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I have a file that I'm parsing that ALWAYS includes an email address. The file is currently laid out with a leading space before the @ and we want to capture the domain.

foo @bar.com more data here
foo @foo.com more data here

We want to pull out @bar.com and @foo.com and I'm just starting to work with regex. I'm trying to pull the pattern "@ at the start of a word boundary inclusive of all following characters up until the next word boundary".

I've tried various iterations of the following, grouping things, square backets for the @ literal...but nothing seems to work.

EDIT - actual code :

import java.util.regex.*;
import java.io.*;
import java.nio.file.*;
import java.lang.*;
//
public class eadd
{
    public static void main(String args[])
    {
        String inputLine = "foo foofoo foo foo @bar.com foofoofoo foo foo foo";
        String eDomain = "";
       // parse eadd
        Pattern p2 = Pattern.compile("(\\b@.*\\b)");
        Matcher m2 = p2.matcher(inputLine);
            if(m2.matches()) {
                eDomain = m2.group(1);
                } else {
                eDomain = "n/a";
            }
        System.out.println(p2+" "+m2+" "+eDomain);
    }
}

And the results when I run it.

(\b@.*\b) java.util.regex.Matcher[pattern=(\b@.*\b) region=0,49 lastmatch=] n/a

All of my problems have been related to the what follows the @ being searched as a literal instead of a pattern (e.g., looking for .* rather than any and all characters). I can't find references to @ being a control character, so I don't think I need to escape out.

There are no similar examples in Oracle's java tutorials or documentation, SO, nor any of the online resources I checked out; I've been unable to find other samples of how people have handled this. Like I said, I'm fairly new with regex, but this looks to me like it should be working to me. What am I missing?

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    You must use "\\b" not "\b" -- which is a control character for backspace. The additional `` is for escaping. Commented Aug 8, 2012 at 15:11
  • Veer is correct; you have to escape the escape character because both Java and regex use it to escape things. Commented Aug 8, 2012 at 15:11
  • @veer When I try that, it appears to be searching for a literal (\b@.*\b)...I'm printing p2 to the console to see the pattern & my matcher isn't getting any hits. Commented Aug 8, 2012 at 15:14
  • matches() tries to match the entire input against your regex - you're looking for a partial match and should use find(). Commented Aug 8, 2012 at 15:33
  • @JacobRaihle I must've missed that distinction in the tutorials. Thanks for that. Commented Aug 8, 2012 at 15:40

3 Answers 3

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2

Java won't treat @ as a word character - thus there is no word boundary at the start of your address. You could replace the word boundary with a simple whitespace match:

"\s(@.+?)\b"

(Or "\\s(@.+?)\\b" since this is Java) should do the trick. It looks for whitespace followed by @ and matches until the next word boundary.

Edit: Oops, ., just like @, isn't a word character (duh). Use

"\\s(@.+?)(?:\\s|$)"

to match until the next whitespace or EOF. (?:\\s|$) is a non-capturing group that will match any whitespace or end of input.

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4 Comments

for the first simple solution that fits the problem. if you KNOW that this is the format of emails you will be getting then why complicate?
@Eugene Right, domains are scary beasts but if they'll always be surrounded by whitespace it makes your life a lot easier :)
@Jacob OK. I get it. That makes a lot of sense! One quick clarification: Does regex (or java) require (@+wildcard) to be grouped separately from (wildcard-space at End) or is that just for better readability?
I separated the groups because you're not interested in the whitespace - this way only the domain itself is in a capturing group while the whitespace is ignored once the match has been confirmed. A capturing group is one you can access with Matcher#group(int) - putting ?: at the start of a group makes it non-capturing.
2 
Pattern p = Pattern.compile("(@(?:[a-z][A-Z0-9_]+)\\.(?:[a-z][A-Z]+))");

This should work for you.

This regex starts looking for the @ . After that it looks for any word followed by the ".", followed by another word. For beeing familiar with the syntax you can take a look at this.

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6 Comments

Could you also explain what this regex is doing for the asker?
Don't you mean [a-z][A-Z] to match both upper and lower case? Also, what about the other legal characters permitted in domain names? Digits [0-9] ...
@HeatfanJohn Thanks, changed the pattern.
Personally I'd go with something less restrictive (as email addresses are notoriously loosely defined), more along the lines of \\b@([^\\b]+) (yes, I think you should even avoid requiring a full stop as I fully expect such email addresses to creep up with the new TLDs coming out).
hyphen is also permitted in urls
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1

try with this: Pattern p = Pattern.compile("(?<=\\s)(@(?:bar|foo)\\.com\\b)");
or a general purpose pattern: "(?<=\\s)(@\\w+(?:\\.\\w+)+\\b)"

Explain:
(?<=\\s): look behind for match leading space before @
\\w: match alphabet, digit, underscore
\\b: word boundary
@\\w+(?:\\.\\w+)+: match @bar.com, @bar.com.au, @bar.com.xyz, @bar.foo.xx.yy.zz

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1 Comment

There really is no need to to use a look-behind, just let your desired result be a group and extract it that way.

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