我如何在c#中生成一个随机的8个字符的字母数字字符串?


当前回答

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)];
    }
}

其他回答

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。

这是我从Dot Net Perls的Sam Allen那里偷来的一个例子

如果你只需要8个字符,那么在系统中使用Path.GetRandomFileName()。IO命名空间。Sam说使用“Path.”这里的GetRandomFileName方法有时更优越,因为它使用RNGCryptoServiceProvider来获得更好的随机性。然而,它被限制为11个随机字符。”

GetRandomFileName总是返回一个12个字符的字符串,第9个字符是句点。所以你需要去掉句点(因为这不是随机的),然后从字符串中取出8个字符。实际上,你可以只取前8个字符而不用考虑句点。

public string Get8CharacterRandomString()
{
    string path = Path.GetRandomFileName();
    path = path.Replace(".", ""); // Remove period.
    return path.Substring(0, 8);  // Return 8 character string
}

PS:谢谢,Sam

我不知道这在密码学上听起来如何,但它比迄今为止(在我看来)更复杂的解决方案更具可读性和简练性,而且它应该比系统更“随机”。Random-based解决方案。

return alphabet
    .OrderBy(c => Guid.NewGuid())
    .Take(strLength)
    .Aggregate(
        new StringBuilder(),
        (builder, c) => builder.Append(c))
    .ToString();

我不知道我认为这个版本还是下一个版本“更漂亮”,但它们给出了完全相同的结果:

return new string(alphabet
    .OrderBy(o => Guid.NewGuid())
    .Take(strLength)
    .ToArray());

当然,它并没有针对速度进行优化,所以如果每秒生成数百万个随机字符串是关键任务,请尝试另一个!

注意:此解决方案不允许字母中符号的重复,并且字母必须等于或大于输出字符串的大小,使得这种方法在某些情况下不太可取,这完全取决于您的用例。

一个包含所有字母字符和数字的解决方案,你可以随心所欲地更改:

public static string RandomString(int length)
{
    Random rand = new Random();
    string charbase = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789";
    return new string(Enumerable.Range(0,length)
           .Select(_ => charbase[rand.Next(charbase.Length)])
           .ToArray());
}

如果你喜欢单行方法;)

public static Random rand = new Random();
public const string charbase = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789";
 

public static string RandomString(int length) =>
        new string(Enumerable.Range(0,length).Select(_ => charbase[rand.Next(charbase.Length)]).ToArray());

下面是Eric J的解决方案的一个变体,即加密声音,用于WinRT (Windows商店应用程序):

public static string GenerateRandomString(int length)
{
    var chars = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890";
    var result = new StringBuilder(length);
    for (int i = 0; i < length; ++i)
    {
        result.Append(CryptographicBuffer.GenerateRandomNumber() % chars.Length);
    }
    return result.ToString();
}

如果性能很重要(特别是当长度很高时):

public static string GenerateRandomString(int length)
{
    var chars = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890";
    var result = new System.Text.StringBuilder(length);
    var bytes = CryptographicBuffer.GenerateRandom((uint)length * 4).ToArray();
    for (int i = 0; i < bytes.Length; i += 4)
    {
        result.Append(BitConverter.ToUInt32(bytes, i) % chars.Length);
    }
    return result.ToString();
}