当我们网站上的用户丢失密码并转到丢失密码页面时,我们需要给他一个新的临时密码。我并不介意这有多随机,或者它是否符合所有“所需的”强密码规则,我想做的只是给他们一个他们以后可以更改的密码。

该应用程序是用c#编写的Web应用程序。所以我想刻薄一点,走一条简单的路线,用Guid的一部分。即。

Guid.NewGuid().ToString("d").Substring(1,8)

Suggesstions吗?想法吗?


当前回答

插入一个定时器:timer1, 2个按钮:button1, button2, 1个textBox: textBox1,和一个comboBox: comboBox1。请务必申报:

int count = 0;

源代码:

 private void button1_Click(object sender, EventArgs e)
    {
    // This clears the textBox, resets the count, and starts the timer
        count = 0;
        textBox1.Clear();
        timer1.Start();
    }

    private void timer1_Tick(object sender, EventArgs e)
    {
    // This generates the password, and types it in the textBox
        count += 1;
            string possible = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890";
            string psw = "";
            Random rnd = new Random { };
            psw += possible[rnd.Next(possible.Length)];
            textBox1.Text += psw;
            if (count == (comboBox1.SelectedIndex + 1))
            {
                timer1.Stop();
            }
    }
    private void Form1_Load(object sender, EventArgs e)
    {
        // This adds password lengths to the comboBox to choose from.
        comboBox1.Items.Add("1");
        comboBox1.Items.Add("2");
        comboBox1.Items.Add("3");
        comboBox1.Items.Add("4");
        comboBox1.Items.Add("5");
        comboBox1.Items.Add("6");
        comboBox1.Items.Add("7");
        comboBox1.Items.Add("8");
        comboBox1.Items.Add("9");
        comboBox1.Items.Add("10");
        comboBox1.Items.Add("11");
        comboBox1.Items.Add("12");
    }
    private void button2_click(object sender, EventArgs e)
    {
        // This encrypts the password
        tochar = textBox1.Text;
        textBox1.Clear();
        char[] carray = tochar.ToCharArray();
        for (int i = 0; i < carray.Length; i++)
        {
            int num = Convert.ToInt32(carray[i]) + 10;
            string cvrt = Convert.ToChar(num).ToString();
            textBox1.Text += cvrt;
        }
    }

其他回答

灵感来自@kitsu的回答。但使用RandomNumberGenerator而不是Random或RNGCryptoServiceProvider(在。net 6中已弃用),并添加了一些特殊字符。

可选参数,用于排除在使用System.Text.Json.JsonSerializer.Serialize时将转义的字符—例如&,它转义为\u0026—以便您可以保证序列化字符串的长度与密码的长度匹配。

适用于。net Core 3.0及以上版本。

public static class PasswordGenerator
{
    const string lower = "abcdefghijklmnopqrstuvwxyz";
    const string upper = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
    const string number = "1234567890";
    const string special = "!@#$%^&*()[]{},.:`~_-=+"; // excludes problematic characters like ;'"/\
    const string specialJsonSafe = "!@#$%^*()[]{},.:~_-="; // excludes problematic characters like ;'"/\ and &`+

    const int lowerLength = 26; // lower.Length
    const int upperLength = 26; // upper.Length;
    const int numberLength = 10; // number.Length;
    const int specialLength = 23; // special.Length;
    const int specialJsonSafeLength = 20; // specialJsonSafe.Length;

    public static string Generate(int length = 96, bool jsonSafeSpecialCharactersOnly = false)
    {
        Span<char> result = length < 1024 ? stackalloc char[length] : new char[length].AsSpan();

        for (int i = 0; i < length; ++i)
        {
            switch (RandomNumberGenerator.GetInt32(4))
            {
                case 0:
                    result[i] = lower[RandomNumberGenerator.GetInt32(0, lowerLength)];
                    break;
                case 1:
                    result[i] = upper[RandomNumberGenerator.GetInt32(0, upperLength)];
                    break;
                case 2:
                    result[i] = number[RandomNumberGenerator.GetInt32(0, numberLength)];
                    break;
                case 3:
                    if (jsonSafeSpecialCharactersOnly)
                    {
                        result[i] = specialJsonSafe[RandomNumberGenerator.GetInt32(0, specialJsonSafeLength)];
                    }
                    else
                    {
                        result[i] = special[RandomNumberGenerator.GetInt32(0, specialLength)];
                    }
                    break;
            }
        }

        return result.ToString();
    }
}

我创建的这个方法类似于会员资格提供程序中可用的方法。如果你不想在某些应用程序中添加web引用,这是很有用的。

效果很好。

public static string GeneratePassword(int Length, int NonAlphaNumericChars)
    {
        string allowedChars = "abcdefghijkmnopqrstuvwxyzABCDEFGHJKLMNOPQRSTUVWXYZ0123456789";
        string allowedNonAlphaNum = "!@#$%^&*()_-+=[{]};:<>|./?";
        Random rd = new Random();

        if (NonAlphaNumericChars > Length || Length <= 0 || NonAlphaNumericChars < 0)
            throw new ArgumentOutOfRangeException();

            char[] pass = new char[Length];
            int[] pos = new int[Length];
            int i = 0, j = 0, temp = 0;
            bool flag = false;

