Regular expression

[Philip Freeman]

 String longString = " Derek Banas CA AK 12345 PA [phone] [e-mail] 412-555-1234 412 555-1234 ";

regexChecker("\\S[^A-Za-z]", longString);

Output: ksAK12345A(412)555-1212h@l.m412-555-1234412555-1234

Question: Why does output contain characters ksAKhlm ?

regexChecker method:

public static void regexChecker(String theRegex, String str2Check)
	{
		Pattern checkRegex = Pattern.compile(theRegex);
		Matcher regexMatcher = checkRegex.matcher(str2Check);
		
		while(regexMatcher.find())
		{
			if(regexMatcher.group().length() != 0)
			{
				System.out.print(regexMatcher.group().trim());
			}
			
		}
	}

[Paweł Baczyński]

\S means non-whitespace character.
k, s, A, K, h, l, m are all non-whitespace characters so they are matched.

[Philip Freeman]

[^A-Za-z] gets me 12345(412)555-1212@.412-555-1234412555-1234, since i said that i dont want any letters.
\S gets me everything just without whitespaces.

What I dont understand is why when I use these two combined \S[^A-Za-z] do I get back some letters? k, s, A, K, h, l, m.
Isnt johnsmith also a non-space characters so why are they not in the output?

[Henry Wong]

Philip Freeman wrote:[^A-Za-z] gets me 12345(412)555-1212@.412-555-1234412555-1234, since i said that i dont want any letters.

This also gets you the spaces. Remember that a space is not a letter... and since you are matching the spaces, it is the "\S" that is matching the letters before the spaces.

Henry

[Paweł Baczyński]

You need to understand that \S[^A-Za-z] doesn't mean anything that is not a whitespace nor a letter (that would be [^\sA-Za-z]).
It means one non-whitespace character followed by one non-letter character.

So it matches:
k_  s_  A_  K_  12  34  5_  A_  (4  12  )5  55  -1  21  2_  h@  l.  m_  41  2-  55  5-  12  34  41  2_  55  5-  12  34

I have replaced spaces with underscores.

[Philip Freeman]

Thanks.