Extract number at end of string in C#

Question

Probably over analysing this a little bit but how would stackoverflow suggest is the best way to return an integer that is contained at the end of a string.

Thus far

Answer 1

Regex pattern like \d+$ is a bit expensive since, by default, a string is parsed from left to right. Once the regex engine finds 1 in 12abc34, it goes on to match 2, and when it encounters a, the match is failed, next position is tried, and so on.

However, in .NET regex, there is a RegexOptions.RightToLeft modifier. It makes the regex engine parse the string from right to left and you may get matches that are known to be closer to the end much quicker.

var result = Regex.Match("000AB22CD1234", @"\d+$", RegexOptions.RightToLeft);
if (result.Success) 
{ 
    Console.Write(result.Value);
}  // => 1234

See the online C# demo.

Answer 2

Am I allowed to go crazy with this?

using System.Text;
using System.Linq;

static string GetNum(string input)
{
    StringBuilder sb = new StringBuilder();
    for (int i = input.Length - 1; i >= 0; i--)
    {
        if (input[i] < 48 || input[i] > 57)
            break;

        sb.Append(input[i]);
    }

    return String.Concat(sb.ToString().ToCharArray().Reverse());
}

Answer 3

Use this regular expression:

\d+$

var result = Regex.Match(input, @"\d+$").Value;

or using Stack, probably more efficient:

var stack = new Stack<char>();

for (var i = input.Length - 1; i >= 0; i--)
{
    if (!char.IsNumber(input[i]))
    {
        break;
    }

    stack.Push(input[i]);
}

var result = new string(stack.ToArray());

Answer 4

[^0-9]+([0-9]+)

should do it I think

Answer 5

Regex would be the easiest, as far as my experience.

Regex ex = new Regex(@"(\d+)$")

This should match it. Just wrap that in a function.

Answer 6

Without using RegEx that can be hard to understand, or Linq and array manipulation that can be slow, a simple loop in an extension method can be used.

It can be used for long or int or ulong or uint or other while adapting the + and - check.

It can be adapted to parse float and double or decimal too.

This method can also be written as a Parse to have exceptions.

Implementation

static public class StringHelper
{
  static public bool TryParseEndAsLong(this string str, out long result)
  {
    result = 0;
    if ( string.IsNullOrEmpty(str) )
      return false;
    int index = str.Length - 1;
    for ( ; index >= 0; index-- )
    {
      char c = str[index];
      bool stop = c == '+' || c == '-';
      if ( !stop && !char.IsDigit(c) )
      {
        index++;
        break;
      }
      if ( stop )
        break;
    }
    return index <= 0 ? long.TryParse(str, out result)
                      : long.TryParse(str.Substring(index), out result);
  }
} 

Test

test(null);
test("");
test("Test");
test("100");
test("-100");
test("100-200");
test("100 - 200");
test("Test 100");
test("Test100");
test("Test+100");
test("Test-100");
test("11111111111111111111");

Action<string> test = str =>
{
  if ( str.TryParseEndAsLong(out var value) )
    Console.WriteLine($"\"{str}\" => {value}");
  else
    Console.WriteLine($"\"{str}\" has not a long at the end");
};

Output

"" has not a long at the end
"" has not a long at the end
"Test" has not a long at the end
"100" => 100
"-100" => -100
"100-200" => -200
"100 - 200" => 200
"Test 100" => 100
"Test100" => 100
"Test+100" => 100
"Test-100" => -100
"11111111111111111111" has not a long at the end