 7, 2010 at 20:46

answered Jun 7, 2010 at 20:40

dplass

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