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
如果你想使用System.Web.Security.Membership.GeneratePassword使用的加密安全随机数生成,但又想将字符集限制为字母数字字符,你可以使用regex过滤结果:
static string GeneratePassword(int characterCount)
{
string password = String.Empty;
while(password.Length < characterCount)
password += Regex.Replace(System.Web.Security.Membership.GeneratePassword(128, 0), "[^a-zA-Z0-9]", string.Empty);
return password.Substring(0, characterCount);
}
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();
}
}
总有system。web。security。membership。GeneratePassword(int length, int numberOfNonAlphanumericCharacters)。
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 = "!@#$%^&*()_-+=[{]};:<>|./?";
Random rd = new Random();
if (NonAlphaNumericChars > Length || Length <= 0 || NonAlphaNumericChars < 0)
throw new ArgumentOutOfRangeException();
char[] pass = new char[Length];
int[] pos = new int[Length];
int i = 0, j = 0, temp = 0;
bool flag = false;
//Random the position values of the pos array for the string Pass
while (i < Length - 1)
{
j = 0;
flag = false;
temp = rd.Next(0, Length);
for (j = 0; j < Length; j++)
if (temp == pos[j])
{
flag = true;
j = Length;
}
if (!flag)
{
pos[i] = temp;
i++;
}
}
//Random the AlphaNumericChars
for (i = 0; i < Length - NonAlphaNumericChars; i++)
pass[i] = allowedChars[rd.Next(0, allowedChars.Length)];
//Random the NonAlphaNumericChars
for (i = Length - NonAlphaNumericChars; i < Length; i++)
pass[i] = allowedNonAlphaNum[rd.Next(0, allowedNonAlphaNum.Length)];
//Set the sorted array values by the pos array for the rigth posistion
char[] sorted = new char[Length];
for (i = 0; i < Length; i++)
sorted[i] = pass[pos[i]];
string Pass = new String(sorted);
return Pass;
}
由于Random是不安全的,RNGCryptoServiceProvider是过时的,我最终这样做:
// possible characters that password can have
private const string passChars =
"ABCDEFGHIJKLMNOPQRSTUVWXYZ" +
"abcdefghijklmnopqrstuvwxyz" +
"0123456789" +
"!@#$%.-_"
;
public static string GetRandomPassword(int length)
{
char[] p = new char[length];
for (int i = 0; i < length; i++)
p[i] = passChars[RandomNumberGenerator.GetInt32(0, passChars.Length)];
return new string(p);
}