获取字符串中包含的所有正数的扩展方法:

    public static List<long> Numbers(this string str)
    {
        var nums = new List<long>();
        var start = -1;
        for (int i = 0; i < str.Length; i++)
        {
            if (start < 0 && Char.IsDigit(str[i]))
            {
                start = i;
            }
            else if (start >= 0 && !Char.IsDigit(str[i]))
            {
                nums.Add(long.Parse(str.Substring(start, i - start)));
                start = -1;
            }
        }
        if (start >= 0)
            nums.Add(long.Parse(str.Substring(start, str.Length - start)));
        return nums;
    }

如果你也想要负数，只需修改这段代码来处理负号(-)

假设输入如下:

"I was born in 1989, 27 years ago from now (2016)"

得到的数字列表将是:

[1989, 27, 2016]

使用正则表达式…

Regex re = new Regex(@"\d+");
Match m = re.Match("test 66");

if (m.Success)
{
    Console.WriteLine(string.Format("RegEx found " + m.Value + " at position " + m.Index.ToString()));
}
else
{
    Console.WriteLine("You didn't enter a string containing a number!");
}

下面是Linq版本:

string s = "123iuow45ss";
var getNumbers = (from t in s
                  where char.IsDigit(t)
                  select t).ToArray();
Console.WriteLine(new string(getNumbers));

使用StringBuilder比在循环中连接字符串的性能稍好一些。如果处理的是大字符串，它的性能要高得多。

    public static string getOnlyNumbers(string input)
    {
        StringBuilder stringBuilder = new StringBuilder(input.Length);
        for (int i = 0; i < input.Length; i++)
            if (input[i] >= '0' && input[i] <= '9')
                stringBuilder.Append(input[i]);

        return stringBuilder.ToString();
    }

注意:上面的例子函数只适用于正数

 string input = "Hello 20, I am 30 and he is 40";
 var numbers = Regex.Matches(input, @"\d+").OfType<Match>().Select(m => int.Parse(m.Value)).ToArray();

你可以像下面这样使用String属性:

 return new String(input.Where(Char.IsDigit).ToArray()); 

它只给出字符串中的数字。