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!");
}

var match=Regex.Match(@"a99b",@"\d+");
if(match.Success)
{
    int val;
    if(int.TryParse(match.Value,out val))
    {
        //val is set
    }
}

遍历字符串并使用Char。IsDigit

string a = "str123";
string b = string.Empty;
int val;

for (int i=0; i< a.Length; i++)
{
    if (Char.IsDigit(a[i]))
        b += a[i];
}

if (b.Length>0)
    val = int.Parse(b);

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。

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]

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);