Captured group in optional part of a regular expression - c#

I want to capture a group in an optional part of a string.
For example:
In the string "firstName:Bill-lastName:Gates", I want to capture 2 groups :
Bill
Gates
I use this regex:
firstName:(.*)-lastName:(.*)
But when the lastName-part is optional, I still want to capture the first
group (firstName).
I used this regex, to make the lastName-part optional (in a non-capturing group):
firstName:(.*)(?:-lastName:(.*))?
Using this updated regex, the resulting groups are:
when the lastName part is not present, for example "firstName:Bill" the captured groups are:
Bill
/empty string/
which is correct,
when the firstName and lastName parts are present: "firstName:Bill-lastName:Gates", the groups are not correct:
Bill-lastName:Gates
/empty/
I think it has to do with greediness of the first capturing group, but how to adjust this regex to make the regex work when the lastName-part is optional?

You are right, it is about greediness. Find a delimiter for the first match group. So, if your firstname "never" contains the dash, only match everything but the dash with the first match group.
firstName:([^-]*)(?:-lastName:(.*))?
firstName:([^-]*)(?:-lastName:(.*))?
Debuggex Demo
If you cannot find such a delimiter you would need to take a different approach. Even if you try to make the first pattern "lazy", the Regex engine always prefers a bigger match over matching an additional optional match.
This is, because lazy matchgroups will match the first string that satisfies the expression (! important wording !)
There might be an option with look arrounds, but you could also use a or -statement without providing optional matches:
firstName:(.*)-lastName:(.*)|firstName:(.*)
This way, the regex engine would match either or, but prefer the pattern with 2 matches since it is listed first. Only if that does not apply, it will try the single match.

Even though you accepted #dognose's answer already, I assure you there are first names with a dash in them (You don't wanna piss off Jean-Claude van Damme). I would advise you to do it like so:
firstName:((?:(?!-lastName:).)*)(?:-lastName:(.*))?
Debuggex Demo
You can see from the visualization that the (?:(?!-lastName:).) says "if the current position is not followed by '-lastName:', capture another character"

Related

C# replace by regular expression

I have below C# code to remove stop words from a string:
public static string RemoveStopWords(string Parameter)
{
Parameter = Regex.Replace(Parameter, #"(?<=(\A|\s|\.|,|!|\?))($|_|0|1|2|3|4|5|6|7|8|9|A|about|after|all|also|an|and|another|any|are|as|at|B|be|because|been|before|being|between|both|but|by|C|came|can|come|could|D|did|do|does|E|each|else|F|for|from|G|get|got|H|had|has|have|he|her|here|him|himself|his|how|I|if|in|into|is|it|its|J|just|K|L|like|M|make|many|me|might|more|most|much|must|my|N|never|no|not|now|O|of|on|only|or|other|our|out|over|P|Q|R|re|S|said|same|see|should|since|so|some|still|such|T|take|than|that|the|their|them|then|there|these|they|this|those|through|to|too|U|under|up|use|V|very|W|want|was|way|we|well|were|what|when|where|which|while|who|will|with|would|X|Y|you|your|Z)(?=(\s|\z|,|!|\?))([^.])", " ", RegexOptions.IgnoreCase);
return Parameter.Trim();
}
But when I run it, it works when the stop word in not at end of the string, for example:
about this book output is book
manager only output is manager only
only manager output is manager
Can anyone please guide?
The capture group at the end of the pattern ([^.]) requires a single char other than a dot. The looakhead preceding that (?=(\s|\z|,|!|\?)) limits that match to only one of the listed alternatives (it can not match a dot already as it is excluded by the lookahead).
If you want to keep that, you could omit that lookahead, and just match what you would allow to match like ([\s,!?]|\z) but it would still require at least 1 of the listed alternatives.
What you could so is only use the positive lookahead, and update it to (?=[\s,!?]|\z)
(?<=\A|[\s.,!?])(?:$|[A-Z0-9_]|about|after|all|also|and?|another|any|are|a[ts]|be|because|been|before|being|between|both|but|by|came|can|come|could|did|do|does|each|else|for|from|get|got|ha[ds]|have|her?|here|him|himself|his|how|i[nf]|into|i[st]|its|just|like|make|many|me|might|more|most|much|must|my|never|not?|now|o[fnr]|only|other|our|out|over|re|said|same|see|should|since|so|some|still|such|take|tha[tn]|the[nm]?|their|there|these|they|this|those|through|too?|under|up|use|very|want|wa[ys]|we|well|were|what|when|where|which|while|who|will|with|would|your?)(?=[\s,!?]|\z)
.NET regex demo
A few notes about the pattern
To shorten the alternation, you can for example a character class a[ts] to either match at or as or make a character optional and? to match either an or and
Inside the lookarounds, you don't have to add another grouping mechanism, so you can use (?=[\s,!?]|\z) instead of (?=(?:[\s,!?]|\z))
If you don't need the values of the capture groups () you can make them non capturing (?:)
The numbers 1|2|3 and the characters A|B|C can be shortened to [A-Z0-9] and also matching the underscore, you might even shorten it to \w

