Finding and replacing an url using regex in java

Question

I am trying to replace a url with regex using String.replace and the code is below

public class Test {
    public static void main(String[] args) {
        String test = "https://google.com";
        //String regex = "\\b(https?|ftp|file)://[-a-zA-Z0-9+&@#/%?=~_|!:,.;]*[-a-zA-Z0-9+&@#/%=~_|]";
        String regex = "(http?|ftp|file)://[-a-zA-Z0-9+&@#/%?=~_|!:,.;]*[-a-zA-Z0-9+&@#/%=~_|]"; // does not match <http://google.com>

        String newText = test.replace(regex, "");
        System.out.println(newText);
    }
}

I have looked into several questions on it in SO but it does not replace the pattern. Can someone please tell me how do i achieve that?

Your regex match google.com , make sure you change http? to https? — Pedro Lobito
– Pedro Lobito, Commented May 9, 2015 at 17:35
I don't see any replacement. You just want to remove google.com ? — Pedro Lobito
– Pedro Lobito, Commented May 9, 2015 at 17:38
@Pedro: am extremely sorry. I was caught up with work so that's why. Thanks a lot for your help. Won't repeat this next time anyways — Aarish Ramesh
– Aarish Ramesh, Commented May 9, 2015 at 19:15

M A · Accepted Answer · 2015-05-09 17:37:01Z

2

String.replace() does not accept a regular expression. Use String.replaceAll instead:

String newText = test.replaceAll(regex, "");

As far as the regex is concerned, you should match the https as well:

String regex = "(https?|ftp|file)://[-a-zA-Z0-9+&@#/%?=~_|!:,.;]*[-a-zA-Z0-9+&@#/%=~_|]";

edited May 9, 2015 at 17:37

answered May 9, 2015 at 17:35

M A

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Comments

Community · Accepted Answer · 2020-06-20 09:12:55Z

2

You cannot use a regex with replace, use replaceAll instead, i.e.:

   String test = "something https://google.com something";
    try {
        String newText = test.replaceAll("(https?|ftp|file)://[-a-zA-Z0-9+&@#/%?=~_|!:,.;]*[-a-zA-Z0-9+&@#/%=~_|]", "");
        System.out.println(newText);
    } catch (PatternSyntaxException ex) {
        // Syntax error in the regular expression
    } catch (IllegalArgumentException ex) {
        // Syntax error in the replacement text (unescaped $ signs?)
    } catch (IndexOutOfBoundsException ex) {
        // Non-existent backreference used the replacement text
    }

Output:

something  something

Live Demo:

http://ideone.com/Yi2hrb

Regex Explanation:

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

Options: Case sensitive; Exact spacing; Dot doesn’t match line breaks; ^$ don’t match at line breaks; Default line breaks; Regex syntax only

Match the regex below and capture its match into backreference number 1 «(https?|ftp|file)»
   Match this alternative «https?»
      Match the character string “http” literally «http»
      Match the character “s” literally «s?»
         Between zero and one times, as many times as possible, giving back as needed (greedy) «?»
   Or match this alternative «ftp»
      Match the character string “ftp” literally «ftp»
   Or match this alternative «file»
      Match the character string “file” literally «file»
Match the character string “://” literally «://»
Match a single character present in the list below «[-a-zA-Z0-9+&@#/%?=~_|!:,.;]*»
   Between zero and unlimited times, as many times as possible, giving back as needed (greedy) «*»
   The literal character “-” «-»
   A character in the range between “a” and “z” «a-z»
   A character in the range between “A” and “Z” «A-Z»
   A character in the range between “0” and “9” «0-9»
   A single character from the list “+&@#/%?=~_|!:,.;” «+&@#/%?=~_|!:,.;»
Match a single character present in the list below «[-a-zA-Z0-9+&@#/%=~_|]»
   The literal character “-” «-»
   A character in the range between “a” and “z” «a-z»
   A character in the range between “A” and “Z” «A-Z»
   A character in the range between “0” and “9” «0-9»
   A single character from the list “+&@#/%=~_|” «+&@#/%=~_|»

edited Jun 20, 2020 at 9:12

CommunityBot

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answered May 9, 2015 at 17:39

Pedro Lobito

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3 Comments

Wiktor Stribiżew Over a year ago

A tip: do not over-format and over-fill your posts with technical "junk". All the "try...catch" blocks in your posts make people believe you yourself doubt you suggested a working regex.

Pedro Lobito Over a year ago

@stribizhev Thank you for the tip, but in java this is the correct way of handling errors.

Wiktor Stribiżew Over a year ago

However, your answer was not accepted... Not the first time. Thus, it is just my thought I wanted to share with you.

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