Can regular expression be utilized to match any string except a specific string constant let us say "ABC" ? Is this possible to exclude just one specific string constant? Thanks your help in advance.
A:
This isn't easy, unless your regexp engine has special support for it. The easiest way would be to use a negative-match option, for example:
$var !~ /^foo$/
or die "too much foo";
If not, you have to do something evil:
$var =~ /^(($)|([^f].*)|(f[^o].*)|(fo[^o].*)|(foo.+))$/
or die "too much foo";
That one basically says "if it starts with non-f, the rest can be anything; if it starts with f, non-o, the rest can be anything; otherwise, if it starts fo, the next character had better not be another o".
derobert
2009-09-08 17:26:41
That won’t allow the empty string, `f`, `fo` and `foo`.
Gumbo
2009-09-08 21:14:09
@Gumbo: It allows the empty string just fine; notice that ($) is the first alternative, so ^$ (empty string) is accepted. I tested it, and at least in perl 5.0.10 the empty string is accepted.
derobert
2009-09-08 21:50:45
... sorry, perl 5.10.0, of course!
derobert
2009-09-08 21:51:17
+5
A:
You have to use a negative lookahead assertion.
(?!^ABC$)
You could for example use the following.
(?!^ABC$)(^.*$)
Daniel Brückner
2009-09-08 17:28:40
This will work if you're looking for a string that does not include ABC. But is that the goal? Or is the goal to match every character except ABC?
Steve Wortham
2009-09-08 18:37:49
Thanks for pointing that out, you are right - my suggestion only avoids strings starting with ABC - I forgot to anchor the assertion. Going to correct that.
Daniel Brückner
2009-09-08 18:56:57
That's still different than what I was thinking. Perhaps the questioner will clarify what they're looking for.
Steve Wortham
2009-09-08 19:04:36
I find it quite clear - "any string except a specific string [constant]" hence any string (including strings containing ABC) except ABC itself.
Daniel Brückner
2009-09-08 19:09:35
Yeah, you may be right. If so then you're answer is perfect. You can see my answer to see how I interpreted it.
Steve Wortham
2009-09-08 19:48:58
not working in Javascript. text = 'hi ABC wow';regex = /(?!^ABC$)/console.log(text.match(regex));
Nadal
2010-03-11 17:13:09
You only used the assertion but forgot the matching expression - use (?!^ABC$)(^.*$) and it works.
Daniel Brückner
2010-03-13 10:33:13
I was helping a friend recently to do something very similar. However, he didn't want to match the string if it contained a string anywhere inside of it. So I wrote a slightly modified version of your expression *(?!.*ABC)^.*$* and this works like a charm.
Steve Wortham
2010-04-22 15:41:55
+1
A:
You could use negative lookahead, or something like this:
^([^A]|A([^B]|B([^C]|$)|$)|$).*$
Maybe it could be simplified a bit.
Adam Crume
2009-09-08 17:31:16
A:
In .NET you can use grouping to your advantage like this:
http://regexhero.net/tester/?id=65b32601-2326-4ece-912b-6dcefd883f31
You'll notice that:
(ABC)|(.)
Will grab everything except ABC in the 2nd group. Parenthesis surround each group. So (ABC) is group 1 and (.) is group 2.
So you just grab the 2nd group like this in a replace:
$2
Or in .NET look at the Groups collection inside the Regex class for a little more control.
You should be able to do something similar in most other regex implementations as well.
UPDATE: I found a much faster way to do this here: http://regexhero.net/tester/?id=997ce4a2-878c-41f2-9d28-34e0c5080e03
It still uses grouping (I can't find a way that doesn't use grouping). But this method is over 10X faster than the first.
Steve Wortham
2009-09-08 18:33:27
A:
Try this regular expression:
^(.{0,2}|([^A]..|A[^B].|AB[^C])|.{4,})$
It describes three cases:
- less than three arbitrary character
- exactly three characters, while either
- the first is not
A, or - the first is
Abut the second is notB, or - the first is
A, the secondBbut the third is notC
- the first is not
- more than three arbitrary characters
Gumbo
2009-09-08 21:07:55