Regex match what did not match

In words, I would describe the problem as "match A or B, match foo, and then match A or B that did NOT match previously".
I can do it with the following regex:
AfooB|BfooA
I'm wondering if there is a more efficient way to do this? I know how to reference a captured group using the "\" and then the group number. In this case I would like to apply something like to say "not the option that matched in the captured group" (and still be restricted to only match the other possible matches for that group).
The reason I'm looking for something more efficient than simply "AfooB|BfooA" is that in my case"foo"is a very long pattern and I would prefer to reduce duplication if possible.
You may use a negative lookahead with a backreference restriction when matching the second A or B:
(A|B)foo(?!\1)(A|B)
Basically, (A|B) matches and captures the value into Group 1, then foo matches foo, (?!\1) makes sure that the text that follows is not the same as the one captured into the first group, and then it can only match the opposite value with (A|B).
See this regex demo
NOTE: if A and B are single characters, use a character class: ([AB])foo(?!\1)([AB])

C# Regular Expression: Search the first 3 letters of each name

Does anyone know how to say I can get a regex (C#) search of the first 3 letters of a full name?
Without the use of (.*)
I used (.**)but it scrolls the text far beyond the requested name, or
if it finds the first condition and after 100 words find the second condition he return a text that is not the look, so I have to limit in number of words.
Example: \s*(?:\s+\S+){0,2}\s*
I would like to ignore names with less than 3 characters if they exist in name.
Search any name that contains the first 3 characters that start with:
'Mar Jac Rey' (regex that performs search)
Should match:
Marck Jacobs L. S. Reynolds
Marcus Jacobine Reys
Maroon Jacqueline by Reyils
Can anyone help me?
The zero or more quantifier (*) is 'greedy' by default—that is, it will consume as many characters as possible in order to finding the remainder of the pattern. This is why Mar.*Jac will match the first Mar in the input and the last Jac and everything in between.
One potential solution is just to make your pattern 'non-greedy' (*?). This will make it consume as few characters as possible in order to match the remainder of the pattern.
Mar.*?Jac.*?Rey
However, this is not a great solution because it would still match the various name parts regardless of what other text appears in between—e.g. Marcus Jacobine Should Not Match Reys would be a valid match.
To allow only whitespace or at most 2 consecutive non-whitespace characters to appear between each name part, you'd have to get more fancy:
\bMar\w*(\s+\S{0,2})*\s+Jac\w*(\s+\S{0,2})*\s+Rey\w*
The pattern (\s+\S{0,2})*\s+ will match any number of non-whitespace characters containing at most two characters, each surrounded by whitespace. The \w* after each name part ensures that the entire name is included in that part of the match (you might want to use \S* instead here, but that's not entirely clear from your question). And I threw in a word boundary (\b) at the beginning to ensure that the match does not start in the middle of a 'word' (e.g. OMar would not match).
I think what you want is this regular expression to check if it is true and is case insensitive
#"^[Mar|Jac|Rey]{3}"
Less specific:
#"^[\w]{3}"
If you want to capture the first three letters of every words of at least three characters words you could use something like :
((?<name>[\w]{3})\w+)+
And enable ExplicitCapture when initializing your Regex.
It will return you a serie of Match named "name", each one of them is a result.
Code sample :
Regex regex = new Regex(#"((?<name>[\w]{3})\w+)+", RegexOptions.ExplicitCapture | RegexOptions.IgnoreCase);
var match = regex.Matches("Marck Jacobs L. S. Reynolds");
If you want capture also 3 characters words, you can replace the last "\w" by a space. In this case think to handle the last word of the phrase.

