这是我的算法

    //Fast, C Language friendly
    public static int GetNumber(string Text)
    {
        int val = 0;
        for(int i = 0; i < Text.Length; i++)
        {
            char c = Text[i];
            if (c >= '0' && c <= '9')
            {
                val *= 10;
                //(ASCII code reference)
                val += c - 48;
            }
        }
        return val;
    }

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

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;
}

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]

如果数字有小数点，可以使用下面的方法

using System;
using System.Text.RegularExpressions;

namespace Rextester
{
    public class Program
    {
        public static void Main(string[] args)
        {
            //Your code goes here
            Console.WriteLine(Regex.Match("anything 876.8 anything", @"\d+\.*\d*").Value);
            Console.WriteLine(Regex.Match("anything 876 anything", @"\d+\.*\d*").Value);
            Console.WriteLine(Regex.Match("$876435", @"\d+\.*\d*").Value);
            Console.WriteLine(Regex.Match("$876.435", @"\d+\.*\d*").Value);
        }
    }
}

结果:

"anything 876.8 anything" ==> 876.8 "anything 876 anything" ==> 876 "$876435" ==> 876435 "$876.435" ==> 876.435

示例:https://dotnetfiddle.net/IrtqVt

这是我的算法

    //Fast, C Language friendly
    public static int GetNumber(string Text)
    {
        int val = 0;
        for(int i = 0; i < Text.Length; i++)
        {
            char c = Text[i];
            if (c >= '0' && c <= '9')
            {
                val *= 10;
                //(ASCII code reference)
                val += c - 48;
            }
        }
        return val;
    }