Hi Lubomir,
First, yes, "\b" (I'm assuming you escaped the backslash for C# syntax)
means "word boundary." However, I believe your understanding of "word
boundary" is the source of your confusion.
Remember that Regular Expressions deal with characters, not language. A
Regular Expression "word" character is an alpha-numeric character (letters
and digits). The shorthand character class "\w" matches all word characters.
All other characters are "non-word" characters. The shorthand character
class "\W" will match all non-word characters. The anchor "\b" indicates
that a match sequence of characters either begins or ends (depending upon
the position of the anchor in the regular expression) at a "word boundary".
This is defined as either a non-word character, or the beginning or end of
the string being evaluated. Therefore, a period being a "non-word"
character, defines the beginning or end of a match.
Now, I'm trying to understand what you meant by what you said, because the
character sequence "one.two" is *not* a match. Only the character sequence
(or substring, if you will) "one" in that character sequence is a match.
From the period on is not part of the match. What confuses me is your
assertion that the regular expression "picks up...the words 'one.two'. I
*think* you are saying that the character sequence "one" in the sequence
"one.two" is matched, and that because these look like 2 words separated by
a period to you. But a string is simply a sequence of characters, with no
significance, other than their individual values and/or character classes. I
hope you understand what I'm trying to explain.
So, going out on a limb just a bit, I'm guessing that you only want
character sequences such as "xyz" (to make the example look less like a
human word) to match if they are at the beginning of a string, at the end of
the string, and, if other characters are in the string, are separated from
these other characters by a white-space character (space, newline, tab,
etc).
If so, your regular expression string would be:
(?m)(?<=^|\s+)xyz(?=$|\s)
Let me explain this:
The "(?m)" modifier indicates that the '^' (Beginnning of string) and '$'
(End of string) also match at the beginning and end of lines within a
string. I included that because I do not know whether your text will have
multiple lines in it.
I replaced your anchors with a positive look-ahead expression and a positive
look-behind expression. The reason for the look-arounds is that an anchor
defines a position, not a match. In other words, an anchor is a
non-capturing expression. Look-arounds are also non-capturing expressions,
which means that anything which matches the expression in the look-around is
not part of the match, but defines the beginning or end of a match.
(?<=^|\s+) is a positive look-behind. It defines the start of a match as
being one of 2 possible alternatives - the beginning of a line, or a
white-space character.
(?=$|\s+) is a positive look-ahead. It defines the end of a match as being
one of 2 possible alternatives - the end of a line, or a white-space
character.
Therefore, for example, consider the following multi-line string:
xyz
xyz.two
xyz two three.xyz
three two xyz zero
zero xyz
In the above string, there are 4 matches using the above regular expression.
In line 1, the entire sequence "xyz" is a match. There is no match in line
2, because the sequence "xyz" is at the beginning of a line, but is not
followed by a white-space character (Both look-arounds must be met). In line
3, the first "xyz" is captured, because it is at the beginning of the line,
and followed by a white-space character. The second "xyz" is *not* a match,
because it is preceded by a non-white-space character. In line 4, "xyz" is a
match because it has a white-space character both before and after it. And
in line 5, "xyz" is a match because it is preceded by a white-space
character, and is at the end of the line.
--
HTH,
Kevin Spencer
Microsoft MVP
Software Composer
http://unclechutney.blogspot.com
A watched clock never boils.
