string verificationCode ="dmdsnjds5344gfgk65585";
            string code = "";
            Regex r1 = new Regex("\\d+");
          Match m1 = r1.Match(verificationCode);
           while (m1.Success)
            {
                code += m1.Value;
                m1 = m1.NextMatch();
            }

另一个使用Regex的简单解决方案 你应该使用这个

using System.Text.RegularExpressions;

代码是

string var = "Hello3453232wor705Ld";
string mystr = Regex.Replace(var, @"\d", "");
string mynumber = Regex.Replace(var, @"\D", "");
Console.WriteLine(mystr);
Console.WriteLine(mynumber);

只需使用一个RegEx来匹配字符串，然后转换:

Match match = Regex.Match(test , @"(\d+)");
if (match.Success) {
   return int.Parse(match.Groups[1].Value);
}

你必须使用Regex作为\d+

\d匹配给定字符串中的数字。

string s = "kg g L000145.50\r\n";
        char theCharacter = '.';
        var getNumbers = (from t in s
                          where char.IsDigit(t) || t.Equals(theCharacter)
                          select t).ToArray();
        var _str = string.Empty;
        foreach (var item in getNumbers)
        {
            _str += item.ToString();
        }
        double _dou = Convert.ToDouble(_str);
        MessageBox.Show(_dou.ToString("#,##0.00"));

基于上一个示例，我创建了一个方法:

private string GetNumberFromString(string sLongString, int iLimitNumbers)
{
    string sReturn = "NA";
    int iNumbersCounter = 0;
    int iCharCounter = 0; 

    string sAlphaChars = string.Empty;
    string sNumbers = string.Empty;
    foreach (char str in sLongString)
    {
        if (char.IsDigit(str))
        {
            sNumbers += str.ToString();
            iNumbersCounter++;
            if (iNumbersCounter == iLimitNumbers)
            {
                return sReturn = sNumbers;
            }
        }
        else
        {
            sAlphaChars += str.ToString();
            iCharCounter++;
            // reset the counter 
            iNumbersCounter = 0; 
        }
    }
    return sReturn;
}