DTB解决方案的一个稍微干净的版本。

    var chars = "ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789";
    var random = new Random();
    var list = Enumerable.Repeat(0, 8).Select(x=>chars[random.Next(chars.Length)]);
    return string.Join("", list);

您的风格偏好可能会有所不同。

I was looking for a more specific answer, where I want to control the format of the random string and came across this post. For example: license plates (of cars) have a specific format (per country) and I wanted to created random license plates. I decided to write my own extension method of Random for this. (this is in order to reuse the same Random object, as you could have doubles in multi-threading scenarios). I created a gist (https://gist.github.com/SamVanhoutte/808845ca78b9c041e928), but will also copy the extension class here:

void Main()
{
    Random rnd = new Random();
    rnd.GetString("1-###-000").Dump();
}

public static class RandomExtensions
{
    public static string GetString(this Random random, string format)
    {
        // Based on http://stackoverflow.com/questions/1344221/how-can-i-generate-random-alphanumeric-strings-in-c
        // Added logic to specify the format of the random string (# will be random string, 0 will be random numeric, other characters remain)
        StringBuilder result = new StringBuilder();
        for(int formatIndex = 0; formatIndex < format.Length ; formatIndex++)
        {
            switch(format.ToUpper()[formatIndex])
            {
                case '0': result.Append(getRandomNumeric(random)); break;
                case '#': result.Append(getRandomCharacter(random)); break;
                default : result.Append(format[formatIndex]); break;
            }
        }
        return result.ToString();
    }

    private static char getRandomCharacter(Random random)
    {
        string chars = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
        return chars[random.Next(chars.Length)];
    }

    private static char getRandomNumeric(Random random)
    {
        string nums = "0123456789";
        return nums[random.Next(nums.Length)];
    }
}

如果你的值不是完全随机的，但实际上可能依赖于某些东西——你可以计算出“某个东西”的md5或sha1哈希，然后将其截断为你想要的任何长度。

你也可以生成和截断一个guid。

尝试将两部分结合起来:独特(序列、计数器或日期)和随机

public class RandomStringGenerator
{
    public static string Gen()
    {
        return ConvertToBase(DateTime.UtcNow.ToFileTimeUtc()) + GenRandomStrings(5); //keep length fixed at least of one part
    }

    private static string GenRandomStrings(int strLen)
    {
        var result = string.Empty;

        using (var gen = new RNGCryptoServiceProvider())
        {
            var data = new byte[1];

            while (result.Length < strLen)
            {
                gen.GetNonZeroBytes(data);
                int code = data[0];
                if (code > 48 && code < 57 || // 0-9
                    code > 65 && code < 90 || // A-Z
                    code > 97 && code < 122   // a-z
                )
                {
                    result += Convert.ToChar(code);
                }
            }

            return result;
        }
    }

    private static string ConvertToBase(long num, int nbase = 36)
    {
        const string chars = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"; //if you wish to make the algorithm more secure - change order of letter here

        // check if we can convert to another base
        if (nbase < 2 || nbase > chars.Length)
            return null;

        int r;
        var newNumber = string.Empty;

        // in r we have the offset of the char that was converted to the new base
        while (num >= nbase)
        {
            r = (int)(num % nbase);
            newNumber = chars[r] + newNumber;
            num = num / nbase;
        }
        // the last number to convert
        newNumber = chars[(int)num] + newNumber;

        return newNumber;
    }
}

测试:

    [Test]
    public void Generator_Should_BeUnigue1()
    {
        //Given
        var loop = Enumerable.Range(0, 1000);
        //When
        var str = loop.Select(x=> RandomStringGenerator.Gen());
        //Then
        var distinct = str.Distinct();
        Assert.AreEqual(loop.Count(),distinct.Count()); // Or Assert.IsTrue(distinct.Count() < 0.95 * loop.Count())
    }

解决方案1 -最大的“范围”与最灵活的长度

string get_unique_string(int string_length) {
    using(var rng = new RNGCryptoServiceProvider()) {
        var bit_count = (string_length * 6);
        var byte_count = ((bit_count + 7) / 8); // rounded up
        var bytes = new byte[byte_count];
        rng.GetBytes(bytes);
        return Convert.ToBase64String(bytes);
    }
}

这个解决方案比使用GUID有更大的范围，因为GUID有几个固定的位，它们总是相同的，因此不是随机的，例如十六进制中的13个字符总是“4”——至少在版本6的GUID中是这样。

这个解决方案还允许您生成任意长度的字符串。

解决方案2 -一行代码-最多22个字符

Convert.ToBase64String(Guid.NewGuid().ToByteArray()).Substring(0, 8);

你不能生成字符串，只要解决方案1和字符串没有相同的范围，由于GUID的固定位，但在很多情况下，这将完成工作。

解决方案3——代码略少

Guid.NewGuid().ToString("n").Substring(0, 8);

主要是为了历史目的。它使用更少的代码，尽管代价是范围更小——因为它使用十六进制而不是base64，所以与其他解决方案相比，它需要更多的字符来表示相同的范围。

这意味着碰撞的可能性更大——用10万次迭代测试8个字符串，生成一个副本。

现在是单行风格。

private string RandomName()
{
        return new string(
            Enumerable.Repeat("ABCDEFGHIJKLMNOPQRSTUVWXYZ", 13)
                .Select(s =>
                {
                    var cryptoResult = new byte[4];
                    using (var cryptoProvider = new RNGCryptoServiceProvider())
                        cryptoProvider.GetBytes(cryptoResult);

                    return s[new Random(BitConverter.ToInt32(cryptoResult, 0)).Next(s.Length)];
                })
                .ToArray());
}