Way to have String.Replace only hit "whole words"

A regex is the easiest approach:

string input = "test, and test but not testing.  But yes to test";
string pattern = @"\btest\b";
string replace = "text";
string result = Regex.Replace(input, pattern, replace);
Console.WriteLine(result);

The important part of the pattern is the \b metacharacter, which matches on word boundaries. If you need it to be case-insensitive use RegexOptions.IgnoreCase:

Regex.Replace(input, pattern, replace, RegexOptions.IgnoreCase);

I've created a function (see blog post here) that wraps regex expression, suggested by Ahmad Mageed

/// <summary>
/// Uses regex '\b' as suggested in https://stackoverflow.com/questions/6143642/way-to-have-string-replace-only-hit-whole-words
/// </summary>
/// <param name="original"></param>
/// <param name="wordToFind"></param>
/// <param name="replacement"></param>
/// <param name="regexOptions"></param>
/// <returns></returns>
static public string ReplaceWholeWord(this string original, string wordToFind, string replacement, RegexOptions regexOptions = RegexOptions.None)
{
    string pattern = String.Format(@"\b{0}\b", wordToFind);
    string ret=Regex.Replace(original, pattern, replacement, regexOptions);
    return ret;
}

As commented by Sga, the regex solution isn't perfect. And I guess not performance friendly too.

Here is my contribution :

public static class StringExtendsionsMethods
{
    public static String ReplaceWholeWord ( this String s, String word, String bywhat )
    {
        char firstLetter = word[0];
        StringBuilder sb = new StringBuilder();
        bool previousWasLetterOrDigit = false;
        int i = 0;
        while ( i < s.Length - word.Length + 1 )
        {
            bool wordFound = false;
            char c = s[i];
            if ( c == firstLetter )
                if ( ! previousWasLetterOrDigit )
                    if ( s.Substring ( i, word.Length ).Equals ( word ) )
                    {
                        wordFound = true;
                        bool wholeWordFound = true;
                        if ( s.Length > i + word.Length )
                        {
                            if ( Char.IsLetterOrDigit ( s[i+word.Length] ) )
                                wholeWordFound = false;
                        }

                        if ( wholeWordFound )
                            sb.Append ( bywhat );
                        else
                            sb.Append ( word );

                        i += word.Length;
                    }

            if ( ! wordFound )
            {
                previousWasLetterOrDigit = Char.IsLetterOrDigit ( c );
                sb.Append ( c );
                i++;
            }
        }

        if ( s.Length - i > 0 )
            sb.Append ( s.Substring ( i ) );

        return sb.ToString ();
    }
}

... With test cases :

String a = "alpha is alpha";
Console.WriteLine ( a.ReplaceWholeWord ( "alpha", "alphonse" ) );
Console.WriteLine ( a.ReplaceWholeWord ( "alpha", "alf" ) );

a = "alphaisomega";
Console.WriteLine ( a.ReplaceWholeWord ( "alpha", "xxx" ) );

a = "aalpha is alphaa";
Console.WriteLine ( a.ReplaceWholeWord ( "alpha", "xxx" ) );

a = "alpha1/alpha2/alpha3";
Console.WriteLine ( a.ReplaceWholeWord ( "alpha", "xxx" ) );

a = "alpha/alpha/alpha";
Console.WriteLine ( a.ReplaceWholeWord ( "alpha", "alphonse" ) );

I just want to add a note about this particular regex pattern (used both in the accepted answer and in ReplaceWholeWord function). It doesn't work if what you are trying to replace isn't a word.

Here a test case:

using System;
using System.Text.RegularExpressions;
public class Test
{
    public static void Main()
    {
        string input = "doin' some replacement";
        string pattern = @"\bdoin'\b";
        string replace = "doing";
        string result = Regex.Replace(input, pattern, replace);
        Console.WriteLine(result);
    }
}

(ready to try code: http://ideone.com/2Nt0A)

This has to be taken into consideration especially if you are doing batch translations (like I did for some i18n work).