            //Random the position values of the pos array for the string Pass
            while (i < Length - 1)
            {
                j = 0;
                flag = false;
                temp = rd.Next(0, Length);
                for (j = 0; j < Length; j++)
                    if (temp == pos[j])
                    {
                        flag = true;
                        j = Length;
                    }

                if (!flag)
                {
                    pos[i] = temp;
                    i++;
                }
            }

            //Random the AlphaNumericChars
            for (i = 0; i < Length - NonAlphaNumericChars; i++)
                pass[i] = allowedChars[rd.Next(0, allowedChars.Length)];

            //Random the NonAlphaNumericChars
            for (i = Length - NonAlphaNumericChars; i < Length; i++)
                pass[i] = allowedNonAlphaNum[rd.Next(0, allowedNonAlphaNum.Length)];

            //Set the sorted array values by the pos array for the rigth posistion
            char[] sorted = new char[Length];
            for (i = 0; i < Length; i++)
                sorted[i] = pass[pos[i]];

            string Pass = new String(sorted);

            return Pass;
    }

使用Random和linq-to-objects来要求每个组都有一个很简单的方法。

随机分组 从第一组中选择随机金额 从以下组中选择剩余的随机金额

 

Random rand = new Random();
int min = 8;
int max = 16;
int totalLen = rand.Next(min, max);
int remainingGroups = 4;
                
string[] allowedLowerChars = "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".Split(',');           
string [] allowedUpperChars = "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".Split(',');
string [] allowedNumbers = "1,2,3,4,5,6,7,8,9,0".Split(',');
string [] allowedSpecialChars = "!,@,#,$,%,&,?".Split(',');


var password = allowedLowerChars.OrderBy(c => rand.Next()).Take(rand.Next(1, totalLen-remainingGroups--)).ToList();
password.AddRange(allowedUpperChars.OrderBy(c => rand.Next()).Take(rand.Next(1, totalLen-password.Count-remainingGroups--)).ToList());
password.AddRange(allowedNumbers.OrderBy(c => rand.Next()).Take(rand.Next(1, totalLen-password.Count-remainingGroups--)).ToList());
password.AddRange(allowedSpecialChars.OrderBy(c => rand.Next()).Take(totalLen-password.Count).ToList());
password = password.OrderBy(c => rand.Next()).ToList(); // randomize groups

我知道这是一个旧线程,但我有什么可能是一个相当简单的解决方案供某人使用。易于实现、易于理解、易于验证。

考虑以下要求:

我需要一个随机密码生成,其中至少有2个小写字母,2个大写字母和2个数字。密码长度至少为8个字符。

下面的正则表达式可以验证这种情况:

^(?=\b\w*[a-z].*[a-z]\w*\b)(?=\b\w*[A-Z].*[A-Z]\w*\b)(?=\b\w*[0-9].*[0-9]\w*\b)[a-zA-Z0-9]{8,}$

这超出了这个问题的范围——但是正则表达式是基于前向/后向和前后向的。

下面的代码将创建一个匹配这个要求的随机字符集:

public static string GeneratePassword(int lowercase, int uppercase, int numerics) {
    string lowers = "abcdefghijklmnopqrstuvwxyz";
    string uppers = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
    string number = "0123456789";

    Random random = new Random();

    string generated = "!";
    for (int i = 1; i <= lowercase; i++)
        generated = generated.Insert(
            random.Next(generated.Length), 
            lowers[random.Next(lowers.Length - 1)].ToString()
        );

    for (int i = 1; i <= uppercase; i++)
        generated = generated.Insert(
            random.Next(generated.Length), 
            uppers[random.Next(uppers.Length - 1)].ToString()
        );

    for (int i = 1; i <= numerics; i++)
        generated = generated.Insert(
            random.Next(generated.Length), 
            number[random.Next(number.Length - 1)].ToString()
        );

    return generated.Replace("!", string.Empty);

}

要满足上述要求,只需调用以下命令:

String randomPassword = GeneratePassword(3, 3, 3);

代码以一个无效字符(“!”)开始——这样字符串就有一个长度,可以向其中注入新字符。

然后,它从1循环到所需的小写字符#,在每次迭代中,从小写列表中抓取一个随机项,并将其注入字符串中的随机位置。

然后对大写字母和数字重复循环。

这将返回长度=小写字母+大写字母+数字的字符串,其中您想要的计数的小写字母、大写字母和数字字符已按随机顺序放置。

public static string GeneratePassword(int passLength) {
        var chars = "abcdefghijklmnopqrstuvwxyz@#$&ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789";
        var random = new Random();
        var result = new string(
            Enumerable.Repeat(chars, passLength)
                      .Select(s => s[random.Next(s.Length)])
                      .ToArray());
        return result;
    }