CODEWITHSUNDEEP
c#regex
SharkyShark
SharkyShark

Reputation: 380

Regex pattern generator

I'm trying to do regex pattern which will match to this:

Name[0]/Something

or 

Name/Something

Verbs Name and Something will be always known. I did for Name[0]/Something, but I want make pattern for this verb in one regex I've tried to sign [0] as optional but it didn't work :

 var regexPattern = "Name" + @"\([\d*\]?)/" + "Something"

Do you know some generator where I will input some verbs and it will make pattern for me?

Upvotes: 3

Views: 3220

Answers (5)

Samvel Petrosov
Samvel Petrosov

Reputation: 7706

The problem is with first '\' in your pattern before '('.
Here is what you need:

var str = "Name[0]/Something or Name/Something";
Regex rg = new Regex(@"Name(\[\d+\])?/Something");
var matches = rg.Matches(str);
foreach(Match a in matches)
{
    Console.WriteLine(a.Value);
}

Upvotes: 1

degant
degant

Reputation: 4981

Use this:

Name(\[\d+\])?\/Something
  • \d+ allows one or more digits
  • \[\d+\] allows one or more digits inside [ and ]. So it will allow [0], [12] etc but reject []
  • (\[\d+\])? allows digit with brackets to be present either zero times or once
  • \/ indicates a slash (only one)
  • Name and Something are string literals

Regex 101 Demo

Upvotes: 4

online Thomas
online Thomas

Reputation: 9381

i think this is what you are looking for:

Name(\[\d+\])?\/Something

Name litteral

([\d+])? a number (1 or more digits) between brackets optional 1 or 0 times

/Something Something litteral

https://regex101.com/r/G8tIHC/1

Upvotes: 0

Andrii Litvinov
Andrii Litvinov

Reputation: 13182

You were close, the regex Name(\[\d+\])?\/Something will do.

Upvotes: 1

user7739271
user7739271

Reputation: 

var string = 'Name[0]/Something';

var regex = /^(Name)(\[\d*\])?\/Something$/;

console.log(regex.test(string));

string = 'Name/Something';

console.log(regex.test(string));

You've tried wrong with this pattern: \([\d*\]?)/

  • No need to use \ before ( (in this case)

  • ? after ] mean: character ] zero or one time

So, if you want the pattern [...] displays zero or one time, you can try: (\[\d*\])?

Hope this helps!

Upvotes: 0

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