Split a string by Regex

Question

I'm currently thinking of how to split this kind of string into regex using c#.

[01,01,01][02,03,00][03,07,00][04,06,00][05,02,00][06,04,00][07,08,00][08,05,00]

Can someone knowledgeable on regex can point me on how to achieved this goal?

sample regex pattern that don't work:

[\dd,\dd,\dd]

sample output:

[01,01,01]
[02,03,00]
[03,07,00]
[04,06,00]
[05,02,00]
[06,04,00]
[07,08,00]
[08,05,00]

I'm thinking if there is a better way rather that splitting it in ']'? — iamcoder, Dec 20 '17 at 06:52
[`\[\d\d,\d\d,\d\d\]` maybe](http://refiddle.com/refiddles/5a3a093475622d4a68030000)? — Uwe Keim, Dec 20 '17 at 06:54
@CinCout and @Backs When splitting at `]`, you loose the `]`. See [this example](https://dotnetfiddle.net/xrNwUV). — Uwe Keim, Dec 20 '17 at 06:59
@UweKeim obviously there's a catch, which is easily fixable. — CinCout, Dec 20 '17 at 07:17

Evan Huang · Accepted Answer · 2017-12-20T07:23:06.830

3

This will do the job in C# (\[.+?\]), e.g.:

var s = @"[01,01,01][02,03,00][03,07,00][04,06,00][05,02,00][06,04,00][07,08,00][08,05,00]";
var reg = new Regex(@"(\[.+?\])");
var matches = reg.Matches(s);
foreach(Match m in matches)
{
    Console.WriteLine($"{m.Value}");
}

EDIT This is how the expression (\[.+?\]) works

first the outter parenthesis, ( and ), means to capture whatever the inside pattern matched
then the escaped square brackets, \[ and \], is to match the [ and ] in the source string
finally the .+? means to match one or more characters, but as few times as possible, so that it won't match all the characters before the first [ and the last ]

edited Dec 20 '17 at 07:23

answered Dec 20 '17 at 06:55

Evan Huang

1,245
9
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can you explain the regex pattern? if so i can marked it as answer. – iamcoder Dec 20 '17 at 07:02
@iamcoder Please see the updated answer – Evan Huang Dec 20 '17 at 07:23

score 3 · Answer 2 · edited Oct 05 '22 at 19:32

I know you stipulated Regex, however it's worth looking at Split again, if for only for academic purposes:

Code

var input = "[01,01,01][02,03,00][03,07,00][04,06,00][05,02,00][06,04,00][07,08,00][08,05,00]";

var output = input.Split(']',StringSplitOptions.RemoveEmptyEntries)
                     .Select(x  => x + "]") // the bracket back
                     .ToList();

foreach(var o in output) 
   Console.WriteLine(o);

Output

[01,01,01]
[02,03,00]
[03,07,00]
[04,06,00]
[05,02,00]
[06,04,00]
[07,08,00]
[08,05,00]

score 1 · Answer 3 · edited May 02 '23 at 09:43

1

The Regex solution below is restricted to 3 values of only 2 digits seperated by comma. Inside the foreach loop you can access the matching value via match.Value.

Remember to include using System.Text.RegularExpressions;

var input = "[01,01,01][02,03,00][03,07,00][04,06,00][05,02,00][06,04,00][07,08,00][08,05,00]";
foreach(var match in Regex.Matches(input, @"(\[\d{2},\d{2},\d{2}\])+"))
{
  // do stuff
}

edited May 02 '23 at 09:43

Thom A

88,727
11
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answered Dec 20 '17 at 06:56

Jules

567
1
6
15

great it can be also done by this. – iamcoder Dec 20 '17 at 07:03

score 0 · Answer 4 · answered Dec 20 '17 at 07:00

Thanks all for the answer i also got it working by using this code

string pattern = @"\[\d\d,\d\d,\d\d]";
Regex rgx = new Regex(pattern, RegexOptions.IgnoreCase);

MatchCollection matches = rgx.Matches(myResult);

Debug.WriteLine(matches.Count);

foreach (Match match in matches)
   Debug.WriteLine(match.Value);

Split a string by Regex

4 Answers4