如何生成随机字符串,并指定所需的长度,或者更好地生成所需的规范上的唯一字符串

Question

有一个库可以生成随机数,那么为什么没有用于生成随机字符串的库？

换句话说,如何生成随机字符串,并指定所需的长度,或更好,在您想要的规范上生成唯一字符串,即指定长度,我的应用程序中的唯一字符串对我来说已经足够了.

我知道我可以创建一个Guid(全球唯一的标识符),但是它们需要更长,更长.

int length = 8;
string s = RandomString.NextRandomString(length)
uniquestringCollection = new UniquestringsCollection(length)
string s2 = uniquestringCollection.GetNext();

Answer 1

我不记得我在哪里得到这个,所以如果你知道谁是最初撰写的,请帮助我给予归属.

private static void Main()
{
    const string AllowedChars = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz#@$^*()";
    Random rng = new Random();

    foreach (string randomString in rng.NextStrings(AllowedChars, (15, 64), 25))
    {
        Console.WriteLine(randomString);
    }

    Console.ReadLine();
}

private static IEnumerable<string> NextStrings(
    this Random rnd,
    string allowedChars,
    (int Min, int Max)length,
    int count)
{
    ISet<string> usedRandomStrings = new HashSet<string>();
    (int min, int max) = length;
    char[] chars = new char[max];
    int setLength = allowedChars.Length;

    while (count-- > 0)
    {
        int stringLength = rnd.Next(min, max + 1);

        for (int i = 0; i < stringLength; ++i)
        {
            chars[i] = allowedChars[rnd.Next(setLength)];
        }

        string randomString = new string(chars, 0, stringLength);

        if (usedRandomStrings.Add(randomString))
        {
            yield return randomString;
        }
        else
        {
            count++;
        }
    }
}

Answer 2

怎么样-

    static Random rd = new Random();
    internal static string CreateString(int stringLength)
    {
        const string allowedChars = "ABCDEFGHJKLMNOPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz0123456789!@$?_-";
        char[] chars = new char[stringLength];

        for (int i = 0; i < stringLength; i++)
        {
            chars[i] = allowedChars[rd.Next(0, allowedChars.Length)];
        }

        return new string(chars);
    }

Answer 3

使用串联 GUID 的随机字符串

此方法使用最少数量的串联 GUID 来返回所需字符数的随机字符串。

    /// <summary>
    /// Uses concatenated then SubStringed GUIDs to get a random string of the
    /// desired length. Relies on the randomness of the GUID generation algorithm.
    /// </summary>
    /// <param name="stringLength">Length of string to return</param>
    /// <returns>Random string of <paramref name="stringLength"/> characters</returns>
    internal static string GetRandomString(int stringLength)
    {
        StringBuilder sb = new StringBuilder();
        int numGuidsToConcat = (((stringLength - 1) / 32) + 1);
        for(int i = 1; i <= numGuidsToConcat; i++)
        {
            sb.Append(Guid.NewGuid().ToString("N"));
        }

        return sb.ToString(0, stringLength);
    }

长度为 8 的输出示例：

39877037
2f1461d8
152ece65
79778fc6
76f426d8
73a27a0d
8efd1210
4bc5b0d2
7b1aa10e
3a7a5b3a
77676839
abffa3c9
37fdbeb1
45736489

长度为 40 的示例输出（请注意重复的“4” - 在 v4 GUID 中，e 是一个十六进制数字，它将始终为 4（有效删除 4 位）——可以通过删除“4”来改进算法为了提高随机性，假设返回的字符串长度可以小于 GUID 的长度（32 个字符）...）：

e5af105b73924c3590e99d2820e3ae7a3086d0e3
e03542e1b0a44138a49965b1ee434e3efe8d063d
c182cecb5f5b4b85a255a397de1c8615a6d6eef5
676548dc532a4c96acbe01292f260a52abdc4703
43d6735ef36841cd9085e56f496ece7c87c8beb9
f537d7702b22418d8ee6476dcd5f4ff3b3547f11
93749400bd494bfab187ac0a662baaa2771ce39d
335ce3c0f742434a904bd4bcad53fc3c8783a9f9
f2dd06d176634c5b9d7083962e68d3277cb2a060
4c89143715d34742b5f1b7047e8107fd28781b39
2f060d86f7244ae8b3b419a6e659a84135ec2bf8
54d5477a78194600af55c376c2b0c8f55ded2ab6
746acb308acf46ca88303dfbf38c831da39dc66e
bdc98417074047a79636e567e4de60aa19e89710
a114d8883d58451da03dfff75796f73711821b02

C# Fiddler 演示：https://dotnetfiddle.net/Y1j6Dw

Answer 4

怎么样

string.Join("", Enumerable.Repeat(0, 100).Select(n => (char)new Random().Next(127)));

或者

var rand = new Random();
string.Join("", Enumerable.Repeat(0, 100).Select(n => (char)rand.Next(127)));