随机字符串生成器返回相同的字符串

Question

我已经开发了一个随机字符串生成器,但它的行为并不像我希望的那样.我的目标是能够运行两次并生成两个不同的四个字符随机字符串.但是,它只生成一个四字符随机字符串两次.

这是代码和输出的示例:

private string RandomString(int size)
{
    StringBuilder builder = new StringBuilder();
    Random random = new Random();
    char ch;
    for (int i = 0; i < size; i++)
    {
        ch = Convert.ToChar(Convert.ToInt32(Math.Floor(26 * random.NextDouble() + 65)));                 
        builder.Append(ch);
    }

    return builder.ToString();
}

// get 1st random string 
string Rand1 = RandomString(4);

// get 2nd random string 
string Rand2 = RandomString(4);

// create full rand string
string docNum = Rand1 + "-" + Rand2;

......输出看起来像这样:UNTE-UNTE ......但看起来应该像UNTE-FWNU

如何确保两个明显随机的字符串？

Answer 1

您正在方法中创建Random实例,这会导致它在快速连续调用时返回相同的值.我会做这样的事情:

private static Random random = new Random((int)DateTime.Now.Ticks);//thanks to McAden
private string RandomString(int size)
    {
        StringBuilder builder = new StringBuilder();
        char ch;
        for (int i = 0; i < size; i++)
        {
            ch = Convert.ToChar(Convert.ToInt32(Math.Floor(26 * random.NextDouble() + 65)));                 
            builder.Append(ch);
        }

        return builder.ToString();
    }

// get 1st random string 
string Rand1 = RandomString(4);

// get 2nd random string 
string Rand2 = RandomString(4);

// creat full rand string
string docNum = Rand1 + "-" + Rand2;

(修改后的代码版本)

Answer 2

您正在实例化Random方法中的对象.

该Random物体是从系统时钟种子,这意味着如果你打电话给你的方法多次快速连续它会每次都使用相同的种子,这意味着它会产生随机数的相同的序列,这意味着你我会得到相同的字符串.

要解决这个问题,请将您的Random实例移出方法本身(当您在它的时候,你可以摆脱那些疯狂的调用Convert和Floor和NextDouble):

private readonly Random _rng = new Random();
private const string _chars = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";

private string RandomString(int size)
{
    char[] buffer = new char[size];

    for (int i = 0; i < size; i++)
    {
        buffer[i] = _chars[_rng.Next(_chars.Length)];
    }
    return new string(buffer);
}

Answer 3

//一个非常简单的实现

using System.IO;   
public static string RandomStr()

{
    string rStr = Path.GetRandomFileName();
    rStr = rStr.Replace(".", ""); // For Removing the .
    return rStr;
}

//现在只需调用RandomStr()方法

Answer 4

只要您使用的是Asp.Net 2.0或更高版本,您也可以使用库调用 System.Web.Security.Membership.GeneratePassword- 但它将包含特殊字符.

获得4个随机字符,最少0个特殊字符 -

Membership.GeneratePassword(4, 0)

Answer 5

只是为了让人们在一行代码中停下来以及如何获得随机字符串

int yourRandomStringLength = 12; //maximum: 32
Guid.NewGuid().ToString("N").Substring(0, yourRandomStringLength);

PS:请记住,最大长度为32 ,yourRandomStringLength不能超过Guid32.

Answer 6

此解决方案是Random课程的扩展.

用法

class Program
{
    private static Random random = new Random(); 

    static void Main(string[] args)
    {
        random.NextString(10); // "cH*%I\fUWH0"
        random.NextString(10); // "Cw&N%27+EM"
        random.NextString(10); // "0LZ}nEJ}_-"
        random.NextString();   // "kFmeget80LZ}nEJ}_-"
    }
}

履行

public static class RandomEx
{
    /// <summary>
    /// Generates random string of printable ASCII symbols of a given length
    /// </summary>
    /// <param name="r">instance of the Random class</param>
    /// <param name="length">length of a random string</param>
    /// <returns>Random string of a given length</returns>
    public static string NextString(this Random r, int length)
    {
        var data = new byte[length];
        for (int i = 0; i < data.Length; i++)
        {
            // All ASCII symbols: printable and non-printable
            // data[i] = (byte)r.Next(0, 128);
            // Only printable ASCII
            data[i] = (byte)r.Next(32, 127);
        }
        var encoding = new ASCIIEncoding();
        return encoding.GetString(data);
    }

    /// <summary>
    /// Generates random string of printable ASCII symbols
    /// with random length of 10 to 20 chars
    /// </summary>
    /// <param name="r">instance of the Random class</param>
    /// <returns>Random string of a random length between 10 and 20 chars</returns>
    public static string NextString(this Random r)
    {
        int length  = r.Next(10, 21);
        return NextString(r, length);
    }
}

Answer 7

另一个版本的字符串生成器.简单,没有花哨的数学和魔术数字.但是有一些魔术字符串指定允许的字符.

更新: 我使生成器静态,因此多次调用时不会返回相同的字符串.但是,此代码不是 线程安全的,绝对不是 加密安全的.

System.Security.Cryptography.RNGCryptoServiceProvider应使用密码生成.

private Random _random = new Random(Environment.TickCount);

public string RandomString(int length)
{
    string chars = "0123456789abcdefghijklmnopqrstuvwxyz";
    StringBuilder builder = new StringBuilder(length);

    for (int i = 0; i < length; ++i)
        builder.Append(chars[_random.Next(chars.Length)]);

    return builder.ToString();
}

Answer 8

还有一个选择:

public System.String GetRandomString(System.Int32 length)
{
    System.Byte[] seedBuffer = new System.Byte[4];
    using (var rngCryptoServiceProvider = new System.Security.Cryptography.RNGCryptoServiceProvider())
    {
        rngCryptoServiceProvider.GetBytes(seedBuffer);
        System.String chars = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789";
        System.Random random = new System.Random(System.BitConverter.ToInt32(seedBuffer, 0));
        return new System.String(Enumerable.Repeat(chars, length).Select(s => s[random.Next(s.Length)]).ToArray());
    }
}

Answer 9

最好的解决方案是使用随机数生成器进行base64转换

public string GenRandString(int length)
{
  byte[] randBuffer = new byte[length];
  RandomNumberGenerator.Create().GetBytes(randBuffer);
  return System.Convert.ToBase64String(randBuffer).Remove(length);
}

Answer 10

一个很好的 LINQ 单线（假设一个private static Random Random）...

public static string RandomString(int length)
{
    return new string(Enumerable.Range(0, length).Select(_ => (char)Random.Next('a', 'z')).ToArray());
}