你必须使用Regex作为\d+

\d匹配给定字符串中的数字。

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]

 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();

正则表达式。Split可以从字符串中提取数字。你会得到在字符串中找到的所有数字。

string input = "There are 4 numbers in this string: 40, 30, and 10.";
// Split on one or more non-digit characters.
string[] numbers = Regex.Split(input, @"\D+");
foreach (string value in numbers)
{
    if (!string.IsNullOrEmpty(value))
    {
    int i = int.Parse(value);
    Console.WriteLine("Number: {0}", i);
    }
}

输出:

数量:4 数量:40 数量:30 数量:10

下面是另一种Linq方法，它从字符串中提取第一个数字。

string input = "123 foo 456";
int result = 0;
bool success = int.TryParse(new string(input
                     .SkipWhile(x => !char.IsDigit(x))
                     .TakeWhile(x => char.IsDigit(x))
                     .ToArray()), out result);

例子:

string input = "123 foo 456"; // 123
string input = "foo 456";     // 456
string input = "123 foo";     // 123

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