C# regex. Everything inside curly brackets{} and mod(%) charaters

Question

I'm trying to get the values between {} and %% in a same Regex. This is what I have till now. I can successfully get values individually for each but I was curious to learn about how can I combine both.

var regex = new Regex(@"%(.*?)%|\{([^}]*)\}");

String s = "This is a {test} %String%. %Stack% {Overflow}";

Expected answer for the above string

test
String
Stack
Overflow

Individual regex

@"%(.*?)%" gives me String and Stack
@"\{([^}]*)\}" gives me test and Overflow

Following is my code.

var regex = new Regex(@"%(.*?)%|\{([^}]*)\}");
var matches = regex.Matches(s); 
foreach (Match match in matches) 
{
    Console.WriteLine(match.Groups[1].Value);
}

Your regex is just fine you need to use match.Captures[0].Value not match.Groups[1].Value — Craig Selbert, Feb 24 '16 at 22:42
http://stackoverflow.com/questions/3320823/whats-the-difference-between-groups-and-captures-in-net-regular-expression — Craig Selbert, Feb 24 '16 at 22:44
Didn't see there is [conditional](http://www.regular-expressions.info/conditional.html) answer already. You can also try [`(?:({)|%)(.*?)(?(1)}|%)`](http://regexstorm.net/tester?p=(%3F%3A(%7B)%7C%25)(.*%3F)(%3F(1)%7D%7C%25)&i=This%20is%20a%20%7Btest%7D%20%25String%25.%20%25Stack%25%20%7BOverflow%7D) — bobble bubble, Feb 24 '16 at 23:29

Eser · Answer 1 · 2016-02-24T22:58:47.330

3

Similar to your regex. You can use Named Capturing Groups

String s = "This is a {test} %String%. %Stack% {Overflow}";
var list = Regex.Matches(s, @"\{(?<name>.+?)\}|%(?<name>.+?)%")
           .Cast<Match>()
           .Select(m => m.Groups["name"].Value)
           .ToList();

edited Feb 24 '16 at 22:58

answered Feb 24 '16 at 22:56

Eser

12,346
1
22
32

@WiktorStribiżew Sorry, I don't get what you are saying. This code returns what OP wants. See `Expected answer for the above string ` and compare with what this code returns.... – Eser Feb 24 '16 at 22:59
You should have explained you are using groups with identical names, and describe the feature. – Wiktor Stribiżew Feb 24 '16 at 23:02
@WiktorStribiżew Seems like "what you dislike in this answer" has changed. No I won't explain anything. This is my way.... And I think this is more readable than your regex... – Eser Feb 24 '16 at 23:04
@WiktorStribiżew thanks for -1, but I won't do the same childish thing for you. – Eser Feb 24 '16 at 23:14
I do not think code only answers deserve ANY upvotes. SO is a place for those who want to learn. Do not give fish, teach to fish. You openly say you don't want to share your knowledge with others - that is why I downvoted. If you downvote my answer, that would be childish, yes. – Wiktor Stribiżew Feb 24 '16 at 23:15
@WiktorStribiżew It is just your opinion..... If you want learn how to use it, You can use the link in my answer :) – Eser Feb 24 '16 at 23:15

Wiktor Stribiżew · Answer 2 · 2016-02-24T23:00:41.307

If you want to learn how conditional expressions work, here is a solution using that kind of .NET regex capability:

(?:(?<p>%)|(?<b>{))(?<v>.*?)(?(p)%|})

See the regex demo

Here is how it works:

(?:(?<p>%)|(?<b>{)) - match and capture either Group "p" with % (percentage), or Group "b" (brace) with {
(?<v>.*?) - match and capture into Group "v" (value) any character (even a newline since I will be using RegexOptions.Singleline) zero or more times, but as few as possible (lazy matching with *? quantifier)
(?(p)%|}) - a conditional expression meaning: if "p" group was matched, match %, else, match }.

C# demo:

var s = "This is a {test} %String%. %Stack% {Overflow}";
var regex = "(?:(?<p>%)|(?<b>{))(?<v>.*?)(?(p)%|})";
var matches = Regex.Matches(s, regex, RegexOptions.Singleline); 
// var matches_list = Regex.Matches(s, regex, RegexOptions.Singleline)
//                 .Cast<Match>() 
//                 .Select(p => p.Groups["v"].Value)
//                 .ToList(); 
// Or just a demo writeline
foreach (Match match in matches) 
    Console.WriteLine(match.Groups["v"].Value);

score 1 · Accepted Answer · answered Feb 24 '16 at 22:40

1

Sometimes the capture is in group 1 and sometimes it's in group 2 because you have two pairs of parentheses.

Your original code will work if you do this instead:

Console.WriteLine(match.Groups[1].Value + match.Groups[2].Value);

because one group will be the empty string and the other will be the value you're interested in.

answered Feb 24 '16 at 22:40

Matthew Strawbridge

19,940
10
72
93

1

I like the adding the two groups together, I was thinking of how to get all the values from group1 and group2 and discard the empty ones. He he... simple and sneaky! – Derek Feb 24 '16 at 22:42
That helps a lot! Thanks! – Ankur Mantri Feb 24 '16 at 23:09

score 0 · Answer 4 · answered Feb 24 '16 at 22:37

0

@"[\{|%](.*?)[\}|%]"

The idea being:

{ or %
anything
} or %

answered Feb 24 '16 at 22:37

Derek

7,615
5
33
58

1

This matches `%something}` – Szabolcs Dézsi Feb 24 '16 at 22:39

score 0 · Answer 5 · answered Feb 24 '16 at 22:50

0

I think you should use a combination of conditional anda nested groups:

((\{(.*)\})|(%(.*)%))

answered Feb 24 '16 at 22:50

tede24

2,304
11
14

C# regex. Everything inside curly brackets{} and mod(%) charaters

5 Answers5