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所希望的那样。我的目标是能够运行两次,并生成两个不同的四字符随机字符串。但是,它只生成一个四个字符的随机字符串两次。

下面是代码和输出示例:

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;

...输出如下:UNTE-UNTE ...但它应该看起来像这个UNTE-FWNU

如何确保两个明显随机的字符串?


当前回答

您正在该方法中创建Random实例,这将导致它在快速连续调用时返回相同的值。我会这样做:

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)

其他回答

还有另一个版本:我在测试中使用这种方法生成随机的伪股票代码:

Random rand = new Random();
Func<char> randChar = () => (char)rand.Next(65, 91); // upper case ascii codes
Func<int,string> randStr = null;
    randStr = (x) => (x>0) ? randStr(--x)+randChar() : ""; // recursive

用法:

string str4 = randStr(4);// generates a random 4 char string
string strx = randStr(rand.next(1,5)); // random string between 1-4 chars in length

你可以重新定义randChar函数,使用一个“允许的”字符数组,而不是ascii码:

char[] allowedchars = {'A','B','C','1','2','3'};
Func<char> randChar = () => allowedchars[rand.Next(0, allowedchars.Length-1)];

我的RandomString()方法来生成一个随机字符串。

private static readonly Random _rand = new Random();

/// <summary>
/// Generate a random string.
/// </summary>
/// <param name="length">The length of random string. The minimum length is 3.</param>
/// <returns>The random string.</returns>
public string RandomString(int length)
{
    length = Math.Max(length, 3);

    byte[] bytes = new byte[length];
    _rand.NextBytes(bytes);
    return Convert.ToBase64String(bytes).Substring(0, length);
}

一个使用Path.GetRandomFileName()的非常简单的实现:

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()。

这是我的解决方案:

private string RandomString(int length)
{
    char[] symbols = { 
                            '0', '1', '2', '3', '4', '5', '6', '7', '8', '9',
                            'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z',
                            'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z'                             
                        };

    Stack<byte> bytes = new Stack<byte>();
    string output = string.Empty;

    for (int i = 0; i < length; i++)
    {
        if (bytes.Count == 0)
        {
            bytes = new Stack<byte>(Guid.NewGuid().ToByteArray());
        }
        byte pop = bytes.Pop();
        output += symbols[(int)pop % symbols.Length];
    }
    return output;
}

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

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

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

对于随机字符串生成器:

#region CREATE RANDOM STRING WORD
        char[] wrandom = {'A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','R','S','T','U','V','X','W','Y','Z'};
        Random random = new Random();
        string random_string = "";
        int count = 12; //YOU WILL SPECIFY HOW MANY CHARACTER WILL BE GENERATE
        for (int i = 0; i < count; i++ )
        {
            random_string = random_string + wrandom[random.Next(0, 24)].ToString(); 
        }
        MessageBox.Show(random_string);
        #endregion