regex pattern number inside brackets (int)

Question

Have a text

hjgv ygkuy (1) erg erg ewr (2)erg erg erg er. (3)

want to get 1,2,3 and if there is only one then just 1

i have tried with (\d)\)

i get matches but i want the result like

1,2,3

if just one then just that digit 1

http://rubular.com/r/xDP7h7IRn7

Looks like you just need `\(([0-9])\)` and grab `match.Groups(1).Value`. Do you need to match `(32)`, `(123)`? — Wiktor Stribiżew, Jul 11 '17 at 13:49
the number can be in 2 or three digits inside the brackets i just need that numbers — skcrpk, Jul 11 '17 at 13:52
It is common practice to accept the correct solution provided first. Moreover, `\d` [matches more than `0-9` digits in .NET regex](https://stackoverflow.com/a/16621778/3832970). Using non-capturing groups gives you no advantage, but hampers regex execution. Tom's regex is equal to `\((\d+)\)` and is basically the same as mine (but less precise). That is why it is downvoted: it brings no additional value to the already posted solution and is less precise. — Wiktor Stribiżew, Jul 11 '17 at 23:50

score 1 · Answer 1 · answered Jul 11 '17 at 13:58

You may use a simple \(([0-9]+)\) regex that matches any ( followed with 1 or more digits and then a ), while capturing the digits into Group 1.

See a VB.NET demo:

Dim res As MatchCollection = Regex.Matches("hjgv ygkuy (1) erg erg ewr (12)erg erg erg er. (321)", "\(([0-9]+)\)")
If res.Count() > 0 Then
    For Each m As Match In res
        Console.WriteLine(m.Groups(1).Value)
    Next
End If

Output:

1
12
321

Pattern details:

\( - a literal (
([0-9]+) - Capturing group 1 matching one or more ASCII digits (as \d in a .NET regex may match more than just 0-9)
\) - a literal ).

score 0 · Accepted Answer · answered Jul 11 '17 at 13:59

0

Try using capturing groups;

(?:\()(\d+)(?:\))

Demo

answered Jul 11 '17 at 13:59

Tom Wyllie

1,840
11
16

Thank you very much – skcrpk Jul 11 '17 at 15:59

regex pattern number inside brackets (int)

2 Answers2