Java regular expression to match ip followed by the date time

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Hi all
I have an input which looks something like 192.168.90.8 - - [321/Jan/2013:00:00:11 -0800]xyz.
I need to extract ip and date. My code is

Java

public class c {
public static void main(String[] args) throws IOException {
	
	input="192.168.90.8 - - [321/Jan/2013:00:00:11 -0800]";
	//System.out.println(input);
	Pattern pattern=Pattern.compile("((?:(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\\.){3}(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?))(?![\\d]) - -([\\d]+/[\\w]+/[\\d]+:[\\d]+:[\\d]+:[\\d]+)");
	Matcher matcher=pattern.matcher(input);
	while(matcher.find()){
		System.out.println("found");
		System.out.println(matcher.group(3));
		
	}
}
}
Iam unable to find issue with my regex

Posted 23-Mar-14 20:52pm

coolRahul_12

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Xiao Ling · Accepted Answer · 2014-03-23T23:05:00

If you want to extract IP and date, why don't you create two patterns?

String input="192.168.90.8 - - [321/Jan/2013:00:00:11 -0800]";
		//System.out.println(input);
		Pattern pattern=Pattern.compile("((?:(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\\.){3}(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?))");
		Matcher matcher=pattern.matcher(input);
		if (matcher.find()){
			System.out.println("found");
			System.out.println(matcher.group());
			
		}
		
		pattern=Pattern.compile("([\\d]+/[\\w]+/[\\d]+:[\\d]+:[\\d]+:[\\d]+)");
		matcher=pattern.matcher(input);
		if (matcher.find()){
			System.out.println("found");
			System.out.println(matcher.group());	
		}

Output:

found
192.168.90.8
found
321/Jan/2013:00:00:11