C# TrimStart with string parameter

To trim all occurrences of the (exactly matching) string, you can use something like the following:

TrimStart

public static string TrimStart(this string target, string trimString)
{
    if (string.IsNullOrEmpty(trimString)) return target;

    string result = target;
    while (result.StartsWith(trimString))
    {
        result = result.Substring(trimString.Length);
    }

    return result;
}

TrimEnd

public static string TrimEnd(this string target, string trimString)
{
    if (string.IsNullOrEmpty(trimString)) return target;

    string result = target;
    while (result.EndsWith(trimString))
    {
        result = result.Substring(0, result.Length - trimString.Length);
    }

    return result;
}

To trim any of the characters in trimChars from the start/end of target (e.g. "foobar'@"@';".TrimEnd(";@'") will return "foobar") you can use the following:

TrimStart

public static string TrimStart(this string target, string trimChars)
{
    return target.TrimStart(trimChars.ToCharArray());
}

TrimEnd

public static string TrimEnd(this string target, string trimChars)
{
    return target.TrimEnd(trimChars.ToCharArray());
}

TrimStart and TrimEnd takes in an array of chars. This means that you can pass in a string as a char array like this:

var trimChars = " .+-";
var trimmed = myString.TrimStart(trimChars.ToCharArray());

So I don't see the need for an overload that takes a string parameter.

I thought the question was trying to trim a specific string from the start of a larger string.

For instance, if I had the string "hellohellogoodbyehello", if you tried to call TrimStart("hello") you would get back "goodbyehello".

If that is the case, you could use code like the following:

string TrimStart(string source, string toTrim)
{
    string s = source;
    while (s.StartsWith(toTrim))
    {
        s = s.Substring(toTrim.Length - 1);
    }
    return s;
}

This wouldn't be super-efficient if you needed to do a lot of string-trimming, but if its just for a few cases, it is simple and gets the job done.