Question

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Validating an IP with regex

I need the regular expression for finding strings like followings in Java:

http://69.39.233.135:8032
http://83.133.127.104
http://119.207.75.238:11120

Answer 1

This pattern would fetch the desired results:

http://\d{1,3}(?:\.\d{1,3}){3}(?::\d{1,5})?

Matches an IP after http:// with an optional port.

Note that this is a simplified validation of an IPv4 and will only match that they are one the correct format and not a valid one. And remember to add an extra backslash to escape each backslash in java.

Here’s an example in java:

String text = "http://69.39.233.135:8032 http://83.133.127.104 http://119.207.75.238:11120";        
String pattern = "(http://\\d{1,3}(?:\\.\\d{1,3}){3}(?::\\d{1,5})?)";

Pattern compiledPattern = Pattern.compile(pattern);
Matcher matcher = compiledPattern.matcher(text);
while(matcher.find()) {
    System.out.println(matcher.group());
}

Output:

http://69.39.233.135:8032
http://83.133.127.104
http://119.207.75.238:11120