How can I implement this condition with regex?

Question

With regex in java, how do one put condition, i.e. if-then-else? Clearly say, I want to put condition if "http" exist, "www" CAN exist or if "http" doesnot exist "www" HAVE TO exist in the string. How can I implement that condition with regex?

What I done

      ((http){0,1}|(www){0,1})

But this is not work

example input :

if string starts with "http" after it, "www" can come.

ex: "httpwww < sometext >" --->should be matched

ex: "http < sometext >" --->should be matched
if string didnot start with "http", string have to start with "www"

ex: "www < sometext >" --->should be matched

ex: "< sometext >" ----> should not be matched by regex because it did not started with nor http and www

So you want to check if a string starts with either http or www, correct? — Joni
– Joni, Commented Aug 13, 2014 at 10:49
@Antony "if string didnot start with "http", string have to start with "www"". In www, it did not start with http, but did start with www (itself)! — Unihedron
– Unihedron, Commented Aug 13, 2014 at 11:02
Now that you've edited to clarify, I rolled-back my solution. My old answer did work what you wanted, after all. — Unihedron
– Unihedron, Commented Aug 13, 2014 at 11:07

Mena · Accepted Answer · 2014-08-13 10:58:32Z

1

A simplistic solution for that would be to use "look around"s, as specified in comments.

You can also use an actual URL object to do that if, as I imagine, you're working with actual URLs.

For instance:

String[] input = { "http://www.google.com", "http://foo.com", "www.foo.com" };
Pattern p = Pattern.compile("(?<=http).*(?=www)");
for (String s : input) {
    Matcher m = p.matcher(s);
    System.out.printf("Found in %s? %b%n", s, m.find());
    try {
        URL u = new URL(s);

        System.out.printf("Authority starts with www and protocol is http for %s? %b%n", s,
                u.getAuthority().startsWith("www") && u.getProtocol().equals("http"));
    }
    catch (MalformedURLException mue) {
        System.out.printf("%s is not interpreted as well-formed URL.%n", s);
    }
}

Output

Found in http://www.google.com? true
Authority starts with www and protocol is http for http://www.google.com? true
Found in http://foo.com? false
Authority starts with www and protocol is http for http://foo.com? false
Found in www.foo.com? false
www.foo.com is not interpreted as well-formed URL.

edited Aug 13, 2014 at 10:58

answered Aug 13, 2014 at 10:51

Mena

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2 Comments

Antony Over a year ago

no I have said before, if it starts with "http" after it "www" can come. But second output is false, it should be true. Wait I will add some input

Mena Over a year ago

@Antony take a look at my answer. It should cover your examples.

TheLostMind · Accepted Answer · 2014-08-13 11:05:24Z

1

Effectively, your String should start with either http or www. You don't care if http is followed by www. All you care about is your string should always start with *http8 or www and nothing else. So,

public static void main(String[] args) {
    String s1 = "http://www.google.com";
    String s2 = "www.google.com";
    String s3 = "sdfwww.google.com";
    System.out.println(s1.matches("^(http|www).*"));
    System.out.println(s2.matches("^(http|www).*"));
    System.out.println(s3.matches("^(http|www).*"));

}

O/P ::

true
true
false

answered Aug 13, 2014 at 11:05

TheLostMind

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Comments

Unihedron · Accepted Answer · 2014-08-13 11:20:52Z

1

Use a simple regex match:

boolean matches = myString.matches("^(?:http|www).*");
// The .* is for Java implementation

Here is a regex demo!

edited Aug 13, 2014 at 11:20

answered Aug 13, 2014 at 10:47

Unihedron

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4 Comments

Antony Over a year ago

your regex means one of the "http" or "www" can exist or both of them. However, there is no have to relation. IF I downvote, I will use it for you

TheLostMind Over a year ago

@Antony - Simply put, your question boils down to - does my regex start with either http or www. right?.

TheLostMind Over a year ago

Why the ?: non-capturing group?

Unihedron Over a year ago

@TheLostMind The Java implementation incurs processing time writing each backreference location and match for each capturing group match. Hence for Java Regexes, I use capturing groups sparingly. YMMV due to JIT compiler optimizations.

Rengasamy · Accepted Answer · 2014-08-13 12:49:31Z

0

Try this Demo,

http://regex101.com/r/dW7jJ0/3

Regex is

^(httpwww|http|www)

answered Aug 13, 2014 at 12:49

Rengasamy

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