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Eric Wilson
Eric Wilson

Reputation: 59335

How to check for a valid URL in Java?

What is the best way to check if a URL is valid in Java?

If tried to call new URL(urlString) and catch a MalformedURLException, but it seems to be happy with anything that begins with http://.

I'm not concerned about establishing a connection, just validity. Is there a method for this? An annotation in Hibernate Validator? Should I use a regex?

Edit: Some examples of accepted URLs are http://*** and http://my favorite site!.

Upvotes: 117

Views: 159883

Answers (10)

blackorbs
blackorbs

Reputation: 625

You can use:

private boolean isValidUrl(String urlString){
        try {
            URL url = new URL(urlString);
            url.toURI();
            return Patterns.WEB_URL.matcher(urlString).matches();
        } catch (MalformedURLException | URISyntaxException e) {
            return false;
        }
    }

Upvotes: 0

Julian
Julian

Reputation: 423

There is alsow a Function in org.apache.xerces.util.URI

isWellFormedAddress(java.lang.String address)

Determine whether a string is syntactically capable of representing a valid IPv4 address, IPv6 reference or the domain name of a network host.

Upvotes: 0

uckelman
uckelman

Reputation: 26194

Judging by the source code for URI, the

public URL(URL context, String spec, URLStreamHandler handler)

constructor does more validation than the other constructors. You might try that one, but YMMV.

Upvotes: 3

user2670200
user2670200

Reputation:

The most "foolproof" way is to check for the availability of URL:

public boolean isURL(String url) {
  try {
     (new java.net.URL(url)).openStream().close();
     return true;
  } catch (Exception ex) { }
  return false;
}

Upvotes: 7

isapir
isapir

Reputation: 23473

I didn't like any of the implementations (because they use a Regex which is an expensive operation, or a library which is an overkill if you only need one method), so I ended up using the java.net.URI class with some extra checks, and limiting the protocols to: http, https, file, ftp, mailto, news, urn.

And yes, catching exceptions can be an expensive operation, but probably not as bad as Regular Expressions:

final static Set<String> protocols, protocolsWithHost;

static {
  protocolsWithHost = new HashSet<String>( 
      Arrays.asList( new String[]{ "file", "ftp", "http", "https" } ) 
  );
  protocols = new HashSet<String>( 
      Arrays.asList( new String[]{ "mailto", "news", "urn" } ) 
  );
  protocols.addAll(protocolsWithHost);
}

public static boolean isURI(String str) {
  int colon = str.indexOf(':');
  if (colon < 3)                      return false;

  String proto = str.substring(0, colon).toLowerCase();
  if (!protocols.contains(proto))     return false;

  try {
    URI uri = new URI(str);
    if (protocolsWithHost.contains(proto)) {
      if (uri.getHost() == null)      return false;

      String path = uri.getPath();
      if (path != null) {
        for (int i=path.length()-1; i >= 0; i--) {
          if ("?<>:*|\"".indexOf( path.charAt(i) ) > -1)
            return false;
        }
      }
    }

    return true;
  } catch ( Exception ex ) {}

  return false;
}

Upvotes: 4

Andrei Volgin
Andrei Volgin

Reputation: 41089

My favorite approach, without external libraries:

try {
    URI uri = new URI(name);

    // perform checks for scheme, authority, host, etc., based on your requirements

    if ("mailto".equals(uri.getScheme()) {/*Code*/}
    if (uri.getHost() == null) {/*Code*/}

} catch (URISyntaxException e) {
}

Upvotes: 5

user123444555621
user123444555621

Reputation: 152956

I'd love to post this as a comment to Tendayi Mawushe's answer, but I'm afraid there is not enough space ;)

This is the relevant part from the Apache Commons UrlValidator source:

/**
 * This expression derived/taken from the BNF for URI (RFC2396).
 */
private static final String URL_PATTERN =
        "/^(([^:/?#]+):)?(//([^/?#]*))?([^?#]*)(\\?([^#]*))?(#(.*))?/";
//         12            3  4          5       6   7        8 9

/**
 * Schema/Protocol (ie. http:, ftp:, file:, etc).
 */
private static final int PARSE_URL_SCHEME = 2;

/**
 * Includes hostname/ip and port number.
 */
private static final int PARSE_URL_AUTHORITY = 4;

private static final int PARSE_URL_PATH = 5;

private static final int PARSE_URL_QUERY = 7;

private static final int PARSE_URL_FRAGMENT = 9;

You can easily build your own validator from there.

Upvotes: 8

Prasanna Pilla
Prasanna Pilla

Reputation: 715

Here is way I tried and found useful,

URL u = new URL(name); // this would check for the protocol
u.toURI(); // does the extra checking required for validation of URI

Upvotes: 69

Tendayi Mawushe
Tendayi Mawushe

Reputation: 26108

Consider using the Apache Commons UrlValidator class

UrlValidator urlValidator = new UrlValidator();
urlValidator.isValid("http://my favorite site!");

There are several properties that you can set to control how this class behaves, by default http, https, and ftp are accepted.

Upvotes: 118

Adam Matan
Adam Matan

Reputation: 136141

validator package:

There seems to be a nice package by Yonatan Matalon called UrlUtil. Quoting its API:

isValidWebPageAddress(java.lang.String address, boolean validateSyntax, 
                      boolean validateExistance) 
Checks if the given address is a valid web page address.

Sun's approach - check the network address

Sun's Java site offers connect attempt as a solution for validating URLs.

Other regex code snippets:

There are regex validation attempts at Oracle's site and weberdev.com.

Upvotes: 2

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