将这两个正则表达式组合为一个

@"^(?=.*[a-zA-Z])(?=.*\d)"

 ^  # From the begining of the string
 (?=.*[a-zA-Z]) # look forward for any number of chars followed by a letter, don't advance pointer
 (?=.*\d) # look forward for any number of chars followed by a digit)

使用两个 positive lookaheads 确保在成功之前找到一个字母和一个数字。您添加 ^ 从字符串开始,只向前看一次。否则,regexp引擎将尝试在字符串中的每个点进行匹配。

对于具有LINQ的C#:

return s.Any(Char.IsDigit) && s.Any(Char.IsLetter);

[a-zA-Z].*[0-9]|[0-9].*[a-zA-Z] ,但我只建议您使用的系统只接受一个正则表达式。我无法想象这会比两个没有交替的简单模式更有效。

这不完全是你想要的,但我有更多的时间。下面的代码应该比正则表达式工作得更快。

    static bool IsAlphaAndNumeric(string str) {
        bool hasDigits = false;
        bool  hasLetters=false;

        foreach (char c in str) {
            bool isDigit = char.IsDigit(c);
            bool isLetter = char.IsLetter(c);
            if (!(isDigit | isLetter))
                return false;
            hasDigits |= isDigit;
            hasLetters |= isLetter;
        }
        return hasDigits && hasLetters;
    }

为什么这么快,让我们看看。 下面是测试字符串生成器。它生成1/3的集合完全正确的字符串和2/3的ad不正确。在2/3中,1/2是所有alphs,另一半是所有数字。

    static IEnumerable<string> GenerateTest(int minChars, int maxChars, int setSize) {
        string letters = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ";
        string numbers = "0123456789";            
        Random rnd = new Random();
        int maxStrLength = maxChars-minChars;
        float probablityOfLetter = 0.0f;
        float probablityInc = 1.0f / setSize;
        for (int i = 0; i < setSize; i++) {
            probablityOfLetter = probablityOfLetter + probablityInc;
            int length = minChars + rnd.Next() % maxStrLength;
            char[] str = new char[length];
            for (int w = 0; w < length; w++) {
                if (probablityOfLetter < rnd.NextDouble())
                    str[w] = letters[rnd.Next() % letters.Length];
                else 
                    str[w] = numbers[rnd.Next() % numbers.Length];                    
            }
            yield return new string(str);
        }
    }

下面是darin的两个解决方案。一个已编译,另一个为非编译版本。

class DarinDimitrovSolution
{
    const string regExpression = @"^(?=.*[a-z])(?=.*[A-Z])(?=.*\d).+$";
    private static readonly Regex _regex = new Regex(
        regExpression, RegexOptions.Compiled);

    public static bool IsAlphaAndNumeric_1(string s) {
        return _regex.IsMatch(s);
    }
    public static bool IsAlphaAndNumeric_0(string s) {
        return Regex.IsMatch(s, regExpression);
    }

下面是测试循环的主要部分

    static void Main(string[] args) {

        int minChars = 3;
        int maxChars = 13;
        int testSetSize = 5000;
        DateTime start = DateTime.Now;
        foreach (string testStr in
            GenerateTest(minChars, maxChars, testSetSize)) {
            IsAlphaNumeric(testStr);
        }
        Console.WriteLine("My solution : {0}", (DateTime.Now - start).ToString());

        start = DateTime.Now;
        foreach (string testStr in
            GenerateTest(minChars, maxChars, testSetSize)) {
            DarinDimitrovSolution.IsAlphaAndNumeric_0(testStr);
        }
        Console.WriteLine("DarinDimitrov  1 : {0}", (DateTime.Now - start).ToString());

        start = DateTime.Now;
        foreach (string testStr in
            GenerateTest(minChars, maxChars, testSetSize)) {
            DarinDimitrovSolution.IsAlphaAndNumeric_1(testStr);
        }
        Console.WriteLine("DarinDimitrov(compiled) 2 : {0}", (DateTime.Now - start).ToString());

        Console.ReadKey();
    }

My solution : 00:00:00.0170017    (Gold)
DarinDimitrov  1 : 00:00:00.0320032  (Silver medal) 
DarinDimitrov(compiled) 2 : 00:00:00.0440044   (Gold)

所以第一个解决方案是最好的。 更多的结果是发布模式和遵循规范

   int minChars = 20;
   int maxChars = 50;
   int testSetSize = 100000;

My solution : 00:00:00.4060406
DarinDimitrov  1 : 00:00:00.7400740
DarinDimitrov(compiled) 2 : 00:00:00.3410341 (now that very fast)

My solution : 00:00:00.4290429 (almost same as before)
DarinDimitrov  1 : 00:00:00.9700970 (it have slowed down )
DarinDimitrov(compiled) 2 : 00:00:00.8440844 ( this as well still fast but look at .3 in last result)

在gnarf提到我的算法有问题后,它检查字符串是否只由字母和数字组成,所以我更改了它,现在它检查字符串显示是否至少有一个字符和一个数字。

    static bool IsAlphaNumeric(string str) {
        bool hasDigits = false;
        bool hasLetters = false;

        foreach (char c in str) {
            hasDigits |= char.IsDigit(c);
            hasLetters |= char.IsLetter(c);
            if (hasDigits && hasLetters)
                return true;
        }
        return false;
    }

My solution : 00:00:00.3900390 (Goody Gold Medal)
DarinDimitrov  1 : 00:00:00.9740974 (Bronze Medal)
DarinDimitrov(compiled) 2 : 00:00:00.8230823 (Silver)

我的很快。

private static readonly Regex _regex = new Regex(
    @"^(?=.*[a-z])(?=.*[A-Z])(?=.*\d).+$", RegexOptions.Compiled);

public static bool IsAlphaAndNumeric(string s)
{
    return _regex.IsMatch(s);
}

如果你想忽略这个案例,你可以使用 RegexOptions.Compiled | RegexOptions.IgnoreCase .

以下内容不仅比其他前瞻性构造更快,而且(在我看来)更接近需求:

[a-zA-Z\d]((?<=\d)[^a-zA-Z]*[a-zA-Z]|[^\d]*\d)

以下是其工作原理(以及原因):

步骤1:它匹配单个字符(我们称之为 )那是一个数字或一个字母。
C
步骤2.1:允许不限数量的非字母字符,后跟单个字母。如果匹配,我们有一个数字( )然后是一封信。
步骤2.2:如果 C C )后面跟着一个数字。