CODEWITHSUNDEEP
c#string
LuckWallace
LuckWallace

Reputation: 133

Get a part of a String using C#

I have to get a little piece of a String, something like this:

20T41

Always are two numbers, a letter and numbers.

But I have many different chains like these:

I have used some codes like this, but I don't know how I can do this for each case. I think it is better than using many IFs.

str = str.Substring (0, str.LastIndexOf ('/') + 1);

Marked in black is the String I want to get. I was thinking of doing something if, but this is not efficient. What do you recommend to do this?

Upvotes: 1

Views: 126

Answers (3)

Pavel Anikhouski
Pavel Anikhouski

Reputation: 23228

The following Regex expression will also help

var regex = new Regex("\\d{2}\\w\\d+");
var input = new List<string>()
    {"PNC 20T17", "PNC 19T1391", "P.N.C 19T1456,", "NC - 19T1099", "PNC 19T323", "19T1512."};

foreach (var item in input)
{
    var match = regex.Match(item);
    Console.WriteLine(match);
}

It produces the following output

20T17
19T1391
19T1456
19T1099
19T323
19T1512

You can have a look at quick reference to see the meaning of every class inside

  • \d{2} decimal digit, two times exactly
  • \w any word character, capital or not
  • \d+ decimal digit one or more times. You can change it to \d{1,} or \d{2,}. It means matching the digit at least one or two times, without upper bounds, which isn't specified in OP. Or use upper bounds, like \d{1,9}

Upvotes: 1

ilkerkaran
ilkerkaran

Reputation: 4344

IF the examples you provided are all combinations then LastIndexOf and Substring would do the work;

var yourInput = "P.N.C 19T1456";
var res = yourInput.Substring(yourInput.LastIndexOf(' ') + 1) ;

Upvotes: 0

Dmitrii Bychenko
Dmitrii Bychenko

Reputation: 186668

You can try matching the fragment with a help of regular expressions:

using System.Text.RegularExpressions;

...

str = Regex.Match(str, "[0-9]{2}T[0-9]+").Value;

Or in case of any (not necessary T) capital letter can appear:

str = Regex.Match(str, "[0-9]{2}[A-Z][0-9]+").Value;

Upvotes: 8

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