How to test if there is left over after using a regular expression?

Example Input:
john#test.com johnny#myserver.com joan#server.com
With an regular expression like [a-z]+#[a-z]+.com we can find alle mail addresses in the given input string. Note: The regular expression is simplified to keep the example easy.
Question: Is there a way to check if the left over of the input string (which did not match the pattern) only consists of whitespace, so that we can test, whether each mail address was regocnized by the pattern or not?
Use grouping and captures.
Use a regex like \s*([a-z]+#[a-z]+.com)\s* (note the extra parenthesis! important!), an then instead of looking at what regex has matched as a whole, get the matchresult object and inspect its groups and captures. From that, you will get not one email, but a list of all emails that were caught (and that were separated by whitespaces)
EDIT:
First, check this article on MSDN for overview about "groups" and "captures".
Then, note that Regex.Match returns an object of class Match which tells you whether there was a Success or not. But, aside from the Success property on that Match object, there are some other properties like Captures or Groups.
Those two properties of Match are collections that keep all strings that were, well, 'captured' by any 'group' (parenthesis) that occurred in the regex.
Depending on how you have structured your regex, the contents of Captures and Groups will differ, but just see them for yourself and it should be clear what/how they work.
For example:
regex: (aaa([b0-9]+)ccc([d0-9]+)eee\s*)*
input: "aaab123cccd456eee aaab789cccd123deee"
will result in 3 groups (because there are 3 sets of parenthesis) and 6 captures (because the regex matched two big strings and small parens matched twice in each of the big string)
groups:
[0] captures: "aaab123cccd456eee ", "aaab789cccd123deee"
[1] captures: "b123", "b789"
[2] captures: "d456", "d123"
Note that there's a space in the "big" capture from the "big" parenthesis, since I've included the \s* token at the end of it to account for the separator.

Regex non-greedy match

If I have the following simplified string:
string331-itemR-icon253,string131-itemA-icon453,
string12131-itemB-icon4535,string22-itemC-icon443
How do I get the following only using only regex?
string12131-itemB-icon4535,
All numbers are unknown. The only known parts are
itemA, itemB, itemC, string and icon
I've tried string.+?itemB.+?, but it also picks up from the first occurrence of string rather than the one adjacent to itemB
I've also tried using [^icon] preceding the itemB in various positions but couldn't get it to work.
Try this:
string[^,]+itemB[^,]+,
Try this regex
string\d+-itemB-icon\d+,
The given solutions that use a restricted set of characters instead of a wildcard are simplest, but to get more at the general question: You got the non-greedy quantifier part right, but being non-greedy doesn't prevent the matcher from taking as many characters as it needs to find a match. You might be looking for the atomic group operator, (?>group). Once the group matches something, it will be treated atomically if the matcher needs to backtrack.
(?>string.+?item)B.+?,
In your example, the group matches string331-item, but the B doesn't match R so the whole group is tossed and the search moves to the next string.
You don't mention the commas separating items as a known part but use it in the example regex so I assume it can be used in a solution. Try excluding the comma as a character set instead of matching against ".".

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